Question

A body thrown vertically upwards with a velocity of 20 ms\(^{-1}\) from the surface of a planet reaches back the surface of the planet in a time of 2 s. The time period of a simple pendulum of length \( \frac{20}{\pi^2} \) m on this planet is

• A. 4 s
• B. 3 s
• C. 2 s
• D. 1 s

Hint:

First find \( g \) using the motion of the body: time to max height is half the total time, then use \( v = u - g t \). For the pendulum, use \( T = 2\pi \sqrt{\frac{l}{g}} \).

Answer 1

The Correct Option is C

First, determine the acceleration due to gravity \( g \) on the planet. The body is thrown upwards with an initial velocity \( u = 20 \, \text{ms}^{-1} \) and returns to the surface in 2 s. The time to reach the maximum height (where velocity becomes zero) is half of the total time, so \( t_{\text{up}} = 1 \, \text{s} \). Using the equation of motion: \[ v = u - g t \quad \text{(at maximum height, } v = 0\text{)} \] \[ 0 = 20 - g \times 1 \quad \Rightarrow \quad g = 20 \, \text{ms}^{-2} \] Now, calculate the time period of the simple pendulum. The time period \( T \) of a simple pendulum is given by: \[ T = 2\pi \sqrt{\frac{l}{g}} \] Given the length of the pendulum \( l = \frac{20}{\pi^2} \, \text{m} \), and \( g = 20 \, \text{ms}^{-2} \), substitute the values: \[ T = 2\pi \sqrt{\frac{\frac{20}{\pi^2}}{20}} = 2\pi \sqrt{\frac{20}{\pi^2 \times 20}} = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \times \frac{1}{\pi} = 2 \, \text{s} \] So, the time period of the simple pendulum is 2 s.

Answer 2

Step 1: Understand the problem
- A body is thrown vertically upward with initial velocity \( u = 20 \, \text{m/s} \).
- Total time of flight is \( T = 2 \, \text{s} \).
- Length of simple pendulum is \( l = \frac{20}{\pi^2} \, \text{m} \).
- Find the time period of the pendulum on this planet.

Step 2: Find acceleration due to gravity \( g \) on the planet
- Time of flight for vertical motion is given by \( T = \frac{2u}{g} \).
- Rearranging, \( g = \frac{2u}{T} = \frac{2 \times 20}{2} = 20 \, \text{m/s}^2 \).

Step 3: Calculate time period \( T_p \) of pendulum
\[ T_p = 2\pi \sqrt{\frac{l}{g}} = 2\pi \sqrt{\frac{20/\pi^2}{20}} = 2\pi \sqrt{\frac{1}{\pi^2}} = 2\pi \times \frac{1}{\pi} = 2 \, \text{s} \]

Final answer:
The time period of the simple pendulum is \( 2 \, \text{s} \).