91Ó°ÊÓ

Un-stretched length is l

Mass of the block is m

Spring constant of the spring is k

Additional length is $∆l$

We use the formula for velocity in terms of force and mass per unit length, and to find time, we use the basic formula for speed which is distance divided by time.

Formula:

The velocity of a transverse wave, $v=\sqrt{\frac{F}{\mathrm{μ}}}........\left(1\right)$

The velocity of a body, $v=\frac{L}{t}........\left(2\right)$

Here,theforce is due tothespring and it is as follows:

And mass per unit length is as follows:

$\mathrm{μ}=\frac{m}{l+∆l}$

So speed of the wave using equation (1) is given as follows:

$$\begin{gathered} v=\sqrt{\frac{k+∆l}{{\displaystyle \frac{m}{l+∆l}}}} \\ =\sqrt{\frac{k×∆l×\left(l+∆l\right)}{m}} \end{gathered}$$

Hence, the value of the speed is $\sqrt{\frac{k∆l\left(l+∆l\right)}{m}}$

Now time using equation (2) and the derived value of the speed is given as follows:

$$\begin{gathered} \sqrt{\frac{k×∆l×\left(l+∆l\right)}{m}}=\frac{2\mathrm{Ï€}\left(\mathrm{l}+∆\mathrm{l}\right)}{t}\left(âˆ\mu dis\tan ce=2\mathrm{Ï€}\left(\mathrm{l}+∆\mathrm{l}\right)\right) \\ \sqrt{\frac{k×∆l×\left(l+∆l\right)}{m}}=\frac{2\mathrm{Ï€}\sqrt{{\left(\mathrm{l}+∆\mathrm{l}\right)}^2}}{t} \\ t=2\mathrm{Ï€}×\sqrt{\frac{m}{k}}×\sqrt{\frac{{\left(l+∆l\right)}^2}{∆l\left(l+∆l\right)}} \\ =2\mathrm{Ï€}×\sqrt{\frac{m}{k}}×\sqrt{\frac{\left(l+∆l\right)}{∆l}} \\ =2\mathrm{Ï€}×\sqrt{\frac{m}{k}}×\sqrt{\frac{l}{∆l}+\frac{∆l}{∆l}} \\ =2\mathrm{Ï€}×\sqrt{\frac{m}{k}}×\sqrt{1+\frac{l}{∆l}} \end{gathered}$$

If we have $\frac{l}{∆l}≫1$then we can write, $t\mathrm{Î\pm }\sqrt{\frac{l}{∆l}}$

If the ratio, $\frac{l}{∆l}≪1$we can neglect it, so we can have, $t=2\mathrm{π}×\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

Hence, the required value is is $\mathrm{Î\pm }\sqrt{\frac{l}{∆l}}$or $2\mathrm{Ï€}=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$.