Question

A sonometer wire of density $p$ and radius $a$ is held between two bridges at a distance $L$ apart. If the tension in the wire is $T$ then the fundamental frequency of the wire will be

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

Answer 1

The correct option is D $\frac{1}{2L}\sqrt{\frac{T}{\pi a^{2}p}}$

For a perfectly flexible, perfectly elastic string, the velocity of wave propagation$\left(v\right)$ is given by:
$v=\sqrt{T/\mu }$. where $\mu$ is mass per unit length.

Now the fundamental wavelength in the wire is:
$\lambda =\frac{2L}{n}$ where, $n=1$.
So the fundamental frequency will be,
$\nu =v/\lambda =\frac{1}{2L}\sqrt{T/\mu }=\frac{1}{2L}\sqrt{T/\pi a^{2}p}$