Wave Motion 5 Question 4

4. The ends of a stretched wire of length L are fixed at x=0 and x=L. In one experiment the displacement of the wire is y1=Asin(πxL)sinωt and energy is E1 and in other experiment its displacement is y2=Asin(2πxL)sin2ωt and energy is E2. Then

(2001, 1M)

(a) E2=E1

(b) E2=2E1

(c) E2=4E1

(d) E2=16E1

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Answer:

Correct Answer: 4. (c)

Solution:

  1. Energy E (amplitude) 2 (frequency )2

Amplitude (A) is same in both the cases, but frequency 2ω in the second case is two times the frequency (ω) in the first case.

Therefore,

E2=4E1



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