The type of rubber band used inside some baseballs and golf balls obeys Hooke’s law over a wide range of elongation of the band. A segment of this material has an un-stretched length land a mass m. When a force Fis applied, the band stretches an additional length$\mathbf{∆}\mathit{l}$. (a) What is the speed (in terms of m, l, and the spring constant k) of transverse waves on this stretched rubber band? (b) Using your answer to (a), show that the time required for a transverse pulse to travel the length of the rubber band is proportional to $\mathbf{1}\mathbf{/}\sqrt{\mathbf{∆}\mathbf{l}}$ if role="math" localid="1660986246683" $\mathbf{∆}\mathit{l}\mathbf{\gg }\mathbf{}\mathit{l}$ and is constant if $\mathbf{∆}\mathit{l}\mathbf{\ll }\mathbf{}\mathit{l}$.

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, $v=\sqrt{\frac{F}{\mu }}........\left(1\right)$

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:

$\mu =\frac{m}{l+∆l}$

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

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

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:

$$\begin{gathered} \sqrt{\frac{k\times ∆l\times \left(l+∆l\right)}{m}}=\frac{2\mathrm{\pi }\left(\mathrm{l}+∆\mathrm{l}\right)}{t}\left(\because dis\tan ce=2\mathrm{\pi }\left(\mathrm{l}+∆\mathrm{l}\right)\right) \\ \sqrt{\frac{k\times ∆l\times \left(l+∆l\right)}{m}}=\frac{2\mathrm{\pi }\sqrt{{\left(\mathrm{l}+∆\mathrm{l}\right)}^2}}{t} \\ t=2\mathrm{\pi }\times \sqrt{\frac{m}{k}}\times \sqrt{\frac{{\left(l+∆l\right)}^2}{∆l\left(l+∆l\right)}} \\ =2\mathrm{\pi }\times \sqrt{\frac{m}{k}}\times \sqrt{\frac{\left(l+∆l\right)}{∆l}} \\ =2\mathrm{\pi }\times \sqrt{\frac{m}{k}}\times \sqrt{\frac{l}{∆l}+\frac{∆l}{∆l}} \\ =2\mathrm{\pi }\times \sqrt{\frac{m}{k}}\times \sqrt{1+\frac{l}{∆l}} \end{gathered}$$

If we have $\frac{l}{∆l}\gg 1$then we can write, $t\alpha \sqrt{\frac{l}{∆l}}$

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

Hence, the required value is is $\alpha \sqrt{\frac{l}{∆l}}$or $2\mathrm{\pi }=\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$.