Physics - Wave on String Question with Solution | TestHub

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

$E=\frac{1}{2}T{\int }_0^L{\left(\frac{dy}{dx}\right)}^{2}dx$

For the first displacement $y_1=A\sin \left(\frac{\pi x}{L}\right)$:

$\frac{d{y}_1}{dx}=\frac{A\pi }{L}\cos \left(\frac{\pi x}{L}\right)$

$E_1=\frac{1}{2}T{\int }_0^L\frac{A^2{\pi }^2}{L^2}{\cos }^2\left(\frac{\pi x}{L}\right)dx$

Since ${\int }_0^L{\cos }^2\left(\frac{\pi x}{L}\right)dx=\frac{L}{2}$:

$E_1=\frac{1}{2}T\left(\frac{A^2{\pi }^2}{L^2}\right)\frac{L}{2}=\frac{T{A}^2{\pi }^2}{4L}$

For the second displacement $y_2=A\sin \left(\frac{2\pi x}{L}\right)$:

$\frac{d{y}_2}{dx}=\frac{2A\pi }{L}\cos \left(\frac{2\pi x}{L}\right)$

$E_2=\frac{1}{2}T{\int }_0^L\frac{4A^2{\pi }^2}{L^2}{\cos }^2\left(\frac{2\pi x}{L}\right)dx$

Since ${\int }_0^L{\cos }^2\left(\frac{2\pi x}{L}\right)dx=\frac{L}{2}$:

$E_2=\frac{1}{2}T\left(\frac{4A^2{\pi }^2}{L^2}\right)\frac{L}{2}=4\left(\frac{T{A}^2{\pi }^2}{4L}\right)$

Therefore, $E_2=4E_1$.

Answer: $E_2=4E_1$