The ends of the stretched wire of length L are fixed at x = 0 and x...

The total energy of a standing wave in a stretched string is equal to its maximum potential energy, given by:

E = \dfrac{1}{2} T \int_0^L \left(\dfrac{dy}{dx}\right)^2 dx$<br><br>For the first displacement$y_1 = A\sin\left(\dfrac{\pi x}{L}\right):

\dfrac{dy_1}{dx} = \dfrac{A\pi}{L}\cos\left(\dfrac{\pi x}{L}\right)$<br><br>$E_1 = \dfrac{1}{2} T \int_0^L \dfrac{A^2\pi^2}{L^2}\cos^2\left(\dfrac{\pi x}{L}\right) dx$<br><br>Since$\int_0^L \cos^2\left(\dfrac{\pi x}{L}\right) dx = \dfrac{L}{2}$:<br><br>$E_1 = \dfrac{1}{2} T \left(\dfrac{A^2\pi^2}{L^2}\right) \dfrac{L}{2} = \dfrac{T A^2 \pi^2}{4L}$<br><br>For the second displacement$y_2 = A\sin $\left(\dfrac{2\pi x}{L}\right)$:

\dfrac{dy_2}{dx} = \dfrac{2A\pi}{L}\cos\left(\dfrac{2\pi x}{L}\right)$<br><br>$E_2 =\dfrac{1}{2}$$T \int_0^L$$\dfrac{4A^2\pi^2}{L^2}\cos^2 $\left(\dfrac{2\pi x}{L}\right)$dx$<br><br>Since $\int_0^L \$\cos^2 $\left(\dfrac{2\pi x}{L}\right)$dx =\dfrac{L}{2}$:<br><br>$E_2 = $$\dfrac{1}{2}$$T $\left(\dfrac{4A^2\pi^2}{L^2}\right) \dfrac{L}{2} =$4 \left(\dfrac{T A² \pi^2}{4L}\right)$<br><br>Therefore,$E_2 = 4E_1$.<br><br>Answer:$E_2 = 4E_1