MasterinqPh,sics Ch15 HW - CoogleChrome 0 Seriire https:/ se sinn.masteringphysi

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MasterinqPh,sics Ch15 HW - CoogleChrome 0 Seriire https:/ se sinn.masteringphysirs.com my t temVif. ? assignment Problem!D=81 83 0465&off; en xt Ch 15 HW Problerm 15.71 Rescurces T previous 14 of 15 next Problem 15.71 Part A A sinusoidal traveling wave has trequency 840 Hz and phase velocity 430 m/s At a given time, find the distance between any two locations that correspond to a difference in phase of rad Express your answer using two significant figures. Submit My Anewer Give Up Incorrect; Try Again; 5 attempts remaining Part B Al a fixed location, by how much does the phase change during a time interval of 1.0x10-4s? Express your answer using two significant figures. rad Submit My Answers Give Up Continue Type here to search 2:58 PM 9/16 2017

Explanation / Answer

15.71)

Part A) There are actually multiple answers for part A), since the phase will repeat with each cycle. You can probably find a formula to do this in one step, but here is a logical sequence you that hopefully you can follow.

1 cycle of 2 pi radians takes 1/840 seconds. In that time, the traveling wave will go 430 * 1/840 = 43/84 meter.
So 2 pi radians = 43/84 meters
pi = 43/168 meters
1/6 pi radian = 0.043 meters.

Since the phase repeats every cycle, not only will two point 0.043 meters be pi/6 phase change, but also points that are 0.043 + n*25/65 meters apart, where n is any positive integer (with n=0 giving the original result). Since the problem doesn't specify what direction of phase angle, there is another whole set of answers, which are -0.043 + n*25/65 meters, where n is any positive integer.

Part B)

The thing to note is that something oscillating at 1 cycle per second or 1 Hz will have a angular frequency of 2 pi radians per second. (There are 2 pi radians in a circle).
840Hz = 840 cycles second
840 cycles/ second * 0.0001 sec = 0.084 cycles in 0.0001 seconds.
0.084 cycles * 2 pi radians/cycle = 0.53 radians of phase change in 0.0001 seconds.

15.54) Part A)

The locations of the nodes satisfy the relation 0.70*xn = k , which tells that

xn = 10k/7   with k = 0,1,2, ......... So the distance between the nodes is

dn,kl = I xk - xl | = 10m/7   with m = 0,1,2, .........

Part B) From the wave function, one can find

Amplitude = DM = 3.2 cm

Part C) Frequency = f = /2 = 48/2 = 7.64 Hz

Part D) v1 = v2 = /k = 48/0.7 = 68.6 cm/s

Part E) The speed of a particle at x at an instant t is

vp = D/t = -153.6*sin(0.7x)sin(48t) cm/s

So, the speed of a particle of the string at x = 2.5 cm when t = 2.2 s is

vp(2.5, 2.2) = -153.6*sin(0.7*2.5)sin(48*2.2) cm/s = 1.42 m/s

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