91Ó°ÊÓ

By keeping the length, l of the string constant, the tension in the wire, T is doubled.

The speed of a wave $\left(v\right)$on a stretched string is directly proportional to the square root of the tension $\left(T\right)$in the string and is inversely proportional to the square root of the linear density $\left(\mathrm{μ}\right)$of the wire.

Thespeed of the wave on a stretched string,$v=\sqrt{\frac{T}{\mathrm{μ}}}$ (i)

Initially, for old wire, the speed of the wave on a stretched string using equation (i) is given by:

$V_0=\sqrt{\frac{t}{\mathrm{μ}}}........................................\left(1\right)$

T is the tension in the string,$\mathrm{μ}=\frac{m}{I}$is the linear density

It is given that the tension in the string is doubled and the length is kept constant.

Hence, for a new situation,$T_2=2T_1,\mathrm{μ}=\frac{m}{I}$will remain constant.

Therefore, the new wave speed using the given data and equation (i) is given by:

$v_n=\sqrt{\frac{2t}{\mathrm{μ}}}....................\left(2\right)$

Therefore, the ratio of the wave speeds of the new and old is given by dividing equation (2) by equation (1), that is:

$$\begin{gathered} \frac{v_n}{v_o}=\sqrt{\frac{2t}{\mathrm{μ}}}×\sqrt{\frac{\mathrm{μ}}{t}} \\ =\sqrt{\frac{2}{1}} \\ =1.44 \end{gathered}$$

Hence, the ratio of the new to the old wave speed for transverse waves traveling along the wire is 1.44