Physics - Wave on String Question with Solution | TestHub

The linear mass density of the string is $\mu =M/L$. The total mass of the system (string + block) is $M+m$. The acceleration of the system is $a=\frac{F}{M+m}$.

Consider a point on the string at a distance $x$ from the block. The tension $T(x)$ at this point is responsible for accelerating the block and the string segment of length $x$. So, $T(x)=(m+\mu x)a=(m+\mu x)\frac{F}{M+m}$. The speed of a transverse wave at this point is $v(x)=\sqrt{\frac{T(x)}{\mu }}=\sqrt{\frac{(m+\mu x)F}{\mu (M+m)}}$.

The time $dt$ for the pulse to travel an infinitesimal distance $dx$ is $dt=\frac{dx}{v(x)}=\sqrt{\frac{\mu (M+m)}{(m+\mu x)F}}dx$. To find the total time, integrate from $x=0$ (block end) to $x=L$ (force end): $t={\int }_0^L\sqrt{\frac{\mu (M+m)}{F}}(m+\mu x{)}^{-1/2}dx$. Evaluating the integral: $t=\sqrt{\frac{\mu (M+m)}{F}}{\left[\frac{2}{\mu }\sqrt{m+\mu x}\right]}_0^L=2\sqrt{\frac{M+m}{\mu F}}(\sqrt{m+\mu L}-\sqrt{m})$. Substitute $\mu =M/L$: $t=2\sqrt{\frac{M+m}{(M/L)F}}(\sqrt{m+M}-\sqrt{m})=2\sqrt{\frac{L(M+m)}{MF}}(\sqrt{M+m}-\sqrt{m})$.