NEET Physics Waves Questions Solved

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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(πx/L)sinωt and energy is E1, and in another experiment its displacement is y2=Asin(2πx/L)sin2ωt and energy is E2. Then 

(1) E2 = E1

(2) E2 = 2E1

(3) E2 = 4E1

(4) E2 = 16E1

(3) 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.

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