A wave moving with constant speed on a uniform string passes the point x = 0 with amplitude $A_{0}$ , angular frequency $ω_{0}$ and average rate of energy transfer $P_{0}$ . As the wave travels down the string it gradually loses energy and at the point x = l, the average rate of energy transfer becomes half. At the point x = l, angular frequency and amplitude are respectively:

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$ω_{0} and \frac{A_{0}}{\sqrt{2}}$

$\frac{ω_{0}}{\sqrt{2}} and \frac{A_{0}}{\sqrt{2}}$

$\frac{ω_{0}}{\sqrt{2}}$

less than $ω_{0} and A_{0}$

answer is A.

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Detailed Solution

$P = \frac{1}{2} {μω}^{2} A^{2} V$ and $v = \sqrt{\frac{T}{μ}}$ willl not change as both T and p are constant. $ω$ will also not change as it is property of the source only that is causing the wave motion. Hence to make power half the amplitude becomes $\frac{A_{0}}{\sqrt{2}}$ .