Question

The fundamental frequency of a sonometer wire increases by $6\text{Hz}$ if the tension in the wire is increased by $44\%$, keeping the length constant. What is the change in the fundamental frequency of the wire when the length is increased by $20\%$, keeping the original tension in the wire constant.

• A. $5\text{Hz}$
• B. $10\text{Hz}$
• C. $2.5\text{Hz}$
• D. $7.5\text{Hz}$

Answer 1

The correct option is A $5\text{Hz}$

Fundamental frequency of vibrations of a string is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

At constant length,
$f\propto \sqrt{T}$ or $\frac{f}{\sqrt{T}}=\text{constant}...\left(1\right)$

It is given that the tension increases by$44$

So, new tension,

$T^{\prime }=T+\frac{44}{100}T=1.44T$

If $f^{\prime }$ is new frequency, then from $\left(1\right)$

$\frac{f^{\prime }}{\sqrt{T^{\prime }}}=\frac{f}{\sqrt{T}}$

$\Rightarrow f^{\prime }=\left(\sqrt{\frac{T^{\prime }}{T}}\right)f...\left(2\right)$

Given that frequency of the wire increases by $6\text{Hz}$ we have,

$f^{\prime }=f+6$

From Eq. $\left(2\right)$,

$f+6=\left(\sqrt{\frac{1.44T}{T}}\right)f$

or $f+6=1.2f$

$\Rightarrow 0.2f=6$ or $f=\frac{6}{0.2}=30\text{Hz}$

Now, when the tension is kept constant,

$f\propto \frac{1}{l}$ or $fl=\text{constant}$

When length increases by $20\%$,

New length $l^{\prime }=l+\frac{20}{100}l=1.2l$

As $fl=\text{constant}$,

$fl=f^{\prime }{l}^{\prime }$

$f^{\prime }=\frac{l}{l^{\prime }}f=\frac{l}{1.2l}\times 30=25\text{Hz}$

Change in fundamental frequency

$\Delta f=f^{\prime }-f=25-30=-5\text{Hz}$

Therefore, $\Delta f=5\text{Hz}$ (decrease)