Question

In a sonometer wire, the tension is maintained by suspended a $50.7\text{kg}$ mass from the free end of the wire. The suspended mass has a volume of $0.0075{\text{m}}^3$. The fundamental frequency of vibration of the wire is $260\text{Hz}$. If the suspended mass is completely submerged in water, the fundamental frequency will become $\text{Hz}$.

Answer 1

Given that
Mass, $m=50.7\text{kg}$

Volume of mass, $\left(V\right)=0.0075{\text{m}}^3$

Fundamental frequency, $f_0=260\text{Hz}$

So initially fundamental frequency,

$f_0=\frac{1}{2L}\sqrt{\frac{T}{\mu }}=\frac{1}{2L}\sqrt{\frac{50.7\times g}{\mu }}$

$\Rightarrow 260=\frac{1}{2L}\sqrt{\frac{50.7\times g}{\mu }}....\left(i\right)$

After mass submerged in water, tension in the wire

$T^{\prime }=50.7\times g-0.0075\times 1000\times g=43.2g$

So,

$f_0^{\prime }=\frac{1}{2L}\sqrt{\frac{T^{\prime }}{\mu }}=\frac{1}{2L}\sqrt{\frac{43.2\times g}{\mu }}------------\left(ii\right)$

On dividing eqn (ii) with (i) we get,

$\frac{f_0^{\prime }}{f_0}=\frac{f_0^{\prime }}{260}=\sqrt{\frac{43.2\text{g}}{50.7\text{g}}}=\sqrt{\frac{43.2}{50.7}}$

$\therefore f_0^{\prime }=260\times 0.923=240\text{Hz}$

Correct answer: $240$