# 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 ${y}_{1}=A\mathrm{sin}\left(\pi x/L\right)\mathrm{sin}\omega t$ and energy is E1, and in another experiment its displacement is ${y}_{2}=A\mathrm{sin}\left(2\pi x/L\right)\mathrm{sin}2\omega t$ 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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