Question

In a sonometer wire the tension is maintained by suspending a $50.7kg$ mass from the free end of the wire. The suspended mass has a volume of $0.0075m^3$. The fundamental frequency of the wire is $260Hz$. Find the new fundamental frequency (in Hz) if the suspended mass is completely submerged in water.

Answer 1

Fundamental Frequency in such a string is directly proportional to square root of tension.
${\nu }_{fundamental}=k\sqrt{T}$
${\nu }_{fundamental}=c\sqrt{m}$ (c is constant of different value, we take out g from tension, T = mg)
i.e.
$\frac{260}{\sqrt{50.7}}=constant$

Now the block is $0.0075m^3$

By Archimedes principle we know that an upward buoyant force of $\rho Vg=1000\times 0.0075\times g=7.5g$ acts.

This is equivalent to the block losing $7.5kg$ in mass. i.e$=50.7-7.5=43.2$

Substituting this.
$\frac{260}{\sqrt{50.7}}=\frac{\nu }{\sqrt{43.2}}$
Solving this we get $\nu =240Hz$ ,which is the new fundamental frequency.