hw12 - al-ameri (aoa434) – Hwk12 – Stokes –...

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Unformatted text preview: al-ameri (aoa434) – Hwk12 – Stokes – (19102)1This print-out should have 19 questions.Multiple-choice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 12. Itis due before 6.00 am on Tuesday 28th April,Abu Dhabi Time (9.00 pm Monday in Texas)00110.0 pointsThe equation of motion of a simple harmonicoscillator isd2xdt2=−9x,wherexis displacement andtis time.What is the period of oscillation?1.T=2π92.T=32π3.T=92π4.T=2π3correct5.T= 6πExplanation:d2xdt2=−ω2x,whereωis the angular frequency, so the pe-riod of oscillation isT=2πω=2π√9=2π3.002(part 1 of 3) 10.0 pointsA 668 g mass is connected to a light springof force constant 2 N/m that is free to oscillateon a horizontal, frictionless track. The massis displaced 8 cm from the equilibrium pointand released from rest.2 N/m668 g8 cmx= 0xFind the period of the motion.Correct answer: 3.63122 s.Explanation:Let :m= 668 g = 0.668 kgandk= 2 N/m.This situation corresponds to the specialcasex(t) =Acosωt,so the frequency isω=radicalbiggkm=radicalBigg2 N/m.668 kg= 1.73032 s−1and the period isT=2πω=2π1.73032 s−1=3.63122 s.003(part 2 of 3) 10.0 pointsWhat is the maximum speed of the mass?Correct answer: 0.138426 m/s.Explanation:Let :A= 8 cm.The velocity as a function of time isv(t) =−ω Asin(ω t),so the maximum speed of the mass isvmax=ω A= (1.73032 s−1) (0.08 m)=.138426 m/s.004(part 3 of 3) 10.0 pointsWhat is the maximum acceleration of themass?Correct answer: 0.239521 m/s2.al-ameri (aoa434) – Hwk12 – Stokes – (19102)2Explanation:The acceleration as a function of time isa(t) =−ω2Acos(ω t),so the maximum acceleration of the mass isamax=ω2A= (1.73032 s−1)2(0.08 m)=.239521 m/s2.005(part 1 of 4) 10.0 pointsA 15.1 kg mass is suspended on a 1×105N/mspring. The mass oscillates up and down fromthe equilibrium positionyeq= 0 according toy(t) =Asin(ωt+φ).Find the angular frequency of the oscillat-ing mass.Correct answer: 81.3788 s−1.Explanation:Let :M= 15.1 kgandk= 1×105N/m.When the mass moves out of equilibrium,it suffers a net restoring forceFnety=Fspring−Mg=−k(y−yeq) =−ky ,and accelerates back towards the equilibriumposition at the rateay=FnetyM=−kMy .Therefore, the mass oscillates harmonicallywith angular frequencyω=radicalbigg−ayy=radicalbiggkM=radicalBigg1×105N/m15.1 kg=81.3788 s−1.006(part 2 of 4) 10.0 pointsAt timet= 0 the mass is at 19.2 cm andmoving upward at velocity +21.1 m/s.Find the amplitude of the oscillating mass.Correct answer: 32.2631 cm.Explanation:Let :y= 19.2 cmandv= 21.1 m/s.The mass oscillates according to the SHMequationy(t) =Asin(ωt+φ),so its velocity isvy(t) =dydt=Aωcos(ωt+φ)....
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This note was uploaded on 02/16/2010 for the course PHYSICS 191262 taught by Professor Najafzadeh during the Spring '09 term at The Petroleum Institute.

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hw12 - al-ameri (aoa434) – Hwk12 – Stokes –...

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