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(B) The power transferred by a transverse wave at a point $P$ is given by the formula $P = T \cdot v_p \cdot \frac{\partial y}{\partial x}$,where $T$

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$20 \ cm/s$

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(B) The power transferred by a transverse wave at a point $P$ is given by the formula $P = T \cdot v_p \cdot \frac{\partial y}{\partial x}$,where $T$ is the tension and $v_p$ is the transverse velocity of the particle. Alternatively,the instantaneous power $P$ can be expressed as $P = F_y \cdot v_p$,where $F_y = T \cdot \frac{\partial y}{\partial x} = T \cdot \frac{v_p}{v_{wave}}$. Thus,$P = T \cdot \frac{v_p^2}{v_{wave}}$. Given: $P = 40 \ mW = 40 	imes 10^{-3} \ W$,$T = 20 \ N$,and $v_{wave} = 20 \ m/s$. Substituting the values: $40 	imes 10^{-3} = 20 \cdot \frac{v_p^2}{20}$. $40 	imes 10^{-3} = v_p^2$. $v_p = \sqrt{40 	imes 10^{-3}} = \sqrt{0.04} = 0.2 \ m/s$. Converting to $cm/s$: $0.2 \ m/s = 20 \ cm/s$.

$A$ transverse wave is passing through a stretched string with a speed of $20 \ m/s$. The tension in the string is $20 \ N$. At a certain point $P$ on the string,it is observed that energy is being transferred at a rate of $40 \ mW$ at a given instant. Find the speed of point $P$.

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