Question

The fundamental frequency of a sonometer wire increases by $6\, Hz$ if its tension is increased by $44 \%$ keeping the length constant. Find the change in the fundamental frequency (in $Hz$) of the sonometer wire, the length of the wire is increased by $20 \%$ keeping the original tension in the wire.

Answer 1

In case of vibration of a string, fundamental frequency is given by
$F=\frac{1}{2 L} \sqrt{\frac{T}{\mu}}$
So if length of given wire is kept constant
$\left(\frac{f^{\prime}}{f}\right)=\left(\frac{T^{\prime}}{T}\right) \frac{1}{2}$
and as here $f^{\prime}=f+6$
and $T^{\prime}=T+0.44 T=1.44 T$
$\frac{(f+6)}{f}=\sqrt{\frac{1.44 T}{T}} \text { or } f=30 \,Hz$
Now if keeping the original tension $(T)$, the length wire is changed, we have
$\frac{f^{\prime \prime}}{f}=\frac{\ell}{\ell^{\prime \prime}}=\frac{1}{1.20}$
$\left[\text { As } \ell^{\prime \prime}=\ell+0.20 \ell=1.20 \ell\right]$
or $f^{\prime \prime}=\frac{30}{1.2}=25\, Hz$
Hence $\Delta f=f^{\prime \prime}-f=25-30=-5 \,Hz$
Thus, fundamental frequency will decrease by $5 \,Hz$.