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Unformatted text preview: alameri (aoa434) – Hwk13 – Stokes – (19102)1This printout should have 24 questions.Multiplechoice questions may continue onthe next column or page – find all choicesbefore answering.This is ONLINE HOMEWORK No 13. Itis due before 6.00 am on Tuesday 5 May, AbuDhabi Time (9.00 pm on Monday in Texas)00110.0 pointsA harmonic wave is traveling along a rope.The oscillator that generates the wave completes 44.0 vibrations in 23.2 s. A given crestof the wave travels 411 cm along the rope in atime period of 10.3 s.What is the wavelength?Correct answer: 0.210397 m.Explanation:Let :n= 44.0 vibrations,t= 23.2 s,Δx= 411 cm = 4.11 m,andΔt= 10.3 s.λ=vf=ΔxΔtnt=tΔxnΔt=(23.2 s)(4.11 m)44 (10.3 s)=.210397 m.002(part 1 of 3) 10.0 pointsA transverse wave of frequency 25 Hz propagates down a string. Two points 15 cm apartare out of phase by7π3.What is the wave length of the wave?Correct answer: 12.8571 cm.Explanation:Let :Δx= 15 cmandδ=7π3.Let the positivexdirection be the directionof propagation of the wave.The phase difference isδ= 2πΔxλλ= 2πΔxδ= 2π15 cm7π/3=12.8571 cm.003(part 2 of 3) 10.0 pointsAt a given point, what is the phase difference(as a multiple ofπ) between two ‘displacements for times 5 ms apart?Correct answer: 0.25π.Explanation:Let :t= 5 ms = 0.005 sandf= 25 Hz.The period isT=1f, so the phase differenceisδ= 2πtT= 2π t f= 2π(0.005 s) (25 Hz)=.25π.004(part 3 of 3) 10.0 pointsWhat is the wave velocity?Correct answer: 3.21429 m/s.Explanation:v=f λ= (25 Hz) (12.8571 cm)parenleftbigg1 m100 cmparenrightbigg=3.21429 m/s.keywords:00510.0 pointsThe figure shows two wave pulses that areapproaching each other.PQalameri (aoa434) – Hwk13 – Stokes – (19102)2Which of the following best shows the shapeof the resultant pulse when the centers of thepulses, pointsPandQ, coincide?1.2.3.correct4.5.Explanation:Notice that the two pulses have the samewidth and amplitude.Choosing the the pointP(the same as pointQwhen the two pulses coincide) as the origin,the two pulses can be described as:P:y1=A ,−d≤x≤dQ:y2=braceleftBiggA,−d≤x <−A,< x < dUsing the principle of superposition, the resultant pulse isy=y1+y2=braceleftbigg2A ,−d≤x <,< x < dPQP+Q00610.0 pointsA 2.5 kg block supported by a string rests ona frictionless incline. The length of the stringis 0.1 m and its mass of 1.6 g≪2.5 kg....
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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.
 Spring '09
 NAJAFZADEH
 mechanics, Work

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