Understanding Simple Harmonic Motion: Key Concepts & Problems
School
St.thomas College Of Education**We aren't endorsed by this school
Course
PHYSICS 231
Subject
Mathematics
Date
Dec 12, 2024
Pages
56
Uploaded by CaptainExploration7396
PHYS 231 - UNIVERSITY PHYSICS I- Simple harmonic motionQuestion Bank - Set 9Liberty UniversityQuestion 1QuestionA block of massmis attached to a spring with spring constantk. The blockis displaced from its equilibrium position and released from rest.Determinethe amplitude of the resulting simple harmonic motion in terms of the initialdisplacementx0.SolutionTo find the amplitude of the resulting simple harmonic motion, we need to con-sider the total mechanical energy of the system, which is conserved throughoutthe motion.Step 1:Calculate the potential energy stored in the spring.The potential energy stored in the spring at a displacementxfrom its equi-librium position is given by:PE=12kx2At the equilibrium position, the potential energy is zero. Therefore, at theinitial displacementx0, the potential energy stored in the spring is:PE=12kx20Step 2:Determine the total mechanical energy of the system.At the equilibrium position, the total mechanical energy is all in the form ofpotential energy:Etot=PE=12kx20Step 3:Express the total mechanical energy in terms of the amplitudeA.
At the amplitudeA, all of the total mechanical energy is in the form ofkinetic energy:Etot=KE=12mv2maxwherevmaxis the maximum speed of the block.Step 4:Relate the amplitudeAto the initial displacementx0.At the amplitudeA, the block momentarily stops before returning, imply-ing that all kinetic energy is converted to potential energy.So,vmax= 0 atamplitudeA.Equating the total mechanical energy at the initial displacement to that atthe amplitude:12kx20=12m·02Thus, we find that the amplitude of the simple harmonic motion is equal tothe initial displacement:A=x0Question 2QuestionA particle is undergoing simple harmonic motion with an amplitude of 5 cmand a period of 2 seconds. If the displacement of the particle is given byx(t) =5 sin(π3t), determine the velocity and acceleration of the particle att= 1 second.SolutionStep 1: To find the velocity function, we differentiate the displacement functionwith respect to time.Step 1:v(t) =dxdt= 5·π3cosπ3tStep 2: To find the acceleration function, we differentiate the velocity func-tion with respect to time.Step 2:a(t) =dvdt=−5·π32sinπ3tStep 3: Now, we can find the velocity and acceleration of the particle att= 1 second.Step 3:v(1) = 5·π3cosπ3(1)= 5·π3cosπ3v(1) = 5·π3·12=5π6cm/s2
Step 3:a(1) =−5·π32sinπ3(1)=−5·π32sinπ3a(1) =−5·π32·√32=−5π9√3 cm/s2Therefore, att= 1 second, the velocity of the particle is5π6cm/s and theacceleration is−5π9√3 cm/s2.Question 3QuestionA particle undergoes simple harmonic motion with an amplitude of 2 m and aperiod of 3 s.At timet= 0, it is at its maximum displacement of 2 m andmoving in the negative direction. Find an expression for the displacementxofthe particle as a function of timet.SolutionStep 1: We are given that the particle moves in the negative direction att= 0,so we can express the equation of motion asx(t) =−2 sin(ωt), whereωis theangular frequency.Step 2: The amplitudeAis equal to 2, thusA= 2.Step 3: The periodTis given as 3 s. We know that the periodTis relatedto the angular frequencyωby the formulaT=2πω. Thus, we can solve forωasω=2πT.Step 4: SubstitutingT= 3 into the formula, we getω=2π3.Step 5: Therefore, the equation of motion for the particle isx(t) =−2 sin(2π3t).Question 4QuestionA mass-spring system executes simple harmonic motion with an amplitude of0.3 m and a frequency of 2 Hz. If the maximum velocity of the mass is 0.6 m/s,determine the maximum acceleration of the mass.SolutionStep 1: Identify the given values and relevant formulas.Given: Amplitude,A= 0.3 mFrequency,f= 2 HzMaximum velocity,vmax= 0.6 m/sWe know that for a mass-spring system undergoing simple harmonic motion,3
the maximum velocity and maximum acceleration are related by the equationvmax=ωA, whereω= 2πfis the angular frequency.Step 2: Find the angular frequency.Sinceω= 2πf, we have:ω= 2π(2) = 4πrad/sStep 3: Calculate the maximum acceleration.We know that the maximum accelerationamax=ω2A.Using the values wehave:amax= (4π)2(0.3) = 37.699 m/s2Therefore, the maximum acceleration of the mass in the simple harmonicmotion is 37.699 m/s2.Question 5QuestionA mass of 0.5 kg is attached to a spring with a spring constant of 20 N/m. If themass is displaced 0.1 m from its equilibrium position and released, determinethe equation of motion describing the subsequent simple harmonic motion.SolutionStep 1: Calculate the angular frequency,ω. Given: Mass,m= 0.5 kg Springconstant,k= 20 N/m Displacement,x= 0.1 mThe angular frequency,ω, is given by:ω=rkmω=r200.5ω=√40ω= 2√10 rad/sStep 2:Write the equation of motion.For simple harmonic motion, theequation of motion is given by:x(t) =Acos(ωt+ϕ)Where:x(t) is the displacement at timet Ais the amplitude of motionϕisthe phase angleSince the mass is displaced 0.1 m from its equilibrium position, the equationbecomes:x(t) = 0.1cos(2√10t+ϕ)Therefore, the equation of motion describing the subsequent simple harmonicmotion of the mass isx(t) = 0.1cos(2√10t+ϕ).4
Question 6QuestionA particle executes simple harmonic motion with an amplitude of 5 cm and afrequency of 2 Hz. If at timet= 0, the particle is at its equilibrium positionand moving in the positive direction, find the displacement and acceleration ofthe particle att=14s.SolutionStep 1: Find the angular frequencyωusing the relationshipω= 2πfwherefis the frequency.ω= 2π×2 = 4πrad/sStep 2: Determine the displacementxof the particle att=14s using theequation for displacement in simple harmonic motion:x=Acos(ωt).x= 5 cos4π×14x= 5 cos(π) =−5 cmStep 3:Calculate the accelerationaof the particle att=14s using theequation for acceleration in simple harmonic motion:a=−ω2x.a=−(4π)2×(−5)a=−100π2cm/s2Question 7QuestionA particle is undergoing simple harmonic motion with an amplitude of 8 cmand a period of 2 seconds.If the particle is at a distance of 6 cm from theequilibrium position at timet= 1 second, determine the displacement, velocity,and acceleration of the particle at that moment.SolutionWe can define the equation of motion for simple harmonic motion as:x(t) =Acos(ωt+ϕ)where: -Ais the amplitude, -ωis the angular frequency (ω=2πT), -ϕis thephase angle, -x(t) is the displacement at timet.Given: - AmplitudeA= 8 cm, - PeriodT= 2 s, - Displacementx(1) = 6cm.5
Step 1: Find the angular frequencyω=2πTω=2π2ω=πrad/sStep 2: Find the phase anglex(1) =Acos(ω×1 +ϕ)6 = 8 cos(π+ϕ)68= cos(π+ϕ)34= cos(π+ϕ)π+ϕ= arccos34ϕ= arccos34−πStep 3: Find the displacement, velocity, and acceleration att= 1sGiven the displacement equation, we have:x(t) =Acos(ωt+ϕ)x(1) = 8 cos(π+ϕ)x(1) = 8 cosπ+ arccos34−πx(1) = 8·34x(1) = 6 cmThe velocity and acceleration at a given time can be found by taking thefirst and second derivatives of the displacement equation. Let’s denote velocityasv(t) and acceleration asa(t).v(t) =−ωAsin(ωt+ϕ)a(t) =−ω2Acos(ωt+ϕ)Substitutet= 1 s into the velocity and acceleration equations to find thevalues at that moment.v(1) =−π×8 sin(π+ϕ)a(1) =−π2×8 cos(π+ϕ)6
Question 8QuestionA particle undergoing simple harmonic motion has an amplitude of 4 cm and aperiod of 2 seconds. If the particle starts from the equilibrium position, find theacceleration of the particle when it is at a distance of 3 cm from the equilibriumposition.SolutionStep 1: Determine the angular frequency Given that the periodT= 2 secondsandω=2πT, we can find the angular frequency:ω=2π2=πrad/sStep 2: Find the position function The general form of the position functionfor simple harmonic motion isx(t) =Acos(ωt), whereAis the amplitude.Substitute the valuesA= 4 cm andω=πinto the position function:x(t) = 4 cos(πt)Step 3: Find the velocity function The velocity function is the derivative ofthe position function,v(t) =−Aωsin(ωt). Differentiate the position functionwith respect tot:v(t) =−4πsin(πt)Step 4:Find the acceleration function The acceleration function is thederivative of the velocity function,a(t) =−Aω2cos(ωt). Differentiate the ve-locity function with respect tot:a(t) =−4π2cos(πt)Step 5: Find the acceleration whenx(t) = 3 cm Given thatx(t) = 3 cm, weneed to find the timetwhenx(t) = 3 cm:4 cos(πt) = 3 =⇒cos(πt) =34Since cosine is positive in the first and fourth quadrants, we haveπt= arccos(34).Solving fort:t=1πarccos34≈0.424 sStep 6: Calculate the acceleration atx= 3 cm Substitutet≈0.424 s intothe acceleration function we found earlier:a(0.424) =−4π2cos(π·0.424)≈ −39.478 cm/s2Therefore, the acceleration of the particle when it is at a distance of 3 cmfrom the equilibrium position is approximately−39.478 cm/s2.7
Question 9QuestionA mass attached to a spring oscillates with simple harmonic motion. The masshas an amplitude of 0.5 m and a period of 2 seconds. If at timet= 0 the mass isat its equilibrium position, find an expression for the displacement of the massas a function of time.SolutionStep 1: First, recall the general equation for simple harmonic motion:x(t) =Acos(ωt+ϕ)where: -x(t) is the displacement of the mass at timet, -Ais the amplitude ofthe motion, -ωis the angular frequency of the motion, -ϕis the phase angle.Step 2:We are given that the amplitudeAis 0.5 m.We are also giventhat the periodTis 2 seconds. Recall that the period is related to the angularfrequency byT=2πω.Step 3: Substituting the given values into the period equation:2 =2πωSolving forω:ω=2π2=πrad/sStep 4: Since the mass is at its equilibrium position att= 0, we haveϕ= 0.Thus, the expression for the displacement of the mass as a function of time is:x(t) = 0.5 cos(πt)Question 10QuestionA mass-spring system with a mass of 0.5 kg is displaced from its equilibriumposition by 0.1 m and then released. If the spring constant is 400 N/m, find theamplitude, period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Find the amplitude (A).A= maximum displacement from equilibrium= 0.1 m8
Step 2: Find the angular frequency (ω).ω=rkm=r4000.5=√800= 20 rad/sStep 3: Find the period (T).T=2πω=2π20=π10= 0.314 sStep 4: Find the frequency (f).f=1T=10.314≈3.18 HzTherefore, the amplitude is 0.1 m, the period is 0.314 s, and the frequencyis approximately 3.18 Hz.Question 11QuestionA particle undergoes simple harmonic motion such that its displacement at timetis given byx(t) = 5 sin(2πt/3). Find the amplitude, period, frequency, andmaximum velocity of the particle.SolutionStep 1: The amplitude of the motion is the coefficient of sin in the equation.Step 2: The amplitude is given byA= 5. So, the amplitude of the motionis 5 units.Step 3: The period of the motion is the time taken for one complete cycleof the motion.9
Step 4: The periodTis related to the angular frequencyωbyT=2πω. Inthis case,ω= 2π/3. So, the period isT=2π2π/3= 3 seconds.Step 5: The frequencyfis the number of cycles per unit time.Step 6: The frequencyfis related to the periodTbyf= 1/T. Therefore,the frequency isf= 1/3 Hz.Step 7: The maximum velocity of the particle can be found by taking thederivative of the displacement function with respect to time.Step 8: The velocityv(t) is given byv(t) =dxdt= 5×(2π/3) cos(2πt/3) =10π3cos(2πt/3).Step 9: To find the maximum velocity, we look for the maximum value ofthe cosine function. The maximum value of cos is 1. Therefore, the maximumvelocity isvmax=10π3units/second.Question 12QuestionA simple harmonic oscillator has an amplitude of 0.1 m and a maximum accel-eration of 2πm/s2. If the oscillator starts from rest at the equilibrium position,determine the maximum speed of the oscillator.SolutionStep 1: Recall the relationship between acceleration and displacement for simpleharmonic motion: The acceleration of an object in simple harmonic motion isgiven bya=−ω2x, whereais the acceleration,ωis the angular frequency, andxis the displacement from the equilibrium position. In this case, we are givenamax= 2πm/s2andxmax= 0.1 m.Step 2: Using the relationship between acceleration and displacement, wecan find the angular frequencyω:2π=−ω2×0.1−ω2=2π0.1ω2=−20πω=√−20πStep 3: The velocity of the oscillator can be determined byv=ω√A2−x2,wherevis the velocity,Ais the amplitude, andxis the displacement from theequilibrium position. We are interested in the maximum velocity, which occurswhenx= 0.Step 4: Calculate the maximum velocity:vmax=√−20π×p0.12−0210
vmax=√−20π×0.1vmax= 0.1√−20πm/sTherefore, the maximum speed of the oscillator is 0.1√−20πm/s.Question 13QuestionA particle undergoes simple harmonic motion with an amplitude of 3 cm and aperiod of 2 seconds. If the particle is at the equilibrium position at timet= 0,find:a) the displacement functionx(t),b) the velocity functionv(t),c) the acceleration functiona(t).Solutiona) To find the displacement functionx(t), we can use the general formula forsimple harmonic motion:x(t) =Acos2πTt,whereAis the amplitude andTis the period.Step 1:Given that the amplitudeA= 3 cm and the periodT= 2 s, wehave:x(t) = 3 cos2π2t.Simplifying:x(t) = 3 cos(πt).b) To find the velocity functionv(t), we differentiate the displacement func-tionx(t) with respect tot:v(t) =dxdt=−3πsin(πt).c) To find the acceleration functiona(t), we differentiate the velocity functionv(t) with respect tot:a(t) =dvdt=−3π2cos(πt).11
Question 14QuestionA block of massmis attached to a spring with spring constantk. The block isdisplaced from its equilibrium position and released. If the maximum speed ofthe block during its motion isv, find the amplitude of the motion.SolutionStep 1: We know that the maximum speed of the block during simple harmonicmotion occurs when the displacement is zero. At this point, all the potentialenergy has been converted to kinetic energy.Step 2:The potential energy stored in the spring at a displacementxisgiven byU=12kx2. At the equilibrium position, all the potential energy hasbeen converted to kinetic energy, soU= 0.Step 3: At the equilibrium position, the total energy of the system is thekinetic energy of the block:KE=12mv2.Step 4: The total energy of the system is constant and is the sum of thekinetic and potential energy:KE+U=12mv2+12kA2=12kA2.Step 5: Solving for the amplitudeA, we haveA=qmv2k. Therefore, theamplitude of the motion isA=rmv2k.Step 6: Thus, the amplitude of the motion when the maximum speed of theblock isvisqmv2k.Question 15QuestionA mass-spring system oscillates with an amplitude of 4 cm and a frequency of2 Hz. If the maximum speed of the mass is 16 cm/s, determine the equation ofmotion for the system.SolutionStep 1: Find the angular frequency,ω. Given that frequencyf= 2 Hz, we have:f=ω2πω= 2πf= 2π×2 = 4πrad/sStep 2: Find the equation of motion. The general equation of motion for amass-spring system undergoing simple harmonic motion is given by:x(t) =Acos(ωt−ϕ)12
where: -Ais the amplitude (given as 4 cm), -ωis the angular frequency (foundto be 4πrad/s), -ϕis the phase constant.Step 3: Find the phase constant,ϕ. To determine the phase constant, weneed to consider the initial conditions of the system. In this case, we know thatthe maximum speed occurs when the displacement is equal to the amplitude.This corresponds to a phase angle ofπ2.x(t) = 4 cos 4πt−π2= 4 cos 4πt+π2Therefore, the equation of motion for the mass-spring system is:x(t) = 4 cos 4πt+π2Question 16QuestionA 0.5 kg object is attached to a spring with a spring constant of 50 N/m. Theobject is pulled 0.1 m away from its equilibrium position and released fromrest.Determine the amplitude, frequency, and period of the resulting simpleharmonic motion.SolutionStep 1: Find the amplitude (A) Given that the object is pulled 0.1 m away fromits equilibrium position, the amplitude is equal to this distance.Therefore,A= 0.1 m.Step 2: Find the angular frequency (ω) The angular frequency can be foundusing the formula:ω=rkmwherekis the spring constant andmis the mass of the object.Substitutek= 50 N/m andm= 0.5 kg:ω=r500.5=√100 = 10 s−1Step 3:Find the frequency The frequency (f) is related to the angularfrequency by the formulaf=ω2π. So,f=102π≈1.59 HzStep 4: Find the period (T) The periodTis the reciprocal of the frequency:T=1f=11.59≈0.63 sTherefore, the amplitude of the simple harmonic motion is 0.1 m, the fre-quency is approximately 1.59 Hz, and the period is approximately 0.63 s.13
Question 17QuestionA particle of massmis attached to a spring with spring constantk. At timet= 0, the particle is displacedaunits from the equilibrium position and releasedfrom rest. Find the amplitude of the resulting simple harmonic motion in termsofa.SolutionStep 1: First, write down the equation of motion for simple harmonic motion:md2xdt2=−kxStep 2: To find the amplitude of the resulting simple harmonic motion, weneed to solve this differential equation. Let’s assume the solution is of the formx(t) =Acos(ωt+ϕ), whereAis the amplitude,ωis the angular frequency, andϕis the phase angle.Step 3: Calculate the first and second derivatives ofx(t):dxdt=−Aωsin(ωt+ϕ)d2xdt2=−Aω2cos(ωt+ϕ)Step 4: Substitutex(t) and its derivatives into the equation of motion:m(−Aω2cos(ωt+ϕ)) =−k(Acos(ωt+ϕ))Step 5: Divide both sides by−Acos(ωt+ϕ):mω2=kStep 6: Solve for the angular frequencyω:ω=rkmStep 7: The amplitudeAis related to the displacementaatt= 0 by:a=Acos(ϕ)Step 8: As the particle is released from rest, we havedxdtt=0= 0. Hence,ϕ=π2.Step 9: Substituteϕ=π2intoa=Acos(ϕ):a=Acosπ214
a=A×0A= 0Step 10: Therefore, the amplitude of the resulting simple harmonic motionin terms ofais0.Question 18QuestionA mass-spring system has a mass of 0.5 kg attached to a spring with a springconstant of 50 N/m. The system is released from rest at its equilibrium posi-tion. Find the amplitude of the resulting simple harmonic motion if the totalmechanical energy of the system is 5 J.SolutionStep 1: The total mechanical energy of the system can be expressed as the sumof the potential energy and the kinetic energy:E=U+KWhereEis the total energy,Uis the potential energy, andKis the kineticenergy.Step 2: At the equilibrium position, all the energy is in the form of potentialenergy, given by:U=12kA2wherekis the spring constant andAis the amplitude of motion.Step 3: At the extreme positions, all the energy is in the form of kineticenergy, given by:K=12mv2maxwheremis the mass of the object andvmaxis the maximum velocity.Step 4: Since the system is released from rest, the velocity at the extremepositions is zero. Thus, the total energy is equal to the potential energy at theequilibrium position:E=U=12kA2Step 5: Given thatE= 5 J,k= 50 N/m, andm= 0.5 kg, we can solve forA:5 =12×50×A2Step 6: Solving forA, we find:A=r550= 0.316 m15
Therefore, the amplitude of the resulting simple harmonic motion is 0.316m.Question 19QuestionA block of massmis attached to a spring with spring constantk. The blockis displaced a distanceAfrom its equilibrium position and released. Find themaximum speed of the block during its motion.SolutionStep 1:The total mechanical energy of the system is conserved.The totalmechanical energy is the sum of the kinetic energy (12mv2) and potential energy(12kx2) of the block-spring system, wherevis the velocity of the block andxis its position from the equilibrium point. Step 2: At the maximum speed, thepotential energy is zero at the equilibrium position. Step 3: At the maximumspeed, all the mechanical energy will be in the form of kinetic energy. Step 4:Using the conservation of energy, we have:12kA2=12mv2maxStep 5: Solving forvmax, we get:vmax=ArkmStep 6: Therefore, the maximum speed of the block during its motion isvmax=Aqkm.Question 20QuestionA particle of mass 0.2 kg is attached to a spring with spring constant 50 N/m.Initially, the particle is at the equilibrium position. The particle is then displaced0.1 m from the equilibrium position and released from rest. Find the amplitude,period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Calculate the amplitude. The amplitude of the simple harmonic motionis the maximum displacement from the equilibrium position.Given that theparticle is initially displaced 0.1 m from the equilibrium position, the amplitudeis also 0.1 m.16
Step 2: Calculate the angular frequency. The angular frequency of a mass-spring system is given by:ω=rkmwherekis the spring constant andmis the mass. Substitutingk= 50 N/m andm= 0.2 kg, we get:ω=r500.2=√250 = 5√10 rad/sStep 3: Calculate the period. The periodTof the simple harmonic motionis related to the angular frequencyωby the formula:T=2πωSubstituteω= 5√10 into the formula to get:T=2π5√10=2π5√10 sStep 4: Calculate the frequency.The frequencyfis the reciprocal of theperiodT. Therefore,f=1T=12π5√10=5√102π≈2.512 HzTherefore, the amplitude of the motion is 0.1 m, the period is2π5√10 seconds,and the frequency is approximately 2.512 Hz.Question 21QuestionA block of massmis attached to a spring with spring constantk. The blockis pulled a distanceAto the right and released from rest. Find an expressionfor the velocity of the block as a function of time during its subsequent simpleharmonic motion.SolutionStep 1: Determine the equation of motion for the block.The equation of motion for a block undergoing simple harmonic motion canbe expressed as:md2xdt2=−kxwherexis the displacement of the block from its equilibrium position.17
Step 2: Find the general solution to the differential equation.The general solution to the differential equationmd2xdt2=−kxcan be writtenas:x(t) =Acos(ωt) +Bsin(ωt)whereAandBare constants to be determined, andω=qkmis the angularfrequency.Step 3: Apply initial conditions to the general solution.Given that the block is released from rest at a distanceAto the right, wehave the initial conditionsx(0) =Aanddxdt(0) = 0.Substitutex(0) =Aanddxdt(0) = 0 into the general solution to solve for theconstantsAandB.Step 4: Determine the velocity of the block as a function of time.Differentiating the equationx(t) =Acos(ωt) +Bsin(ωt) with respect totgives the velocity function:dxdt=−Aωsin(ωt) +Bωcos(ωt)Therefore, the velocity of the block as a function of time is:v(t) =−Aωsin(ωt) +Bωcos(ωt)Question 22QuestionA particle undergoes simple harmonic motion with an angular frequency ofω= 3 rad/s and an amplitude of 0.1 m. If the particle starts from rest at theequilibrium positionx= 0, find the displacement of the particle at timet=π3s.SolutionStep 1: The general equation for simple harmonic motion is given byx(t) =Acos(ωt) +Bsin(ωt)whereAis the amplitude,ωis the angular frequency, andBis a constantdetermined by the initial conditions.Step 2: We are given that the amplitudeA= 0.1 m and the angular fre-quencyω= 3 rad/s. Since the particle starts from rest atx= 0, we can usethis information to determineB.Step 3: Att= 0, the equation becomes0 =Acos(0) +Bsin(0) =A·1 +B·0 =A= 0.118
Step 4: Now we have the equation for the motion of the particle:x(t) = 0.1 cos(3t) +Bsin(3t)Step 5: To find the value ofB, we can differentiatex(t) with respect totand sett= 0 (to represent the initial condition of starting from rest):v(t) =dxdt=−0.3 sin(3t) + 3Bcos(3t)v(0) =−0.3 sin(0) + 3Bcos(0) = 0Step 6: From this, we find thatB= 0.1.Step 7: Therefore, the equation for the displacement of the particle isx(t) = 0.1 cos(3t) + 0.1 sin(3t)Step 8: To find the displacement of the particle att=π3s, substitutet=π3intox(t):xπ3= 0.1 cos3·π3+ 0.1 sin3·π3Step 9: Simplifying givesxπ3= 0.1 cos(π) + 0.1 sin(π) = 0.1·(−1) + 0 =−0.1 mStep 10: Therefore, the displacement of the particle at timet=π3s is−0.1m.Question 23QuestionAn object is attached to a spring and undergoes simple harmonic motion withan amplitude of 0.2 m. If the object’s velocity is 2 m/s when it is 0.1 m fromthe equilibrium position, determine the object’s position after 0.1 seconds.SolutionStep 1: Find the angular frequencyωof the motion using the amplitude andvelocity. Given that the amplitudeA= 0.2 m and the velocityv= 2 m/s, wehave:v=Aω⇒ω=vA=20.2= 10 rad/sStep 2:Determine the object’s position after 0.1 seconds.The generalequation for the object’s position as a function of time in SHM isx(t) =Acos(ωt+ϕ), whereϕis the phase angle.19
Given that the object’s velocity is 2 m/s when it is 0.1 m from the equilibriumposition, we can determine the phase angleϕ:v=−Aωsin(ϕ) = 2⇒sin(ϕ) =−20.2×10=−1As sin(ϕ) =−1 in the second or third quadrant, we haveϕ=−π2.So, the object’s position after 0.1 seconds is:x(0.1) = 0.2 cos10×0.1−π2= 0.2 cosπ2= 0 mTherefore, the object’s position after 0.1 seconds is 0 meters from the equi-librium position.Question 24QuestionAn object of massmis attached to a spring with spring constantk. The objectis displaced a distanceAfrom its equilibrium position and released from rest.Find the period of the resulting simple harmonic motion.SolutionStep 1: Write the formula for the period of simple harmonic motion.The period (T) of simple harmonic motion is given by:T= 2πrmkStep 2: Find the angular frequency of the system.The angular frequency (ω) is given by:ω=rkmStep 3: Find the period of the simple harmonic motion.Substituteω=rkminto the period formula to get:T=2πω=2πqkm= 2πrmk20
Step 4: Conclusion.Therefore, the period of the simple harmonic motion of the object will be:T= 2πrmkQuestion 25QuestionA particle undergoes simple harmonic motion with an amplitude of 4 cm and aperiod of 2πseconds. Att= 0, the particle is at its maximum displacement of4 cm. Find the equation that describes the position of the particle at timet.SolutionStep 1: The general equation for simple harmonic motion is given byx(t) =Acos(ωt+ϕ), where: -Ais the amplitude, -ωis the angular frequency, -ϕisthe phase constant.Step 2: Given that the amplitudeA= 4 cm, we havex(t) = 4 cos(ωt+ϕ).Step 3: The periodTof the simple harmonic motion is related to the angularfrequencyωbyT=2πω.Step 4: Since the periodT= 2πseconds, we have2πω= 2π, which impliesω= 1 rad/s.Step 5: The equation becomesx(t) = 4 cos(t+ϕ).Step 6: Att= 0, the particle is at its maximum displacement of 4 cm. Thismeansx(0) = 4 = 4 cos(ϕ).Step 7: Solving forϕ: cos(ϕ) =44= 1, which impliesϕ= 0.Step 8:The equation describing the position of the particle at timetisx(t) = 4 cos(t).Question 26QuestionA particle undergoes simple harmonic motion with an amplitude of 4 cm and aperiod of 2 seconds. If the displacement of the particle is 3 cm at timet= 1second, find an equation for the particle’s motion.SolutionStep 1: Find the angular frequency.The angular frequency,ω, of a particle undergoing simple harmonic motion is21
given byω=2πTwhereTis the period. In this case, the periodT= 2 seconds, soω=2π2=πrad/sStep 2: Find the equation for the particle’s motion.The equation for the displacement of a particle undergoing simple harmonicmotion isx(t) =Acos(ωt−ϕ)where -Ais the amplitude, -ωis the angular frequency, -tis the time, and -ϕis the phase angle.Given that the amplitudeA= 4 cm, the angular frequencyω=π, and thedisplacement att= 1 second isx(1) = 3 cm, we can find the phase angleϕ.Step 3: Find the phase angleϕ.Substitutet= 1 andx= 3 into the equation and solve forϕ:3 = 4 cos(π−ϕ)34= cos(π−ϕ)It follows thatπ−ϕ= arccos34π−ϕ=π3ϕ=π−π3=2π3Step 4: Write the equation for the particle’s motion.Therefore, the equation for the particle’s motion isx(t) = 4 cosπt−2π3Question 27QuestionA particle is undergoing simple harmonic motion along the x-axis with an am-plitude of 5 cm and a period of 2 seconds.If the particle is at its maximumdisplacement (positive) and the velocity is also positive, determine the positionof the particle after 1 second.22
SolutionStep 1: First, we need to determine the angular frequency of the motion. Theangular frequencyωcan be calculated using the formulaω=2πT, whereTisthe period.ω=2π2=πrad/sStep 2: The position of the particle at any timetis given by the equationx(t) =Acos(ωt), whereAis the amplitude.Given that the particle is at itsmaximum displacement (positive) att= 0, the initial conditionx(0) =Agivesus the value ofA.x(0) =Acos(0) =A= 5 cmStep 3: Now we have the position functionx(t) = 5 cos(πt).To find theposition of the particle after 1 second, we substitutet= 1 into the equation.x(1) = 5 cos(π) = 5×(−1) =−5 cmTherefore, the position of the particle after 1 second is−5 cm.Question 28QuestionA particle undergoes simple harmonic motion with an amplitude of 2 cm anda frequency of 3 Hz.If the displacement of the particle is given byx(t) =2 sin(6πt), determine the velocity and acceleration of the particle when the dis-placement is 1 cm.SolutionStep 1:Find the velocity function.Given that the displacement function isx(t) = 2 sin(6πt), we can find the velocity function by taking the derivative ofthe displacement function with respect to time.v(t) =dxdt= 2(6π) cos(6πt) = 12πcos(6πt)Step 2: Find the acceleration function. Similarly, we can find the accelerationfunction by taking the derivative of the velocity function with respect to time.a(t) =dvdt=−12π2sin(6πt) =−12π2sin(6πt)Step 3: Determine the velocity and acceleration when the displacement is 1cm. To find the time when the particle’s displacement is 1 cm, we substitutex(t) = 1 into the displacement function:2 sin(6πt) = 123
sin(6πt) =12This occurs when 6πt=π6or 6πt=5π6. Thus,t=136ort=536seconds.Step 4: Calculate the velocity and acceleration att=136seconds. Substitutet=136into the velocity and acceleration functions:v136= 12πcosπ6= 6πa136=−12π2sinπ6=−6π2Therefore, when the particle’s displacement is 1 cm, the velocity is 6πcm/sand the acceleration is−6π2cm/s2.Question 29QuestionA particle undergoes simple harmonic motion with an amplitude of 5 cm anda period of 2 seconds.If att= 0, the particle is at its equilibrium position,determine the position functionx(t) for the particle.SolutionStep 1: Determine the angular frequencyωGiven that the periodT= 2 seconds, we have the relationT=2πω.Hence,ω=2π2=πrad/s.Step 2: Write the position function in terms of the amplitude and angularfrequencyThe position function for a particle undergoing simple harmonic motion is givenby:x(t) =Acos(ωt+ϕ)whereAis the amplitude,ωis the angular frequency, andϕis the phase angle.Step 3: Determine the phase angleϕSince att= 0 the particle is at its equilibrium position, we havex(0) =Acos(ϕ) = 0. This implies that cos(ϕ) = 0, which occurs whenϕ=π2.Step 4: Write the final position functionSubstitute the values ofA= 5 cm,ω=πrad/s, andϕ=π2into the positionfunction, we have:x(t) = 5 cosπt+π224
Question 30QuestionA mass-spring system has a period of oscillation of 2 seconds.If the mass isreplaced with one that is 4 times as heavy, what will be the new period ofoscillation?SolutionStep 1: Recall the formula for the period of oscillation of a mass-spring system:T= 2πrmkwhere:T= period of oscillation,m= mass of the object, andk= springconstant.Step 2: LetT1be the period of oscillation when the original massmis used,andT2be the period of oscillation when the mass is increased to 4m.Step 3: For the original system, we have:T1= 2πrmkStep 4: For the system with the increased mass, we have:T2= 2πr4mk= 2πr4mk= 2·2πrmk= 2T1Step 5: Thus, when the mass is replaced with one that is 4 times as heavy,the new period of oscillation will be twice the original period. The new periodwill be 4 seconds.Question 31QuestionA particle undergoes simple harmonic motion with an amplitude of 5 cm and aperiod of 2 seconds. If at timet= 0, the particle is at its equilibrium positionand moving upwards with a speed of 4 cm/s, determine the displacement of theparticle after 1 second.SolutionStep 1: Find the angular frequencyωusing the periodT. Given thatT= 2seconds, the angular frequencyωcan be calculated as:ω=2πT25
ω=2π2=πStep 2: Determine the displacementx(t) of the particle at timetusing theequation for simple harmonic motion:x(t) =Asin(ωt+ϕ)whereAis the amplitude,ωis the angular frequency,tis the time, andϕis thephase angle.Given that the amplitudeA= 5 cm, the angular frequencyω=π, and theparticle is at its equilibrium position att= 0, we have:x(t) = 5 sin(πt+ϕ)Step 3: Use the initial conditions to determine the phase angleϕ. Att= 0,the particle is at its equilibrium position and moving upwards with a speed of4 cm/s. This implies that att= 0, the particle is at its maximum displacementand moving upwards, which corresponds to the equation:x(0) = 5 sin(ϕ) = 5sin(ϕ) = 1ϕ=π2Step 4:Find the displacement of the particle after 1 second.Substitutet= 1 second into the equation for displacement:x(1) = 5 sinπ+π2x(1) = 5 sin3π2x(1) = 5×(−1) =−5Therefore, the displacement of the particle after 1 second is−5 cm.Question 32QuestionA mass-spring system oscillates with a frequency of 10 Hz and an amplitude of0.1 m. Determine the maximum acceleration of the mass during its motion.26
SolutionTo find the maximum acceleration, we first need to find the angular frequencyω. From the given frequencyf= 10 Hz, we haveω= 2πf= 2π×10 = 20πrad/s.Step 1:The equation for the acceleration of an object in simple harmonicmotion is given bya(t) =−ω2x(t), wherex(t) is the displacement of the objectat timet. Since acceleration is maximal when the displacement is maximal, weneed to find the maximum displacement. The equation for the displacement ofan object in simple harmonic motion is given byx(t) =Acos(ωt), whereAisthe amplitude.Step 2:At the maximum displacement (A), the displacement equationbecomesxmax=A. SubstitutingA= 0.1 m into the equation, we getxmax= 0.1m.Step 3:Now, we can find the maximum acceleration of the mass. Substi-tutingxmax= 0.1 m into the acceleration equation, we getamax=−ω2xmax=−(20π)2×0.1 m/s2.Step 4:Calculating the maximum acceleration givesamax=−400π2≈−1256.64 m/s2.Therefore, the maximum acceleration of the mass during itsmotion is approximately 1256.64 m/s2.Question 33QuestionA mass attached to a spring with spring constantk= 5 N/m undergoes simpleharmonic motion with a period of 2πs. If the amplitude of the motion is 0.1 m,determine the maximum speed of the mass.SolutionStep 1: Find the angular frequency Given that the period isT= 2πs, we canfind the angular frequencyωusing the formulaω=2πT.ω=2π2π= 1 s−1Step 2:Calculate the maximum speed The maximum speed of the massoccurs at the equilibrium position where the displacement is zero. At this point,the velocity is at its maximum. The maximum speed can be determined usingthe formulavmax=ωA, whereAis the amplitude of the motion.vmax= 1 s−1×0.1 mvmax= 0.1 m/sTherefore, the maximum speed of the mass is 0.1 m/s.27
Question 34QuestionA particle oscillates with simple harmonic motion given by the equationx(t) =0.1 cos(2t). Determine the amplitude, period, frequency, and maximum velocityof the particle.SolutionThe given equation for simple harmonic motion isx(t) = 0.1 cos(2t).Step 1:Find the amplitude of the motion.The amplitude of the motion is given by the coefficient of the cosine function,so the amplitude is 0.1.Step 2:Find the period of the motion.The period of the motion is the time taken for one complete oscillation. Inthis case, the periodTcan be found using the formulaT=2π2=π.Step 3:Find the frequency of the motion.The frequencyfof the motion is the reciprocal of the period, sof=1T=1π.Step 4:Find the maximum velocity of the particle.The velocity of the particle at any timetis given by the derivative of theposition functionx(t).v(t) =dxdt=−0.2 sin(2t)The maximum velocity occurs when sin(2t) = 1, so the maximum velocity is|vmax|= 0.2.Therefore, the amplitude is 0.1, the period isπ, the frequency is1π, and themaximum velocity of the particle is 0.2.Question 35QuestionA mass-spring system has a mass of 0.5 kg attached to a spring with a springconstant of 100 N/m. Initially, the mass is at its equilibrium position and is thendisplaced 0.1 m to the right and released from rest. Determine the amplitude,period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Calculate the angular frequency (ω).Given that the spring constantk= 100 N/m and the massm= 0.5 kg, we can find the angular frequency usingthe formula:ω=rkm28
At the amplitudeA, all of the total mechanical energy is in the form ofkinetic energy:Etot=KE=12mv2maxwherevmaxis the maximum speed of the block.Step 4:Relate the amplitudeAto the initial displacementx0.At the amplitudeA, the block momentarily stops before returning, imply-ing that all kinetic energy is converted to potential energy.So,vmax= 0 atamplitudeA.Equating the total mechanical energy at the initial displacement to that atthe amplitude:12kx20=12m·02Thus, we find that the amplitude of the simple harmonic motion is equal tothe initial displacement:A=x0Question 2QuestionA particle is undergoing simple harmonic motion with an amplitude of 5 cmand a period of 2 seconds. If the displacement of the particle is given byx(t) =5 sin(π3t), determine the velocity and acceleration of the particle att= 1 second.SolutionStep 1: To find the velocity function, we differentiate the displacement functionwith respect to time.Step 1:v(t) =dxdt= 5·π3cosπ3tStep 2: To find the acceleration function, we differentiate the velocity func-tion with respect to time.Step 2:a(t) =dvdt=−5·π32sinπ3tStep 3: Now, we can find the velocity and acceleration of the particle att= 1 second.Step 3:v(1) = 5·π3cosπ3(1)= 5·π3cosπ3v(1) = 5·π3·12=5π6cm/s2
Step 3:a(1) =−5·π32sinπ3(1)=−5·π32sinπ3a(1) =−5·π32·√32=−5π9√3 cm/s2Therefore, att= 1 second, the velocity of the particle is5π6cm/s and theacceleration is−5π9√3 cm/s2.Question 3QuestionA particle undergoes simple harmonic motion with an amplitude of 2 m and aperiod of 3 s.At timet= 0, it is at its maximum displacement of 2 m andmoving in the negative direction. Find an expression for the displacementxofthe particle as a function of timet.SolutionStep 1: We are given that the particle moves in the negative direction att= 0,so we can express the equation of motion asx(t) =−2 sin(ωt), whereωis theangular frequency.Step 2: The amplitudeAis equal to 2, thusA= 2.Step 3: The periodTis given as 3 s. We know that the periodTis relatedto the angular frequencyωby the formulaT=2πω. Thus, we can solve forωasω=2πT.Step 4: SubstitutingT= 3 into the formula, we getω=2π3.Step 5: Therefore, the equation of motion for the particle isx(t) =−2 sin(2π3t).Question 4QuestionA mass-spring system executes simple harmonic motion with an amplitude of0.3 m and a frequency of 2 Hz. If the maximum velocity of the mass is 0.6 m/s,determine the maximum acceleration of the mass.SolutionStep 1: Identify the given values and relevant formulas.Given: Amplitude,A= 0.3 mFrequency,f= 2 HzMaximum velocity,vmax= 0.6 m/sWe know that for a mass-spring system undergoing simple harmonic motion,3
the maximum velocity and maximum acceleration are related by the equationvmax=ωA, whereω= 2πfis the angular frequency.Step 2: Find the angular frequency.Sinceω= 2πf, we have:ω= 2π(2) = 4πrad/sStep 3: Calculate the maximum acceleration.We know that the maximum accelerationamax=ω2A.Using the values wehave:amax= (4π)2(0.3) = 37.699 m/s2Therefore, the maximum acceleration of the mass in the simple harmonicmotion is 37.699 m/s2.Question 5QuestionA mass of 0.5 kg is attached to a spring with a spring constant of 20 N/m. If themass is displaced 0.1 m from its equilibrium position and released, determinethe equation of motion describing the subsequent simple harmonic motion.SolutionStep 1: Calculate the angular frequency,ω. Given: Mass,m= 0.5 kg Springconstant,k= 20 N/m Displacement,x= 0.1 mThe angular frequency,ω, is given by:ω=rkmω=r200.5ω=√40ω= 2√10 rad/sStep 2:Write the equation of motion.For simple harmonic motion, theequation of motion is given by:x(t) =Acos(ωt+ϕ)Where:x(t) is the displacement at timet Ais the amplitude of motionϕisthe phase angleSince the mass is displaced 0.1 m from its equilibrium position, the equationbecomes:x(t) = 0.1cos(2√10t+ϕ)Therefore, the equation of motion describing the subsequent simple harmonicmotion of the mass isx(t) = 0.1cos(2√10t+ϕ).4
Question 6QuestionA particle executes simple harmonic motion with an amplitude of 5 cm and afrequency of 2 Hz. If at timet= 0, the particle is at its equilibrium positionand moving in the positive direction, find the displacement and acceleration ofthe particle att=14s.SolutionStep 1: Find the angular frequencyωusing the relationshipω= 2πfwherefis the frequency.ω= 2π×2 = 4πrad/sStep 2: Determine the displacementxof the particle att=14s using theequation for displacement in simple harmonic motion:x=Acos(ωt).x= 5 cos4π×14x= 5 cos(π) =−5 cmStep 3:Calculate the accelerationaof the particle att=14s using theequation for acceleration in simple harmonic motion:a=−ω2x.a=−(4π)2×(−5)a=−100π2cm/s2Question 7QuestionA particle is undergoing simple harmonic motion with an amplitude of 8 cmand a period of 2 seconds.If the particle is at a distance of 6 cm from theequilibrium position at timet= 1 second, determine the displacement, velocity,and acceleration of the particle at that moment.SolutionWe can define the equation of motion for simple harmonic motion as:x(t) =Acos(ωt+ϕ)where: -Ais the amplitude, -ωis the angular frequency (ω=2πT), -ϕis thephase angle, -x(t) is the displacement at timet.Given: - AmplitudeA= 8 cm, - PeriodT= 2 s, - Displacementx(1) = 6cm.5
Step 1: Find the angular frequencyω=2πTω=2π2ω=πrad/sStep 2: Find the phase anglex(1) =Acos(ω×1 +ϕ)6 = 8 cos(π+ϕ)68= cos(π+ϕ)34= cos(π+ϕ)π+ϕ= arccos34ϕ= arccos34−πStep 3: Find the displacement, velocity, and acceleration att= 1sGiven the displacement equation, we have:x(t) =Acos(ωt+ϕ)x(1) = 8 cos(π+ϕ)x(1) = 8 cosπ+ arccos34−πx(1) = 8·34x(1) = 6 cmThe velocity and acceleration at a given time can be found by taking thefirst and second derivatives of the displacement equation. Let’s denote velocityasv(t) and acceleration asa(t).v(t) =−ωAsin(ωt+ϕ)a(t) =−ω2Acos(ωt+ϕ)Substitutet= 1 s into the velocity and acceleration equations to find thevalues at that moment.v(1) =−π×8 sin(π+ϕ)a(1) =−π2×8 cos(π+ϕ)6
Question 8QuestionA particle undergoing simple harmonic motion has an amplitude of 4 cm and aperiod of 2 seconds. If the particle starts from the equilibrium position, find theacceleration of the particle when it is at a distance of 3 cm from the equilibriumposition.SolutionStep 1: Determine the angular frequency Given that the periodT= 2 secondsandω=2πT, we can find the angular frequency:ω=2π2=πrad/sStep 2: Find the position function The general form of the position functionfor simple harmonic motion isx(t) =Acos(ωt), whereAis the amplitude.Substitute the valuesA= 4 cm andω=πinto the position function:x(t) = 4 cos(πt)Step 3: Find the velocity function The velocity function is the derivative ofthe position function,v(t) =−Aωsin(ωt). Differentiate the position functionwith respect tot:v(t) =−4πsin(πt)Step 4:Find the acceleration function The acceleration function is thederivative of the velocity function,a(t) =−Aω2cos(ωt). Differentiate the ve-locity function with respect tot:a(t) =−4π2cos(πt)Step 5: Find the acceleration whenx(t) = 3 cm Given thatx(t) = 3 cm, weneed to find the timetwhenx(t) = 3 cm:4 cos(πt) = 3 =⇒cos(πt) =34Since cosine is positive in the first and fourth quadrants, we haveπt= arccos(34).Solving fort:t=1πarccos34≈0.424 sStep 6: Calculate the acceleration atx= 3 cm Substitutet≈0.424 s intothe acceleration function we found earlier:a(0.424) =−4π2cos(π·0.424)≈ −39.478 cm/s2Therefore, the acceleration of the particle when it is at a distance of 3 cmfrom the equilibrium position is approximately−39.478 cm/s2.7
Question 9QuestionA mass attached to a spring oscillates with simple harmonic motion. The masshas an amplitude of 0.5 m and a period of 2 seconds. If at timet= 0 the mass isat its equilibrium position, find an expression for the displacement of the massas a function of time.SolutionStep 1: First, recall the general equation for simple harmonic motion:x(t) =Acos(ωt+ϕ)where: -x(t) is the displacement of the mass at timet, -Ais the amplitude ofthe motion, -ωis the angular frequency of the motion, -ϕis the phase angle.Step 2:We are given that the amplitudeAis 0.5 m.We are also giventhat the periodTis 2 seconds. Recall that the period is related to the angularfrequency byT=2πω.Step 3: Substituting the given values into the period equation:2 =2πωSolving forω:ω=2π2=πrad/sStep 4: Since the mass is at its equilibrium position att= 0, we haveϕ= 0.Thus, the expression for the displacement of the mass as a function of time is:x(t) = 0.5 cos(πt)Question 10QuestionA mass-spring system with a mass of 0.5 kg is displaced from its equilibriumposition by 0.1 m and then released. If the spring constant is 400 N/m, find theamplitude, period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Find the amplitude (A).A= maximum displacement from equilibrium= 0.1 m8
Step 2: Find the angular frequency (ω).ω=rkm=r4000.5=√800= 20 rad/sStep 3: Find the period (T).T=2πω=2π20=π10= 0.314 sStep 4: Find the frequency (f).f=1T=10.314≈3.18 HzTherefore, the amplitude is 0.1 m, the period is 0.314 s, and the frequencyis approximately 3.18 Hz.Question 11QuestionA particle undergoes simple harmonic motion such that its displacement at timetis given byx(t) = 5 sin(2πt/3). Find the amplitude, period, frequency, andmaximum velocity of the particle.SolutionStep 1: The amplitude of the motion is the coefficient of sin in the equation.Step 2: The amplitude is given byA= 5. So, the amplitude of the motionis 5 units.Step 3: The period of the motion is the time taken for one complete cycleof the motion.9
Step 4: The periodTis related to the angular frequencyωbyT=2πω. Inthis case,ω= 2π/3. So, the period isT=2π2π/3= 3 seconds.Step 5: The frequencyfis the number of cycles per unit time.Step 6: The frequencyfis related to the periodTbyf= 1/T. Therefore,the frequency isf= 1/3 Hz.Step 7: The maximum velocity of the particle can be found by taking thederivative of the displacement function with respect to time.Step 8: The velocityv(t) is given byv(t) =dxdt= 5×(2π/3) cos(2πt/3) =10π3cos(2πt/3).Step 9: To find the maximum velocity, we look for the maximum value ofthe cosine function. The maximum value of cos is 1. Therefore, the maximumvelocity isvmax=10π3units/second.Question 12QuestionA simple harmonic oscillator has an amplitude of 0.1 m and a maximum accel-eration of 2πm/s2. If the oscillator starts from rest at the equilibrium position,determine the maximum speed of the oscillator.SolutionStep 1: Recall the relationship between acceleration and displacement for simpleharmonic motion: The acceleration of an object in simple harmonic motion isgiven bya=−ω2x, whereais the acceleration,ωis the angular frequency, andxis the displacement from the equilibrium position. In this case, we are givenamax= 2πm/s2andxmax= 0.1 m.Step 2: Using the relationship between acceleration and displacement, wecan find the angular frequencyω:2π=−ω2×0.1−ω2=2π0.1ω2=−20πω=√−20πStep 3: The velocity of the oscillator can be determined byv=ω√A2−x2,wherevis the velocity,Ais the amplitude, andxis the displacement from theequilibrium position. We are interested in the maximum velocity, which occurswhenx= 0.Step 4: Calculate the maximum velocity:vmax=√−20π×p0.12−0210
vmax=√−20π×0.1vmax= 0.1√−20πm/sTherefore, the maximum speed of the oscillator is 0.1√−20πm/s.Question 13QuestionA particle undergoes simple harmonic motion with an amplitude of 3 cm and aperiod of 2 seconds. If the particle is at the equilibrium position at timet= 0,find:a) the displacement functionx(t),b) the velocity functionv(t),c) the acceleration functiona(t).Solutiona) To find the displacement functionx(t), we can use the general formula forsimple harmonic motion:x(t) =Acos2πTt,whereAis the amplitude andTis the period.Step 1:Given that the amplitudeA= 3 cm and the periodT= 2 s, wehave:x(t) = 3 cos2π2t.Simplifying:x(t) = 3 cos(πt).b) To find the velocity functionv(t), we differentiate the displacement func-tionx(t) with respect tot:v(t) =dxdt=−3πsin(πt).c) To find the acceleration functiona(t), we differentiate the velocity functionv(t) with respect tot:a(t) =dvdt=−3π2cos(πt).11
Question 14QuestionA block of massmis attached to a spring with spring constantk. The block isdisplaced from its equilibrium position and released. If the maximum speed ofthe block during its motion isv, find the amplitude of the motion.SolutionStep 1: We know that the maximum speed of the block during simple harmonicmotion occurs when the displacement is zero. At this point, all the potentialenergy has been converted to kinetic energy.Step 2:The potential energy stored in the spring at a displacementxisgiven byU=12kx2. At the equilibrium position, all the potential energy hasbeen converted to kinetic energy, soU= 0.Step 3: At the equilibrium position, the total energy of the system is thekinetic energy of the block:KE=12mv2.Step 4: The total energy of the system is constant and is the sum of thekinetic and potential energy:KE+U=12mv2+12kA2=12kA2.Step 5: Solving for the amplitudeA, we haveA=qmv2k. Therefore, theamplitude of the motion isA=rmv2k.Step 6: Thus, the amplitude of the motion when the maximum speed of theblock isvisqmv2k.Question 15QuestionA mass-spring system oscillates with an amplitude of 4 cm and a frequency of2 Hz. If the maximum speed of the mass is 16 cm/s, determine the equation ofmotion for the system.SolutionStep 1: Find the angular frequency,ω. Given that frequencyf= 2 Hz, we have:f=ω2πω= 2πf= 2π×2 = 4πrad/sStep 2: Find the equation of motion. The general equation of motion for amass-spring system undergoing simple harmonic motion is given by:x(t) =Acos(ωt−ϕ)12
where: -Ais the amplitude (given as 4 cm), -ωis the angular frequency (foundto be 4πrad/s), -ϕis the phase constant.Step 3: Find the phase constant,ϕ. To determine the phase constant, weneed to consider the initial conditions of the system. In this case, we know thatthe maximum speed occurs when the displacement is equal to the amplitude.This corresponds to a phase angle ofπ2.x(t) = 4 cos 4πt−π2= 4 cos 4πt+π2Therefore, the equation of motion for the mass-spring system is:x(t) = 4 cos 4πt+π2Question 16QuestionA 0.5 kg object is attached to a spring with a spring constant of 50 N/m. Theobject is pulled 0.1 m away from its equilibrium position and released fromrest.Determine the amplitude, frequency, and period of the resulting simpleharmonic motion.SolutionStep 1: Find the amplitude (A) Given that the object is pulled 0.1 m away fromits equilibrium position, the amplitude is equal to this distance.Therefore,A= 0.1 m.Step 2: Find the angular frequency (ω) The angular frequency can be foundusing the formula:ω=rkmwherekis the spring constant andmis the mass of the object.Substitutek= 50 N/m andm= 0.5 kg:ω=r500.5=√100 = 10 s−1Step 3:Find the frequency The frequency (f) is related to the angularfrequency by the formulaf=ω2π. So,f=102π≈1.59 HzStep 4: Find the period (T) The periodTis the reciprocal of the frequency:T=1f=11.59≈0.63 sTherefore, the amplitude of the simple harmonic motion is 0.1 m, the fre-quency is approximately 1.59 Hz, and the period is approximately 0.63 s.13
Question 17QuestionA particle of massmis attached to a spring with spring constantk. At timet= 0, the particle is displacedaunits from the equilibrium position and releasedfrom rest. Find the amplitude of the resulting simple harmonic motion in termsofa.SolutionStep 1: First, write down the equation of motion for simple harmonic motion:md2xdt2=−kxStep 2: To find the amplitude of the resulting simple harmonic motion, weneed to solve this differential equation. Let’s assume the solution is of the formx(t) =Acos(ωt+ϕ), whereAis the amplitude,ωis the angular frequency, andϕis the phase angle.Step 3: Calculate the first and second derivatives ofx(t):dxdt=−Aωsin(ωt+ϕ)d2xdt2=−Aω2cos(ωt+ϕ)Step 4: Substitutex(t) and its derivatives into the equation of motion:m(−Aω2cos(ωt+ϕ)) =−k(Acos(ωt+ϕ))Step 5: Divide both sides by−Acos(ωt+ϕ):mω2=kStep 6: Solve for the angular frequencyω:ω=rkmStep 7: The amplitudeAis related to the displacementaatt= 0 by:a=Acos(ϕ)Step 8: As the particle is released from rest, we havedxdtt=0= 0. Hence,ϕ=π2.Step 9: Substituteϕ=π2intoa=Acos(ϕ):a=Acosπ214
a=A×0A= 0Step 10: Therefore, the amplitude of the resulting simple harmonic motionin terms ofais0.Question 18QuestionA mass-spring system has a mass of 0.5 kg attached to a spring with a springconstant of 50 N/m. The system is released from rest at its equilibrium posi-tion. Find the amplitude of the resulting simple harmonic motion if the totalmechanical energy of the system is 5 J.SolutionStep 1: The total mechanical energy of the system can be expressed as the sumof the potential energy and the kinetic energy:E=U+KWhereEis the total energy,Uis the potential energy, andKis the kineticenergy.Step 2: At the equilibrium position, all the energy is in the form of potentialenergy, given by:U=12kA2wherekis the spring constant andAis the amplitude of motion.Step 3: At the extreme positions, all the energy is in the form of kineticenergy, given by:K=12mv2maxwheremis the mass of the object andvmaxis the maximum velocity.Step 4: Since the system is released from rest, the velocity at the extremepositions is zero. Thus, the total energy is equal to the potential energy at theequilibrium position:E=U=12kA2Step 5: Given thatE= 5 J,k= 50 N/m, andm= 0.5 kg, we can solve forA:5 =12×50×A2Step 6: Solving forA, we find:A=r550= 0.316 m15
Therefore, the amplitude of the resulting simple harmonic motion is 0.316m.Question 19QuestionA block of massmis attached to a spring with spring constantk. The blockis displaced a distanceAfrom its equilibrium position and released. Find themaximum speed of the block during its motion.SolutionStep 1:The total mechanical energy of the system is conserved.The totalmechanical energy is the sum of the kinetic energy (12mv2) and potential energy(12kx2) of the block-spring system, wherevis the velocity of the block andxis its position from the equilibrium point. Step 2: At the maximum speed, thepotential energy is zero at the equilibrium position. Step 3: At the maximumspeed, all the mechanical energy will be in the form of kinetic energy. Step 4:Using the conservation of energy, we have:12kA2=12mv2maxStep 5: Solving forvmax, we get:vmax=ArkmStep 6: Therefore, the maximum speed of the block during its motion isvmax=Aqkm.Question 20QuestionA particle of mass 0.2 kg is attached to a spring with spring constant 50 N/m.Initially, the particle is at the equilibrium position. The particle is then displaced0.1 m from the equilibrium position and released from rest. Find the amplitude,period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Calculate the amplitude. The amplitude of the simple harmonic motionis the maximum displacement from the equilibrium position.Given that theparticle is initially displaced 0.1 m from the equilibrium position, the amplitudeis also 0.1 m.16
Step 2: Calculate the angular frequency. The angular frequency of a mass-spring system is given by:ω=rkmwherekis the spring constant andmis the mass. Substitutingk= 50 N/m andm= 0.2 kg, we get:ω=r500.2=√250 = 5√10 rad/sStep 3: Calculate the period. The periodTof the simple harmonic motionis related to the angular frequencyωby the formula:T=2πωSubstituteω= 5√10 into the formula to get:T=2π5√10=2π5√10 sStep 4: Calculate the frequency.The frequencyfis the reciprocal of theperiodT. Therefore,f=1T=12π5√10=5√102π≈2.512 HzTherefore, the amplitude of the motion is 0.1 m, the period is2π5√10 seconds,and the frequency is approximately 2.512 Hz.Question 21QuestionA block of massmis attached to a spring with spring constantk. The blockis pulled a distanceAto the right and released from rest. Find an expressionfor the velocity of the block as a function of time during its subsequent simpleharmonic motion.SolutionStep 1: Determine the equation of motion for the block.The equation of motion for a block undergoing simple harmonic motion canbe expressed as:md2xdt2=−kxwherexis the displacement of the block from its equilibrium position.17
Step 2: Find the general solution to the differential equation.The general solution to the differential equationmd2xdt2=−kxcan be writtenas:x(t) =Acos(ωt) +Bsin(ωt)whereAandBare constants to be determined, andω=qkmis the angularfrequency.Step 3: Apply initial conditions to the general solution.Given that the block is released from rest at a distanceAto the right, wehave the initial conditionsx(0) =Aanddxdt(0) = 0.Substitutex(0) =Aanddxdt(0) = 0 into the general solution to solve for theconstantsAandB.Step 4: Determine the velocity of the block as a function of time.Differentiating the equationx(t) =Acos(ωt) +Bsin(ωt) with respect totgives the velocity function:dxdt=−Aωsin(ωt) +Bωcos(ωt)Therefore, the velocity of the block as a function of time is:v(t) =−Aωsin(ωt) +Bωcos(ωt)Question 22QuestionA particle undergoes simple harmonic motion with an angular frequency ofω= 3 rad/s and an amplitude of 0.1 m. If the particle starts from rest at theequilibrium positionx= 0, find the displacement of the particle at timet=π3s.SolutionStep 1: The general equation for simple harmonic motion is given byx(t) =Acos(ωt) +Bsin(ωt)whereAis the amplitude,ωis the angular frequency, andBis a constantdetermined by the initial conditions.Step 2: We are given that the amplitudeA= 0.1 m and the angular fre-quencyω= 3 rad/s. Since the particle starts from rest atx= 0, we can usethis information to determineB.Step 3: Att= 0, the equation becomes0 =Acos(0) +Bsin(0) =A·1 +B·0 =A= 0.118
Step 4: Now we have the equation for the motion of the particle:x(t) = 0.1 cos(3t) +Bsin(3t)Step 5: To find the value ofB, we can differentiatex(t) with respect totand sett= 0 (to represent the initial condition of starting from rest):v(t) =dxdt=−0.3 sin(3t) + 3Bcos(3t)v(0) =−0.3 sin(0) + 3Bcos(0) = 0Step 6: From this, we find thatB= 0.1.Step 7: Therefore, the equation for the displacement of the particle isx(t) = 0.1 cos(3t) + 0.1 sin(3t)Step 8: To find the displacement of the particle att=π3s, substitutet=π3intox(t):xπ3= 0.1 cos3·π3+ 0.1 sin3·π3Step 9: Simplifying givesxπ3= 0.1 cos(π) + 0.1 sin(π) = 0.1·(−1) + 0 =−0.1 mStep 10: Therefore, the displacement of the particle at timet=π3s is−0.1m.Question 23QuestionAn object is attached to a spring and undergoes simple harmonic motion withan amplitude of 0.2 m. If the object’s velocity is 2 m/s when it is 0.1 m fromthe equilibrium position, determine the object’s position after 0.1 seconds.SolutionStep 1: Find the angular frequencyωof the motion using the amplitude andvelocity. Given that the amplitudeA= 0.2 m and the velocityv= 2 m/s, wehave:v=Aω⇒ω=vA=20.2= 10 rad/sStep 2:Determine the object’s position after 0.1 seconds.The generalequation for the object’s position as a function of time in SHM isx(t) =Acos(ωt+ϕ), whereϕis the phase angle.19
Given that the object’s velocity is 2 m/s when it is 0.1 m from the equilibriumposition, we can determine the phase angleϕ:v=−Aωsin(ϕ) = 2⇒sin(ϕ) =−20.2×10=−1As sin(ϕ) =−1 in the second or third quadrant, we haveϕ=−π2.So, the object’s position after 0.1 seconds is:x(0.1) = 0.2 cos10×0.1−π2= 0.2 cosπ2= 0 mTherefore, the object’s position after 0.1 seconds is 0 meters from the equi-librium position.Question 24QuestionAn object of massmis attached to a spring with spring constantk. The objectis displaced a distanceAfrom its equilibrium position and released from rest.Find the period of the resulting simple harmonic motion.SolutionStep 1: Write the formula for the period of simple harmonic motion.The period (T) of simple harmonic motion is given by:T= 2πrmkStep 2: Find the angular frequency of the system.The angular frequency (ω) is given by:ω=rkmStep 3: Find the period of the simple harmonic motion.Substituteω=rkminto the period formula to get:T=2πω=2πqkm= 2πrmk20
Step 4: Conclusion.Therefore, the period of the simple harmonic motion of the object will be:T= 2πrmkQuestion 25QuestionA particle undergoes simple harmonic motion with an amplitude of 4 cm and aperiod of 2πseconds. Att= 0, the particle is at its maximum displacement of4 cm. Find the equation that describes the position of the particle at timet.SolutionStep 1: The general equation for simple harmonic motion is given byx(t) =Acos(ωt+ϕ), where: -Ais the amplitude, -ωis the angular frequency, -ϕisthe phase constant.Step 2: Given that the amplitudeA= 4 cm, we havex(t) = 4 cos(ωt+ϕ).Step 3: The periodTof the simple harmonic motion is related to the angularfrequencyωbyT=2πω.Step 4: Since the periodT= 2πseconds, we have2πω= 2π, which impliesω= 1 rad/s.Step 5: The equation becomesx(t) = 4 cos(t+ϕ).Step 6: Att= 0, the particle is at its maximum displacement of 4 cm. Thismeansx(0) = 4 = 4 cos(ϕ).Step 7: Solving forϕ: cos(ϕ) =44= 1, which impliesϕ= 0.Step 8:The equation describing the position of the particle at timetisx(t) = 4 cos(t).Question 26QuestionA particle undergoes simple harmonic motion with an amplitude of 4 cm and aperiod of 2 seconds. If the displacement of the particle is 3 cm at timet= 1second, find an equation for the particle’s motion.SolutionStep 1: Find the angular frequency.The angular frequency,ω, of a particle undergoing simple harmonic motion is21
given byω=2πTwhereTis the period. In this case, the periodT= 2 seconds, soω=2π2=πrad/sStep 2: Find the equation for the particle’s motion.The equation for the displacement of a particle undergoing simple harmonicmotion isx(t) =Acos(ωt−ϕ)where -Ais the amplitude, -ωis the angular frequency, -tis the time, and -ϕis the phase angle.Given that the amplitudeA= 4 cm, the angular frequencyω=π, and thedisplacement att= 1 second isx(1) = 3 cm, we can find the phase angleϕ.Step 3: Find the phase angleϕ.Substitutet= 1 andx= 3 into the equation and solve forϕ:3 = 4 cos(π−ϕ)34= cos(π−ϕ)It follows thatπ−ϕ= arccos34π−ϕ=π3ϕ=π−π3=2π3Step 4: Write the equation for the particle’s motion.Therefore, the equation for the particle’s motion isx(t) = 4 cosπt−2π3Question 27QuestionA particle is undergoing simple harmonic motion along the x-axis with an am-plitude of 5 cm and a period of 2 seconds.If the particle is at its maximumdisplacement (positive) and the velocity is also positive, determine the positionof the particle after 1 second.22
SolutionStep 1: First, we need to determine the angular frequency of the motion. Theangular frequencyωcan be calculated using the formulaω=2πT, whereTisthe period.ω=2π2=πrad/sStep 2: The position of the particle at any timetis given by the equationx(t) =Acos(ωt), whereAis the amplitude.Given that the particle is at itsmaximum displacement (positive) att= 0, the initial conditionx(0) =Agivesus the value ofA.x(0) =Acos(0) =A= 5 cmStep 3: Now we have the position functionx(t) = 5 cos(πt).To find theposition of the particle after 1 second, we substitutet= 1 into the equation.x(1) = 5 cos(π) = 5×(−1) =−5 cmTherefore, the position of the particle after 1 second is−5 cm.Question 28QuestionA particle undergoes simple harmonic motion with an amplitude of 2 cm anda frequency of 3 Hz.If the displacement of the particle is given byx(t) =2 sin(6πt), determine the velocity and acceleration of the particle when the dis-placement is 1 cm.SolutionStep 1:Find the velocity function.Given that the displacement function isx(t) = 2 sin(6πt), we can find the velocity function by taking the derivative ofthe displacement function with respect to time.v(t) =dxdt= 2(6π) cos(6πt) = 12πcos(6πt)Step 2: Find the acceleration function. Similarly, we can find the accelerationfunction by taking the derivative of the velocity function with respect to time.a(t) =dvdt=−12π2sin(6πt) =−12π2sin(6πt)Step 3: Determine the velocity and acceleration when the displacement is 1cm. To find the time when the particle’s displacement is 1 cm, we substitutex(t) = 1 into the displacement function:2 sin(6πt) = 123
sin(6πt) =12This occurs when 6πt=π6or 6πt=5π6. Thus,t=136ort=536seconds.Step 4: Calculate the velocity and acceleration att=136seconds. Substitutet=136into the velocity and acceleration functions:v136= 12πcosπ6= 6πa136=−12π2sinπ6=−6π2Therefore, when the particle’s displacement is 1 cm, the velocity is 6πcm/sand the acceleration is−6π2cm/s2.Question 29QuestionA particle undergoes simple harmonic motion with an amplitude of 5 cm anda period of 2 seconds.If att= 0, the particle is at its equilibrium position,determine the position functionx(t) for the particle.SolutionStep 1: Determine the angular frequencyωGiven that the periodT= 2 seconds, we have the relationT=2πω.Hence,ω=2π2=πrad/s.Step 2: Write the position function in terms of the amplitude and angularfrequencyThe position function for a particle undergoing simple harmonic motion is givenby:x(t) =Acos(ωt+ϕ)whereAis the amplitude,ωis the angular frequency, andϕis the phase angle.Step 3: Determine the phase angleϕSince att= 0 the particle is at its equilibrium position, we havex(0) =Acos(ϕ) = 0. This implies that cos(ϕ) = 0, which occurs whenϕ=π2.Step 4: Write the final position functionSubstitute the values ofA= 5 cm,ω=πrad/s, andϕ=π2into the positionfunction, we have:x(t) = 5 cosπt+π224
Question 30QuestionA mass-spring system has a period of oscillation of 2 seconds.If the mass isreplaced with one that is 4 times as heavy, what will be the new period ofoscillation?SolutionStep 1: Recall the formula for the period of oscillation of a mass-spring system:T= 2πrmkwhere:T= period of oscillation,m= mass of the object, andk= springconstant.Step 2: LetT1be the period of oscillation when the original massmis used,andT2be the period of oscillation when the mass is increased to 4m.Step 3: For the original system, we have:T1= 2πrmkStep 4: For the system with the increased mass, we have:T2= 2πr4mk= 2πr4mk= 2·2πrmk= 2T1Step 5: Thus, when the mass is replaced with one that is 4 times as heavy,the new period of oscillation will be twice the original period. The new periodwill be 4 seconds.Question 31QuestionA particle undergoes simple harmonic motion with an amplitude of 5 cm and aperiod of 2 seconds. If at timet= 0, the particle is at its equilibrium positionand moving upwards with a speed of 4 cm/s, determine the displacement of theparticle after 1 second.SolutionStep 1: Find the angular frequencyωusing the periodT. Given thatT= 2seconds, the angular frequencyωcan be calculated as:ω=2πT25
ω=2π2=πStep 2: Determine the displacementx(t) of the particle at timetusing theequation for simple harmonic motion:x(t) =Asin(ωt+ϕ)whereAis the amplitude,ωis the angular frequency,tis the time, andϕis thephase angle.Given that the amplitudeA= 5 cm, the angular frequencyω=π, and theparticle is at its equilibrium position att= 0, we have:x(t) = 5 sin(πt+ϕ)Step 3: Use the initial conditions to determine the phase angleϕ. Att= 0,the particle is at its equilibrium position and moving upwards with a speed of4 cm/s. This implies that att= 0, the particle is at its maximum displacementand moving upwards, which corresponds to the equation:x(0) = 5 sin(ϕ) = 5sin(ϕ) = 1ϕ=π2Step 4:Find the displacement of the particle after 1 second.Substitutet= 1 second into the equation for displacement:x(1) = 5 sinπ+π2x(1) = 5 sin3π2x(1) = 5×(−1) =−5Therefore, the displacement of the particle after 1 second is−5 cm.Question 32QuestionA mass-spring system oscillates with a frequency of 10 Hz and an amplitude of0.1 m. Determine the maximum acceleration of the mass during its motion.26
SolutionTo find the maximum acceleration, we first need to find the angular frequencyω. From the given frequencyf= 10 Hz, we haveω= 2πf= 2π×10 = 20πrad/s.Step 1:The equation for the acceleration of an object in simple harmonicmotion is given bya(t) =−ω2x(t), wherex(t) is the displacement of the objectat timet. Since acceleration is maximal when the displacement is maximal, weneed to find the maximum displacement. The equation for the displacement ofan object in simple harmonic motion is given byx(t) =Acos(ωt), whereAisthe amplitude.Step 2:At the maximum displacement (A), the displacement equationbecomesxmax=A. SubstitutingA= 0.1 m into the equation, we getxmax= 0.1m.Step 3:Now, we can find the maximum acceleration of the mass. Substi-tutingxmax= 0.1 m into the acceleration equation, we getamax=−ω2xmax=−(20π)2×0.1 m/s2.Step 4:Calculating the maximum acceleration givesamax=−400π2≈−1256.64 m/s2.Therefore, the maximum acceleration of the mass during itsmotion is approximately 1256.64 m/s2.Question 33QuestionA mass attached to a spring with spring constantk= 5 N/m undergoes simpleharmonic motion with a period of 2πs. If the amplitude of the motion is 0.1 m,determine the maximum speed of the mass.SolutionStep 1: Find the angular frequency Given that the period isT= 2πs, we canfind the angular frequencyωusing the formulaω=2πT.ω=2π2π= 1 s−1Step 2:Calculate the maximum speed The maximum speed of the massoccurs at the equilibrium position where the displacement is zero. At this point,the velocity is at its maximum. The maximum speed can be determined usingthe formulavmax=ωA, whereAis the amplitude of the motion.vmax= 1 s−1×0.1 mvmax= 0.1 m/sTherefore, the maximum speed of the mass is 0.1 m/s.27
Question 34QuestionA particle oscillates with simple harmonic motion given by the equationx(t) =0.1 cos(2t). Determine the amplitude, period, frequency, and maximum velocityof the particle.SolutionThe given equation for simple harmonic motion isx(t) = 0.1 cos(2t).Step 1:Find the amplitude of the motion.The amplitude of the motion is given by the coefficient of the cosine function,so the amplitude is 0.1.Step 2:Find the period of the motion.The period of the motion is the time taken for one complete oscillation. Inthis case, the periodTcan be found using the formulaT=2π2=π.Step 3:Find the frequency of the motion.The frequencyfof the motion is the reciprocal of the period, sof=1T=1π.Step 4:Find the maximum velocity of the particle.The velocity of the particle at any timetis given by the derivative of theposition functionx(t).v(t) =dxdt=−0.2 sin(2t)The maximum velocity occurs when sin(2t) = 1, so the maximum velocity is|vmax|= 0.2.Therefore, the amplitude is 0.1, the period isπ, the frequency is1π, and themaximum velocity of the particle is 0.2.Question 35QuestionA mass-spring system has a mass of 0.5 kg attached to a spring with a springconstant of 100 N/m. Initially, the mass is at its equilibrium position and is thendisplaced 0.1 m to the right and released from rest. Determine the amplitude,period, and frequency of the resulting simple harmonic motion.SolutionStep 1: Calculate the angular frequency (ω).Given that the spring constantk= 100 N/m and the massm= 0.5 kg, we can find the angular frequency usingthe formula:ω=rkm28
ω=s100 N/m0.5 kgω=√200ω= 10√2 rad/sStep 2: Calculate the amplitude (A). The amplitude of the motion is equalto the initial displacement, soA= 0.1 m.Step 3: Calculate the period (T). The period of the motion can be foundusing the formula:T=2πωT=2π10√2T=π5√2sStep 4: Calculate the frequency (f).The frequency of the motion is thereciprocal of the period, so:f=1T=5√2πHzTherefore, the amplitude of the resulting simple harmonic motion is 0.1 m,the period isπ5√2s, and the frequency is5√2πHz.29