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: The maximum tension in the string occurs when the bob is at the lowest point of the swing. At this point, the tension is the sum of the gravitational force and the centripetal force. The formula for tension is: \[ T = mg + \frac{mv^2}{L} \] where:
- \( m = 0.2 \, \text{kg} \) (mass of the bob),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( L = 1 \, \text{m} \) (length of the pendulum),
- \( v \) is the speed at the lowest point, which can be found using conservation of mechanical energy.
Step 2: The total mechanical energy in the system is conserved. At the highest point (amplitude), the potential energy is maximum and the kinetic energy is zero. At the lowest point, all potential energy has been converted into kinetic energy.
The potential energy at the amplitude is: \[ U = mgh \] where \( h = L - L \cos(\theta) \approx L \) for small oscillations, and \( \theta \) is the angle at maximum displacement. The velocity at the lowest point can be calculated by setting the total energy equal to the potential energy: \[ \frac{1}{2}mv^2 = mgh \] Thus, solving for \( v \), we get: \[ v = \sqrt{2gh} \] Substituting \( h = 0.1 \, \text{m} \) (the amplitude), we get: \[ v = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} \approx 1.414 \, \text{m/s} \]
Step 3: Now we can find the maximum tension: \[ T = mg + \frac{mv^2}{L} = (0.2 \times 10) + \frac{0.2 \times (1.414)^2}{1} = 2 + \frac{0.2 \times 2}{1} = 2 + 0.4 = 2.4 \, \text{N} \]