A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi imes10^{-2}$ s, where the value of $y$ is (Acceleration due to gravity, $g=10\ \mathrm{m/s^2}$, density of water $=10^3\ \mathrm{kg/m^3}$)

Question

A light hollow cube of side length 10 cm and mass 10 g, is floating in water. It is pushed down and released to execute simple harmonic oscillations. The time period of oscillations is $y\pi	imes10^{-2}$ s, where the value of $y$ is (Acceleration due to gravity, $g=10\ \mathrm{m/s^2}$, density of water $=10^3\ \mathrm{kg/m^3}$)

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] The restoring force is the extra buoyancy $ho gAx$, so $\omega^2=\frac{ho gA}m=\frac{10^3	imes10	imes10^{-2}}{10^{-2}}=10^4$, $\omega=100$ rad/s. $T=\frac{2\pi}{100}=2\pi	imes10^{-2}$ s, so $y=2$. Answer (2).

Explanation

Reviewed by Dr. Abubacker

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