Question

The time period of a simple pendulum of length \( \sqrt{5} \) m suspended in a car moving with uniform acceleration of \( 5 \, \text{m s}^{-2} \) in a horizontal straight road is (\( g = 10 \, \text{m s}^{-2} \))

• A. $\frac{2\pi}{\sqrt{5}}$ s
• B. $\frac{\pi}{\sqrt{5}}$ s
• C. $5\pi$ s
• D. $4\pi$ s
• E. $3\pi$ s

Hint:

In accelerating frames, always replace $g$ with $g_{\text{eff}} = \sqrt{g^2 + a^2}$.

Answer 1

The Correct Option is A

Concept: When a pendulum is inside an accelerating frame (like a car), effective gravity changes: \[ g_{\text{eff}} = \sqrt{g^2 + a^2} \] Time period: \[ T = 2\pi \sqrt{\frac{l}{g_{\text{eff}}}} \]

Step 1: Given: \[ l = \sqrt{5}, \quad g = 10, \quad a = 5 \]

Step 2: Effective gravity: \[ g_{\text{eff}} = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \]

Step 3: Substitute: \[ T = 2\pi \sqrt{\frac{\sqrt{5}}{5\sqrt{5}}} \] \[ = 2\pi \sqrt{\frac{1}{5}} = \frac{2\pi}{\sqrt{5}} \]

Step 4: Simplify: \[ T = \frac{2\pi}{\sqrt{5}} \] Final Answer: \[ \boxed{\frac{2\pi}{\sqrt{5}} \text{ s}} \]