Question

A bob of mass 'm' suspended by a thread of 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 thread is increased by $1/3^{rd}$ of the original length, then the time period of the simple harmonic oscillations will be :

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

Hint:

When a bob is immersed in liquid, $g$ is replaced by $g_{eff} = g(1 - \text{relative density of liquid})$. Keep track of length changes carefully as they often appear together in such problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The time period of a simple pendulum is determined by its effective length and the effective acceleration due to gravity ($g_{eff}$). When immersed in a liquid, the buoyant force reduces the effective weight, thereby reducing $g_{eff}$.
Step 2: Key Formula or Approach:
1. Original Time Period: \( T = 2\pi \sqrt{\frac{l}{g}} \)
2. Effective gravity in liquid: \( g' = g \left( 1 - \frac{\rho_{liquid}}{\rho_{bob}} \right) \)
Step 3: Detailed Explanation:
Given: - Density of liquid \(\sigma = \frac{1}{4} \rho\) (where \(\rho\) is bob density). - New length \(l' = l + \frac{l}{3} = \frac{4l}{3}\).
First, calculate the effective acceleration due to gravity (\(g'\)) in the liquid: \[ g' = g \left( 1 - \frac{\sigma}{\rho} \right) = g \left( 1 - \frac{1/4 \rho}{\rho} \right) = g \left( 1 - \frac{1}{4} \right) = \frac{3g}{4} \]
Now, calculate the new time period \(T'\): \[ T' = 2\pi \sqrt{\frac{l'}{g'}} = 2\pi \sqrt{\frac{4l/3}{3g/4}} \] \[ T' = 2\pi \sqrt{\frac{4l}{3} \times \frac{4}{3g}} = 2\pi \sqrt{\frac{16l}{9g}} \] \[ T' = \frac{4}{3} \times 2\pi \sqrt{\frac{l}{g}} \] Since \(T = 2\pi \sqrt{l/g}\), we get: \[ T' = \frac{4}{3} T \]
Step 4: Final Answer:
The new time period is \(\frac{4}{3} T\).