The tension in a wire clamped at both ends is doubled without appreciably changing the wire’s length between the clamps. What is the ratio of the new to the old wave speed for transverse waves traveling along this wire?

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 $\left(\mu \right)$of the wire.

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

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

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

T is the tension in the string,$\mu =\frac{m}{I}$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,$T_2=2T_1,\mu =\frac{m}{I}$will remain constant.

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

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

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

$$\begin{gathered} \frac{v_n}{v_o}=\sqrt{\frac{2t}{\mu }}\times \sqrt{\frac{\mu }{t}} \\ =\sqrt{\frac{2}{1}} \\ =1.44 \end{gathered}$$

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