Physics - Wave on String Question with Solution | TestHub

The linear mass density of the string is $\mu =M/L$. Consider a point at a distance $y$ from the lower (free) end. The tension $T(y)$ at this point is due to the weight of the string segment below it: $T(y)=\mu yg$. The speed of a transverse wave at this point is $v(y)=\sqrt{\frac{T(y)}{\mu }}=\sqrt{\frac{\mu yg}{\mu }}=\sqrt{yg}$.

The time $dt$ for the pulse to travel an infinitesimal distance $dy$ is $dt=\frac{dy}{v(y)}=\frac{dy}{\sqrt{yg}}$. Integrating from $y=0$ (lower end) to $y=L$ (upper end): $t={\int }_0^L\frac{dy}{\sqrt{yg}}=\frac{1}{\sqrt{g}}{\int }_0^L{y}^{-1/2}dy=\frac{1}{\sqrt{g}}[2y^{1/2}{]}_0^L=\frac{2}{\sqrt{g}}(\sqrt{L}-0)=2\sqrt{\frac{L}{g}}$.