Understanding Oscillations: Multiple-Choice Questions for Physics
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shawarira (as239924) – shm3 – fox – (APCfox)1Thisprint-outshouldhave35questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.00110.0pointsA non-uniform stick of massMand lengthLhas a rotational inertiaI=45M L2aboutone end with its center of mass a distancercm=56Lfrom the same end. If the stick ishung from that end and swings freely in smalloscillations, its period isT=2π√b.pivotθM vectorgWhat is the value of the constantb?1.25g24L2.25g36L3.2g3L4.Zero; it does not oscillate at all sincethe period is proportional to the amplitude(assumed to be vanishingly small).5.9M g10L6.M g L257.3g4L8.g2L9.4g6L M10.g L35M002(part1of2)10.0pointsTwo massless springs with spring constants363 N/m and 726 N/m are hung from a hor-izontal support.A block of mass 2 kg issuspended from the pair of springs, as shown.When the block is in equilibrium, each springis stretched an additional Δx. Then the blockoscillates with an amplitude of 95 cm, and itpasses through its equilibrium point with aspeed ofv.363 N/m726 N/m2 kgWhat is the angular velocity of this system?The acceleration due to gravity is 9.8 m/s.1.ω= 42.ω= 11 rad/s3.ω= 54.ω= 155.ω= 176.ω= 167.ω= 128.ω= 29.ω= 810.ω= 10
shawarira (as239924) – shm3 – fox – (APCfox)2003(part2of2)10.0pointsWhat is the speed of the mass at its equilib-rium position?1.v0= 1.482.v0= 9.123.v0= 10.45 m/s4.v0= 9.795.v0= 10.786.v0= 9.47.v0= 15.778.v0= 8.329.v0= 1.110.v0= 1.3200410.0pointsA solid sphere withIcm=25M R2is attachedto a horizontal spring with negligible massand spring constantk.If the sphere rollswithout slipping on a horizontal tabletop, itssimple harmonic oscillations will have periodT= 2π√e.What is the value ofe?1.5k7M2.Mk3.7M5k4.7M4k5.5M2k6.2M5k7.The sphere will not be a harmonic oscil-lator because the force due to static frictionfsis a constant independent of the sphere’scenter of mass positionx.8.Zero9.3M4k005(part1of4)10.0pointsA block of mass 1.1 kg hangs without vi-brating at the end of a spring (with constant287 N/m) that is attached to the ceiling ofan elevator car. The car is rising with an up-ward acceleration ofg8when the accelerationsuddenly ceases (att= 0 s).What is the angular frequency of oscillationof the block after the elevator car’s accelera-tion ceases?The acceleration of gravity is9.8 m/s2.Answer in units of rad/s.006(part2of4)10.0pointsBy what amount is the spring stretched whenelevator car is accelerating?Answer in units of cm.007(part3of4)10.0pointsWhat is the amplitude of the oscillation ob-served by a rider in the car?Answer in units of cm.008(part4of4)10.0pointsIn the sine function description of the motion,x-x0=Asin(ω t+φ0) where the positivedirection ofxis upward, what is the initialphaseφ0?1.φ0=-π42.φ0=-π2
shawarira (as239924) – shm3 – fox – (APCfox)33.φ0=π44.φ0= 15.φ0=-16.φ0=-π7.φ0=π8.φ0= 09.φ0=π210.φ0= 2π009(part1of3)10.0pointsIn an experiment conducted on the spaceshuttle;i.e., in free fall), a horizontal rod ofmass 63 kg and length 63 m is pivoted abouta point 15 m from one end, while the oppositeend is attached to a spring of force constant32 N/m and negligible mass.15 m11 m63 kg63 mθ32 N/mCalculate the moment of inertiaIaboutthe pivot point.Answer in units of kgm2.010(part2of3)10.0pointsThe “torque” equation of motion for simpleharmonic motion is of the formτ=-κ θ .When the rod is displaced by a small angleθfrom the horizontal and released, calculateκ .Assumeθis small so the small angle ap-proximation may be used.Answer in units of Nm.011(part3of3)10.0pointsCalculate the angular velocity of the resultingsimple harmonic motion.Answer in units of rad/s.01210.0pointsA 49 kg person steps into a car of mass2347 kg, causing it to sink 4.43 cm on itssprings.Assumingnodamping,withwhatfre-quency will the car and passenger vibrate onthe springs?The acceleration of gravity is9.8 m/s2.Answer in units of Hz.013(part1of2)10.0pointsAmassm1=12.5kgisinequilibriumwhile connected to a light spring of constant129 N/m that is fastened to a wall (see a). Asecond mass,m2= 5.3 kg, is slowly pushedup against massm1, compressing the springby the amountAi= 0.15 m (see b). The sys-tem is then released, causing both masses tostart moving to the right on the frictionlesssurface. Whenm1is at the equilibrium point,m2loses contact withm1(see c) and movesto the right with speedvmax.m1m1m1m1m2m2m2kkkkAiAfDvv(a)(b)(c)(d)Determine the value ofvmax.Answer in units of m/s.014(part2of2)10.0pointsHow far apart are the masses when the springis fully stretched for the first time (see d)?First determine the period of oscillation andthe amplitude of them1spring system afterm2loses contact withm1.Answer in units of cm.
shawarira (as239924) – shm3 – fox – (APCfox)401510.0pointsThe graph below represents the potential en-ergyUas a function of displacementxforan object on the end of a spring (F=-k x)oscillating in simple harmonic motion withamplitudexmax.+xmax-xmaxxUUmaxWhich graph represents the kinetic energyKof the object as a function of displacementx?1.+xmax-xmaxxKKmax2.+xmax-xmaxxKKmax3.+xmax-xmaxxKKmax4.+xmax-xmaxxKKmax5.+xmax-xmaxxKKmax6.+xmax-xmaxxKKmax7.+xmax-xmaxxKKmax01610.0pointsThe graph below represents the potential en-ergyUas a function of displacementxforan object on the end of a spring (F=-k x)oscillating in simple harmonic motion withamplitudexmax.+xmax-xmaxxUUmaxWhich graph represents the kinetic energyKof the object as a function of displacementx?1.+xmax-xmaxxKKmax
shawarira (as239924) – shm3 – fox – (APCfox)52.+xmax-xmaxxKKmax3.+xmax-xmaxxKKmax4.+xmax-xmaxxKKmax5.+xmax-xmaxxKKmax01710.0pointsIf almost any system in stable equilibrium isslightly disturbed, it will then exhibit simpleharmonic motion because1.momentum and energy are both con-served.2.the potential energy of a system neara state of stable equilibrium is proportionalto the square of the displacement from theequilibrium position.3.the kinetic energy on a system in stableequilibrium is zero.4.energy is conserved.5.momentum is conserved.6.the momentum of a system in stable equi-librium is zero.7.the force on a system in stable equilibriumis zero.8.the force on a system in unstable equilib-rium is zero.9.the potential energy of a system near astate of stable equilibrium is proportional tothe cube of the displacement from the equi-librium position.10.the potential energy of a system near astate of stable equilibrium is linearly propor-tional to the displacement from the equilib-rium position.01810.0pointsThe cylindrical disk has mass 2.35 kg andouter radius 3.85 cm with a radial mass dis-tribution (not necessarily uniform) so that itsmoment of inertia is78m r2.The disk rollsaround a horizontal axis through its centerwithout slipping in a cylindrical trough of ra-dius 34.4575 cm.The acceleration of gravity is 9.8 m/s2.34.4575 cm3.85 cmDetermine (for small displacements fromequilibrium) the period of harmonic oscilla-tion which the disk undergoes.Answer in units of s.019(part1of2)10.0pointsA 0.362 kg mass is attached to a spring andexecutes simple harmonic motion with a pe-riod of 0.99 s. The total energy of the systemis 2.6 J.Find the force constant of the spring.Answer in units of N/m.
shawarira (as239924) – shm3 – fox – (APCfox)6020(part2of2)10.0pointsFind the amplitude of the motion.Answer in units of m.02110.0pointsThe displacement in simple harmonic motionis maximum when the1.velocity is zero.2.linear momentum is a maximum.3.acceleration is zero.4.velocity is a maximum.5.kinetic energy is a maximum.02210.0pointsA mass is placed on a spring and oscillateswith a period of 1 second.Now a heavier mass is placed on the samespring.Which statements are true?I. Theheaviermassoscillateswithashorter period because the gravitationalforce on it is greater.II. Theheaviermassoscillateswiththesame period because gravitational accel-eration is constant.III. The heavier mass oscillates with a longerperiod because of its greater inertia.IV. The heavier mass must have greater me-chanical energy than the first because itis heavier.V. The heavier mass must have less me-chanical energy than the first because itmoves more slowly.VI. One cannot reach any conclusion aboutmechanical energy without knowing theamplitude of motion in each case.1.I and IV only2.II and VI only3.II and V only4.III and VI only5.III and IV only6.III and V only7.I and VI only8.II and IV only9.I and V only02310.0pointsAn automobile having a mass of 880 kg isdriven into a brick wall in a safety test.The bumper behaves like a spring of con-stant 5.18×106N/m and compresses 3 cm asthe car is brought to rest.What was the speed of the car before im-pact, assuming no energy is lost during impactwith the wall?Answer in units of m/s.024(part1of3)10.0pointsConsider a uniform stick with massm1=mand lengthℓ. A massm2=m3is attached toits end.It is pivoted at P which is14of thedistance from the other end.XX12PmmFind the moment of inertia about a rotationaxis passing through P perpendicular to thepaper.1.512m ℓ22.712m ℓ23.12m ℓ2
shawarira (as239924) – shm3 – fox – (APCfox)74.23m ℓ25.112m ℓ26.34m ℓ27.14m ℓ28.13m ℓ29.16m ℓ210.56m ℓ2025(part2of3)10.0pointsThe total kinetic energy as the mass-rod sys-tem passes the vertical after being releasedfrom rest in the horizontal position is givenby1.712m ℓ g2.56m ℓ g3.23m ℓ g4.16m ℓ g5.13m ℓ g6.112m ℓ g7.14m ℓ g8.34m ℓ g9.12m ℓ g10.512m ℓ g026(part3of3)10.0pointsThe period of a simple pendulum with lengthℓisT0= 2πradicalBiggℓg.What is the period of oscillation for thesmall angle approximation about the verticaldirection?1.12T02.radicalbigg56T03.radicalbigg23T04.radicalbigg512T05.radicalbigg13T06.radicalbigg12T07.radicalbigg34T08.radicalbigg712T09.radicalbigg16T010.radicalbigg112T0027(part1of2)10.0pointsConsider a light rod of negligible mass andlength 6 m pivoted on a frictionless horizontalbearing at a pointO .Attached to the end ofthe rod is a massM1= 8 kg.Also, a secondmassM2= 8 kg of equal size is attached tothe rodparenleftbigg37Lfrom the lower endparenrightbigg, as shownin the figure below.
shawarira (as239924) – shm3 – fox – (APCfox)837LLM2M1OθWhat is the moment of inertiaIaboutO?The acceleration of gravity is 9.8 m/s2.Answer in units of kg m2.028(part2of2)10.0pointsWhat is the period of this pendulum in thesmall angle approximation?Answer in units of s.02910.0pointsA uniform circular disk is pivoted at its edgeA.RAFind its period of the small oscillations.Use small angle approximation.1.T= 2πradicalBiggRg2.T= 2πradicalBigg11R6g3.T= 2πradicalBigg3R2g4.T= 2πradicalBigg5R3g5.T= 2πradicalBigg2Rg6.T= 2πradicalBigg5R2g7.T= 2πradicalBiggR3g8.T= 2πradicalBigg4R3g9.T= 2πradicalBigg7R6g10.T= 2πradicalBiggR2g030(part1of2)10.0pointsA hoop has a radiusRand massm.It ispivoted atA, a point on the circumferenceof the hoop.Denote the moment of inertiaabout the center byI0and about the pivotpoint A byIA.The acceleration of gravity is 9.8 m/s2.AOO'θThe equation of motion aboutAin thesmall angle approximation (sinθ≈θ) is givenby
shawarira (as239924) – shm3 – fox – (APCfox)91.I0d2θdt2=-12m g R θ .2.IAd2θdt2= 2m g R θ .3.IAd2θdt2=-m g R θ .4.I0d2θdt2= 2m g R θ .5.I0d2θdt2=-2m g R θ .6.I0d2θdt2=m g R θ .7.IAd2θdt2=-12m g R θ .8.IAd2θdt2=m g R θ .9.IAd2θdt2=-2m g R θ .10.I0d2θdt2=-m g R θ .031(part2of2)10.0pointsLetm= 0.9 kg,R= 0.64 m,g= 9.8 m/s2,I0=m R2.Determine the period of small oscillation ofthe hoop.Answer in units of s.03210.0pointsA physical pendulum in the form of a planarbody moves in simple harmonic motion witha frequency of 0.184 Hz.The acceleration of gravity is 9.8 m/s2.If the pendulum has a mass of 2.87 kg andthe pivot is located 0.259 m from the center ofmass, determine the moment of inertia of thependulum.Answer in units of kg m2.033(part1of2)10.0pointsHint:The moment of inertia of a uniform rodabout its center-of-mass is112M L2.The acceleration of gravity is 9.8 m/s2.Consider a uniform rod with a massM=2 kg and lengthL= 7.5 m pivoted on africtionless horizontal bearing at a pointOparenleftbigg89Lfrom the lower endparenrightbigg,as shown in thefigure.89LLOθThe moment of inertiaIof the rod aboutthe pivot pointOis given byAnswer in units of kg m2.034(part2of2)10.0pointsBased on the equation of the motion givenin the previous question, the period of thispendulum in the small angle approximation isgiven byAnswer in units of s.03510.0pointsHint:The moment of inertia of a uniform rodabout its center-of-mass is112M L2.The acceleration of gravity is 9.8 m/s2.Consider a uniform rod with a massM=4 kg and lengthL= 6.2 m pivoted on africtionless horizontal bearing at a pointOparenleftbigg67Lfrom the lower endparenrightbigg,as shown in thefigure.
shawarira (as239924) – shm3 – fox – (APCfox)1067LLOθThe period of this pendulum in the smallangle approximation is given byAnswer in units of s.