Question

A bob of mass $m$ suspended by a length $l$ undergoes simple harmonic oscillations with time period $T$. If the bob is immersed in a liquid that has density $\frac{1}{4}$ times that of the bob and the length of the threat is increased by $(1/3{)}^{rd}$ of the original length, then the time period of the simple harmonic oscillation will be

• A. $\frac{3}{4}T$
• B. $\frac{3}{2}T$
• C. $T$
• D. $\frac{4}{3}T$

Answer 1

The correct option is D $\frac{4}{3}T$

Time period of simple pendulum in air is $T=2\pi \sqrt{\frac{l}{g}}$
When bob is immersed in liquid, time period of simple pendulum is $T^{\prime }=2\pi \sqrt{\frac{l^{\prime }}{g(1-\frac{{\rho }_l}{{\rho }_b})}}$

Given that
${\rho }_l=\frac{{\rho }_b}{4}\&l^{\prime }=\frac{4l}{3}$

Substituting in above equation, we get
$T^{\prime }=2\pi \sqrt{\frac{4l/3}{3g/4}}=\frac{4}{3}\left(2\pi \sqrt{\frac{l}{g}}\right)=\frac{4}{3}T$