Question

The fundamental frequency of a sonometer wire increases by $6Hz$ 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

Answer: 5.0

In case of vibration of a string, fundamental frequency is given by
$F=\frac{1}{2L}\sqrt{\frac{T}{\mu }}$
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 $T^{\prime }=T+0.44T=1.44T$
$\frac{(f+6)}{f}=\sqrt{\frac{1.44T}{T}}\text{ or }f=30Hz$
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 $f^{\prime \prime }=\frac{30}{1.2}=25Hz$
Hence $\Delta f=f^{\prime \prime }-f=25-30=-5Hz$
Thus, fundamental frequency will decrease by $5Hz$.