Question

The frequency of fundamental note of a sonometer wire is $120Hz$. The two bridges of sonometer are displaced away from each other, so that the length of the wire is doubled and the weight suspended to the wire is increased to $4$ times, then the frequency of fundamental note is

• A. $120Hz$
• B. $180Hz$
• C. $240Hz$
• D. $480Hz$

Answer 1

The correct option is A $120Hz$

Fundamental frequency = $\frac{v}{2l}=\frac{1}{2}\frac{\sqrt{\frac{T}{\mu }}}{l}=120HZ$
when the tension is increased to $4T$ and length is made double
($\mu$ is mass per unit length)
The new fundamental frequency = $\frac{1}{2}\frac{\sqrt{\frac{4T}{\mu }}}{2l}=\frac{1}{2}\frac{2\sqrt{\frac{T}{\mu }}}{2l}$
$=\frac{1}{2}\frac{\sqrt{\frac{T}{\mu }}}{l}=120Hz.$