Question

A single pendulum hanging vertically in an elevator has a time period 𝑇₀ when the elevator is stationary. If the elevator moves upward with an acceleration of 𝑎 = 0.2𝑔, the time period of oscillations is 𝑇₁. Here 𝑔 is the acceleration due to gravity. The ratio\( \frac{𝑇_0}{ 𝑇_1}\) is (rounded off to two decimal places).

Answer 1

Correct Answer: 1.09 - 1.11

Given: 
Time period when elevator is stationary: \( T_0 \)
Elevator accelerates upward with \( a = 0.2g \)
Effective gravity: \( g_{\text{eff}} = g + a = g + 0.2g = 1.2g \)

Time period of a simple pendulum: \[ T = 2\pi \sqrt{\frac{L}{g}} \] Thus, \[ T_0 = 2\pi \sqrt{\frac{L}{g}}, \qquad T_1 = 2\pi \sqrt{\frac{L}{1.2g}} \] Required ratio: \[ \frac{T_0}{T_1} = \frac{2\pi\sqrt{\frac{L}{g}}}{2\pi\sqrt{\frac{L}{1.2g}}} = \sqrt{\frac{1.2g}{g}} = \sqrt{1.2} \] Calculate: \[ \sqrt{1.2} \approx 1.095 \] Final Answer:
\[ \frac{T_0}{T_1} \approx 1.10 \]