Question

A cylindrical block of wood of mass 'm' and cross section 'A' is floating in a liquid of density $\sigma$ with its axis vertical. It is depressed a little and then released. Then its frequency is

• A. $f=\frac{1}{\pi }\sqrt{\frac{A\sigma g}{m}}$
• B. $f=\frac{1}{2\pi }\sqrt{\frac{A\sigma g}{m}}$
• C. $f=\frac{1}{2\pi }\sqrt{\frac{m}{A\sigma g}}$
• D. $f=\frac{1}{\pi }\sqrt{\frac{m}{A\sigma g}}$

Answer 1

The correct option is B $f=\frac{1}{2\pi }\sqrt{\frac{A\sigma g}{m}}$

Let the height upto which the block depressed remains in equilibrium be $h$.

Then $\left(Ah\right)\sigma g=mg$

Hence when depressed by a distance $x$, the restoring force becomes

$\left(A\right(h+x\left)\right)\sigma g-mg$

$=Ax\sigma g=m{\omega }^{2}x$

$⟹\omega =\sqrt{\frac{A\sigma g}{m}}$

Hence the frequency of oscillation=$\frac{\omega }{2\pi }$

$=\frac{1}{2\pi }\sqrt{\frac{A\sigma g}{m}}$