Question

When the length of the vibrating segment of a sonometer wire is increased by $1\%$ the percentage change in its frequency is(in $\%$ )

• A. $3$
• B. $4$
• C. $1$
• D. $2$

Answer 1

Answer: (C)

Given,
$\frac{\Delta l}{l}=1\%$
The frequency of sonometer wire is,
$v=\frac{1}{2l}\sqrt{\frac{T}{m}}$ , where $T$ is the tension in the wire, $m$ is the linear mass density, and $l$ is the length of the wire.
Taking logarithm and differentiating, we get
$\frac{\Delta v}{v}=-\frac{\Delta l}{l}+\frac{1}{2}\frac{\Delta T}{T}-\frac{1}{2}\frac{\Delta m}{m}$
$\therefore \frac{\Delta v}{v}=-\frac{\Delta l}{l}=-1\%$
Hence, the frequency will decrease by $1\%$ .