Question

A sonometer wire of density $\rho$ and radius 'a' is held between two bridges at a distance 'L' apart. Tension in the wire is 'T' the fundamental frequency of the wire will be

• A. $\frac{1}{2L}\sqrt\frac{\pi\,a^2}{T\rho}$
• B. $\frac{1}{2L}\sqrt\frac{T\rho}{\pi\,a^2}$
• C. $\frac{1}{2L}\sqrt{T}{\pi\,a^2}$
• D. $\frac{1}{2L}\sqrt\frac{T}{\pi\,a^2\rho}$

Answer 1

Answer: (D)

For a perfectly flexible, perfectly elastic string,
the velocity of wave propagation $(v)$ is given by:
$v =\sqrt{ T / \mu}$ where $\mu$ is mass per unit length.
Now the fundamental wavelength in the wire is:
$\lambda=\frac{2 L }{ n }$ where, $n =1$.
So the fundamental frequency will be,
$v=v / \lambda=\frac{1}{2 L} \sqrt{T} / \mu$
$=\frac{1}{2 L} \sqrt{T} / \pi a^{2} p$