Simple Harmonic Motion 3 Question 10
10. A solid sphere of radius $R$ is floating in a liquid of density $\rho$ with half of its volume submerged. If the sphere is slightly pushed and released, it starts performing simple harmonic motion. Find the frequency of these oscillations.
(2004, 4M)
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Answer:
Correct Answer: 10. $\frac{1}{2 \pi} \sqrt{\frac{3 g}{2 R}}$
Solution:
- Half of the volume of sphere is submerged.
For equilibrium of sphere, weight $=$ upthrust
$$ \begin{aligned} \therefore \quad V \rho _s g & =\frac{V}{2}\left(\rho _L\right)(g) \\ \rho _s & =\frac{\rho _L}{2} \end{aligned} $$
When slightly pushed downwards by $x$, weight will remain as it is while upthrust will increase. The increased upthrust will become the net restoring force (upwards).
$$ \begin{aligned} & F=-(\text { extra upthrust }) \\ & =-(\text { extra volume immersed })\left(\rho _L\right)(g) \\ & \text { or } \quad m a=-\left(\pi R^{2}\right) x \rho _L g \quad(a=\text { acceleration }) \\ & \therefore \quad \frac{4}{3} \pi R^{3} \quad \frac{\rho}{2} \quad a=-\left(\pi R^{2} \rho g\right) x \\ & \therefore \quad a=-\frac{3 g}{2 R} x \end{aligned} $$
as $a \propto-x$, motion is simple harmonic.
Frequency of oscillation, $f=\frac{1}{2 \pi} \sqrt{\frac{a}{x}}$
$$ =\frac{1}{2 \pi} \sqrt{\frac{3 g}{2 R}} $$
