Question

A seconds pendulum is placed in a space laboratory orbiting round the earth at a height '3R' from the earth's surface. The time period of the pendulum will be (R = radius of earth)

• A. zero
• B. $\frac{2}{3}$ s
• C. 4 s
• D. infinite

Hint:

Logic Tip: Without gravity to pull it back, a displaced pendulum will never return to its equilibrium position.

Answer 1

The Correct Option is D

Concept: The time period of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{l}{g \] where $l$ is the length of the pendulum and $g$ is the acceleration due to gravity.
Step 1: Condition in an orbiting laboratory In an orbiting laboratory (such as a satellite), the system is in a state of free fall. Although gravity still acts, both the pendulum and its support accelerate equally under gravity. As a result, there is no effective restoring force, and the pendulum experiences weightlessness. Thus, the effective gravitational acceleration becomes: \[ g_{\text{eff = 0 \]
Step 2: Substitute into the time period formula \[ T = 2\pi \sqrt{\frac{l}{g_{\text{eff \] As $g_{\text{eff \to 0$: \[ T \to \infty \]
Step 3: Physical interpretation Since there is no restoring force acting on the bob, it does not oscillate. Hence, the pendulum cannot execute periodic motion. Conclusion: \[ \boxed{T = \infty} \] The pendulum does not oscillate in an orbiting laboratory.