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WileyPLUS: MyWileyPLUS| Help I Contact Us I Log Out WileyPLUS Halliday, Fundamentals of Physics, 10e PHYSICS WITH CALC (PHY 2048/2049) Home Read, Study & Pract ice Assignment Gradebook ORION Assignment > Open Assignment FULL SCREEN PRINTER VERSION BACK NEXT ASSIGNMENT RESOURCES Your answer is partially correct. Try again. omework These two waves travel along the same string: y1 = (3.72 mm) sin(2.15Tx-400t) Y2 = (5.35 mm) sin(2.15Tx-400 + 0.789/trad). What are (a) the amplitude and (b) the phase angle (relative to wave 1) of the resultant wave? (c) If a third wave of amplitude 5.25 mm is also to be sent along the string in the same direction as the first two waves, Problem 009 what should be its phase angle in order to maximize the amplitude of the new resultant wave? (a) Number 3.32960287 (b) Number -1.42096 (c) Numbe1.42096 Units mm UnitsT rad Units T rad Chapter 16 Problem 037 en the tolerance is +/-2% SHOW HINTG GO TUTORIAL er LINK TO TEXT LINK TO SAMPLE PROBLEM LINK TO SAMPLE PROBLEM VIDEO MINI-LECTURE eview Score Review Results b Study Obiective Question Attempts: 3 of 5 used SAVE FOR LATER SUBMIT ANSWER Copyright © 2000-2015 by John Wiley & Sons, Inc. or related companies. All rights reserved emen ns Version 4.16.3.2

Explanation / Answer

Let's say
y1= a1sin(kx + t + 1)
y2 = a2sin(kx + t + 2)
The resulting wave is the sum of the two waves. You get a wave of same frequency and wave number:
y_r = y1 + y2 = a_rsin(kx + t + _r)
ist amplitude and phase are given by:
a_r = ( a1² + a2² + 2a1a2cos(1-2) )
tan(_r) = (a1sin(1) + a2sin(2)) / (a1cos(1) + a2cos(2)

y1 = (3.72 mm) sin(2.15x - 400t)
y2 = (5.35 mm) sin(2.15x - 400t + 0.789rad).

A)
a1 = 3.72 mm
a2 = 5.35 mm
1 = 0
2 = 0.789
=>
a_r = ( 3.72² + 5.35² + 2*3.72*5.35*cos(0-0.789) ) = 3.32962 mm

B)
tan(3) = (3.72sin(0*) + 5.35sin(0.789)) / (3.72cos(0) + 5.35cos(0.789)) = -6.62429
=>
_r = -1.421 = -0.45
=>
phase shift with respect to wave 1 is:
= 3 - 1 = -0.45 - 0 = -0.45

C)
The addition of the third wave can be treated as superposition of the resultant of the first two and the third wave:
y_r' = y_r + y3
Assuming third wave has the same frequency and wave number as the others, you get for the resulting amplitude:
a_r' = ( a_r + a² + 2a_ra3cos(_r-3) )
The only factor influenced by the phase of the wave the cosine function, which takes values between -1 and +1. In order to maximize the
overall amplitude is should take the value 1, i.e.
cos(_r-3) = 1
<=>
_r - 3 = 0 (or more general n where n is an arbitrary integer)
Hence the phase angle which maximizes the resultant of all three waves is:
3 = _r = -0.45

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