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