Question

A uniform string of length $20\mathrm{m}$ is suspended from a rigid support. A short wave pulse is introduced at the lowest end. It starts moving up the string. The time taken to reach the support is?

Answer 1

Step 1: Given information

Length of uniform string, $\mathrm{L}=20\mathrm{m}$

Step 2: To find

We have to find the total time taken to reach the support.

Step 3: Calculation

Consider the figure given below.

The tension in a string is given as:

$\mathrm{T}=\frac{\mathrm{mgx}}{\mathrm{l}}$

Here,

function of length $\mathrm{x}$.

g is the acceleration due to gravity.

$\mathrm{L}$ is the length.

Now, the velocity at point P will be:

$$\begin{gathered} \mathrm{v}=\sqrt{\frac{{\displaystyle \frac{\mathrm{mgx}}{\mathrm{l}}}}{{\displaystyle \frac{\mathrm{m}}{\mathrm{l}}}}} \\ \mathrm{v}=\sqrt{\mathrm{gx}} \\ \frac{\mathrm{dx}}{\mathrm{dt}}=\sqrt{\mathrm{gx}} \end{gathered}$$

By integrating the terms,

$$\begin{gathered} {\int }_0^{20}\frac{\mathrm{dxt}}{\sqrt{\mathrm{x}}}={\int }_0^{\mathrm{t}}\sqrt{\mathrm{g}}\mathrm{dt} \\ {\left[2\sqrt{\mathrm{x}}\right]}_0^{20}=\sqrt{10}\mathrm{t} \\ 2\sqrt{20}=\sqrt{10}\mathrm{t} \\ \mathrm{t}=2\sqrt{2}\mathrm{s} \end{gathered}$$

Therefore, the time taken will be $2\sqrt{2}\mathrm{s}$.