The ends of a stretched wire of length L are fixed at x = 0 and x… - Vidyadip

MCQ

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 ${y_1} = A\sin (\pi x/L)\sin \omega t$ and energy is ${E_1}$, and in another experiment its displacement is ${y_2} = A\sin (2\pi x/L)\sin 2\omega t$ and energy is ${E_2}$. Then

A
${E_2} = {E_1}$
B
${E_2} = 2{E_1}$
✓
${E_2} = 4{E_1}$
D
${E_2} = 16{E_1}$

✓

Answer

Correct option: C.

${E_2} = 4{E_1}$

c
(c) Energy $(E) \propto$ (Amplitude)$^2$ (Frequency)$^2$
Amplitude is same in both the cases, but frequency $2 \omega $ in the second case is two times the frequency $(\omega )$ in the first case. Hence ${E_2} = 4{E_1}$.

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