Question

A simple pendulum is made of a metal wire of length \( L \), area of cross-section \( A \), material of Young's modulus \( Y \), and a bob of mass \( m \). This pendulum is hung in a bus moving with a uniform speed \( V \) on a horizontal circular road of radius \( R \). The elongation in the wire is: 

• A. \( \frac{mL}{RAY} \sqrt{g^2 R^2 + V^4} \)
• B. \( \frac{mgL}{AY} \)
• C. \( \frac{mL V^2}{RAY} \)
• D. \( \frac{L}{AY} \sqrt{mg + \frac{mV^2}{R}} \)

Hint:

For pendulum problems in moving frames, resolve forces along tension and apply Hooke’s Law: \[ \Delta L = \frac{T L}{A Y} \] where \( T \) is the resultant tension.

Answer 1

The Correct Option is A

Step 1: Identify the Forces Acting on the Pendulum Bob The bob of the pendulum experiences: 1. The gravitational force acting downward: \[ F_g = mg \] 2. The centripetal force due to circular motion of the bus: \[ F_c = \frac{mV^2}{R} \] These forces result in a net force along the string, which causes an elongation in the wire. 

Step 2: Find the Resultant Tension in the Wire The net tension \( T \) in the wire is given by: \[ T = \sqrt{F_g^2 + F_c^2} = \sqrt{(mg)^2 + \left( \frac{mV^2}{R} \right)^2} \] \[ = m \sqrt{g^2 + \frac{V^4}{R^2}} \] 

Step 3: Apply Hooke’s Law for Elongation Using the formula for elongation in a wire: \[ \Delta L = \frac{T L}{A Y} \] Substituting \( T \): \[ \Delta L = \frac{m L}{A Y} \sqrt{g^2 + \frac{V^4}{R^2}} \] Multiplying by \( R/R \): \[ \Delta L = \frac{m L}{RAY} \sqrt{g^2 R^2 + V^4} \] Thus, the elongation in the wire is: \[ \mathbf{\frac{mL}{RAY} \sqrt{g^2 R^2 + V^4}} \]