Question
A wire having length $\mathrm{L}$ is kept under tension between $\mathrm{x}=0$ and $\mathrm{x}=\mathrm{L}$. In one experiment, the equation of the wave and energy is given by $\mathrm{y}_{1}=\mathrm{A} \sin (\pi \mathrm{x} / \mathrm{L}) \sin \omega \mathrm{t}$ and $\mathrm{E}_{1}$respectively. In another experiment, it is $\mathrm{y}_{2}=\mathrm{A} \sin$ $\{(2 \pi \mathrm{x}) / \mathrm{L}\} \sin 2 \omega \mathrm{t}$ and $\mathrm{E}_{2}$. Then........(A) $E_{2}=E_{1}$(B) $E_{2}=2 \mathrm{E}_{1}$(C) $\mathrm{E}_{2}=4 \mathrm{E}_{1}$(D) $E_{2}=16 \mathrm{E}_{1}$
Step 1
Step 1: The energy of a wave is given by the formula $E = \frac{1}{2} \mu v^2 A^2$, where $\mu$ is the linear mass density, $v$ is the velocity of the wave, and $A$ is the amplitude of the wave. Show more…
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