The extension in a string obeying Hooke’s law is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be

The correct option is 1.22 v.

Step-by-Step Explanation:

The speed of sound (transverse wave) in a stretched string is given by the formula:

$v=\sqrt{\frac{T}{\mu }}$

where:
- $T$ is the tension in the string.
- $\mu$ is the linear mass density of the string (mass per unit length).

Assuming the change in the linear mass density is negligible during stretching, the speed of sound is directly proportional to the square root of the tension in the string:

$v\propto \sqrt{T}$

According to Hooke's law, the tension $T$ developed in the string is directly proportional to its extension $x$:

$T=kx$

where $k$ is the force constant of the string.

Substituting this relationship into the velocity proportionality, we get:

$v\propto \sqrt{x}$

Let the initial speed of sound be $v$ for an extension $x$, and the new speed of sound be $v^{\prime }$ when the extension is increased to $x^{\prime }=1.5x$. We can set up the ratio of the two speeds:

$\frac{v^{\prime }}{v}=\sqrt{\frac{x^{\prime }}{x}}$

Substitute the value of $x^{\prime }$ into the equation:

$\frac{v^{\prime }}{v}=\sqrt{\frac{1.5x}{x}}$

$\frac{v^{\prime }}{v}=\sqrt{1.5}$

Calculating the square root of 1.5:

$\sqrt{1.5}\approx 1.2247$

Thus:

$v^{\prime }\approx 1.22v$

Therefore, if the extension in the string is increased to 1.5x, the speed of sound will be approximately 1.22 v.