The ends of a stretched wire of length L are fixed at x = 0 and x = L. ln one experiment, the displacement of the wire is y1 = A sin(πxL) and energy is E1, and in another experiment its displacement is y2 = A sin(2πxL)sin 2 ωt and energy is E2. Then
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a
E2 = E1
b
E2 = 2E1
c
E2 = 4E1
d
E2 = 16E1
answer is C.
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Detailed Solution
Energy (E) ∝ (Amplitude)2 (Frequency)2 Amplitude is same in both the cases, but frequency 2ω in the second case is two times the frequency (ω) in the first case. Hence E2 = 4E1