Question

A simple pendulum of length $L$ is suspended from the roof of a trolley. The trolley moves in horizontal direction with an acceleration $a$. What would be the period of oscillation of a simple pendulum? [$g$ is acceleration due to gravity]

• A. $2\pi \sqrt{L}\,(a^2+g^2)^{-1/4}$
• B. $2\pi \sqrt{L}\,(a^2+g^2)^{-1/2}$
• C. $2\pi \sqrt{\dfrac{L}{g+a}}$
• D. $2\pi \sqrt{\dfrac{L}{g-a}}$

Hint:

In a non-inertial frame, replace $g$ by the effective gravity to find the time period.

Answer 1

The Correct Option is A

Step 1: Effective gravity in an accelerating frame.
In a frame accelerating horizontally with acceleration $a$, the pendulum experiences an effective gravity:
\[ g_{\text{eff}}=\sqrt{g^2+a^2} \]

Step 2: Time period of a simple pendulum.
For small oscillations, the time period is:
\[ T = 2\pi \sqrt{\frac{L}{g_{\text{eff}}}} \]

Step 3: Substitute effective gravity.
\[ T = 2\pi \sqrt{\frac{L}{\sqrt{g^2+a^2}}} = 2\pi \sqrt{L}\,(g^2+a^2)^{-1/4} \]

Step 4: Conclusion.
The period of oscillation is $2\pi \sqrt{L}\,(a^2+g^2)^{-1/4}$.