Question

A sonometer wire has a length $l$ and tension $T$. If on reducing the tension to half of its original value and changing the length, the second harmonic becomes equal to the fundamentals frequency of the first case, then the new length of the wire is

• A. $l/\sqrt{2}$
• B. $\sqrt{2}l$
• C. $l/2$
• D. $2l$

Answer 1

Answer: (B)

$\sqrt{2}l$

For sonometer the working formula is:
$f_n=\frac{n}{2l}\sqrt{\frac{T}{m}}$
where, $f_n$ is the freq of nth mode, $n$ is the mode number, $l$ is the length of the wire, $T$ is the tension in the wire, $m$ is the linear mass density or mass per unit length of the wire.

If the tension reduced to half, and length becomes $l_1$ then from the above condition,
$\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{2}{2l_1}\sqrt{\frac{T}{2m}}$

$\Rightarrow l_1=l\sqrt{2}$