Question

The frequency of the $1st$ harmonic of a sonometer wire is $160Hz$. If the length of the wire is increased by $50\%$ and the tension in the wire is decreased by $19\%$, the frequency of its first overtone is :

( Assume linear mass density to be constant )

• A. $180Hz$
• B. $192Hz$
• C. $220Hz$
• D. $232Hz$

Answer 1

The correct option is B $192Hz$

The natural frequency of nth mode is
$f_n=\frac{n}{2L}\sqrt{\frac{T}{\mu }}$
where, $L$ is the length of the sonometer wire,
$\mu$ is the mass per unit length of the wire and $T$ is the tension in the string.

Frequency of the first harmonic before :$f=\frac{1}{2L}\sqrt{\frac{T}{\mu }}......................................\left(1\right)$
Frequency of the first harmonic after :$f^{\prime }=\frac{1}{2L^{\prime }}\sqrt{\frac{T^{\prime }}{\mu }}.........................................\left(2\right)$
Given $L^{\prime }=1.5L$
$T^{\prime }=0.81T$
Dividing (1) by (2)
$\frac{f}{f^{\prime }}=\frac{L^{\prime }}{L}\sqrt{\frac{T}{T^{\prime }}}=\frac{1.5L}{L}\sqrt{\frac{T}{0.81T}}=\frac{1.5}{0.9}$
$\therefore \frac{160}{f^{\prime }}=\frac{1.5}{0.9}$
$\Rightarrow f^{\prime }=160\times \frac{0.9}{1.5}=96Hz$
Frequency of the 1st overtone is $2f^{\prime }=2\times 96=192Hz$