Question

A simple pendulum of length 1 m is oscillating with an amplitude of 0.1 m. What is the maximum tension in the string if the mass of the bob is 0.2 kg? (Assume \( g = 10 \, \text{m/s}^2 \))

• A. \( 2.2 \, \text{N} \)
• B. \( 2.4 \, \text{N} \)
• C. \( 2.0 \, \text{N} \)
• D. \( 2.6 \, \text{N} \)

Hint:

For simple pendulums, remember that the maximum tension occurs at the lowest point, where the speed is highest. Use conservation of energy to relate potential and kinetic energy and find the speed at the lowest point.

Answer 1

The Correct Option is B

Step 1: Maximum string tension occurs at the lowest swing point. Tension equals gravitational force plus centripetal force: \[ T = mg + \frac{mv^2}{L} \] where:
- \( m = 0.2 \, \text{kg} \) (bob mass),
- \( g = 10 \, \text{m/s}^2 \) (gravity),
- \( L = 1 \, \text{m} \) (pendulum length),
- \( v \) is the speed at the lowest point, derived from mechanical energy conservation.
Step 2: Total mechanical energy is conserved. At the highest point, potential energy is maximal, kinetic energy is zero. At the lowest point, all potential energy converts to kinetic energy.
Potential energy at amplitude is: \[ U = mgh \] where \( h = L - L \cos(\theta) \approx L \) for small oscillations, and \( \theta \) is the maximum displacement angle. Velocity at the lowest point is calculated by equating total energy to potential energy: \[ \frac{1}{2}mv^2 = mgh \] Solving for \( v \): \[ v = \sqrt{2gh} \] With \( h = 0.1 \, \text{m} \) (amplitude): \[ v = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} \approx 1.414 \, \text{m/s} \]
Step 3: Calculate maximum tension: \[ T = mg + \frac{mv^2}{L} = (0.2 \times 10) + \frac{0.2 \times (1.414)^2}{1} = 2 + \frac{0.2 \times 2}{1} = 2 + 0.4 = 2.4 \, \text{N} \]