Question

A simple pendulum is taken at a place where its distance from the Earth's surface is equal to the radius of the Earth. Calculate the time period of small oscillations if the length of the string is 4.0 m. (Take \( g = 9 \, \text{m/s}^2 \) at the surface of the Earth.)

• A. 4 s
• B. 6 s
• C. 8 s
• D. 2 s

Hint:

When the pendulum is at a height equal to the Earth's radius, the effective gravity is reduced, which increases the time period of oscillation.

Answer 1

The Correct Option is C

Step 1: Formula for Time Period of a Simple Pendulum The time period \( T \) of a simple pendulum is given by the formula: \[ T = 2\pi \sqrt{\frac{L}{g}} \] Where: - \( L \) is the length of the string, - \( g \) is the acceleration due to gravity at the surface of the Earth. Step 2: Adjusting for the Height of the Pendulum The pendulum is at a height equal to the Earth's radius, meaning the effective value of \( g \) will change. The acceleration due to gravity at a height \( h \) above the Earth's surface (where \( h = R_{\text{earth}} \)) is given by the formula: \[ g' = \frac{g}{(1 + \frac{h}{R_{\text{earth}}})^2} \] Since \( h = R_{\text{earth}} \), the expression simplifies to: \[ g' = \frac{g}{4} \] Thus, the effective value of gravity at this height is \( \frac{g}{4} \). Step 3: Calculating the Time Period Now, substitute the effective gravity \( g' = \frac{g}{4} \) into the formula for the time period: \[ T = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4L}{g}} \] Given that \( L = 4 \, \text{m} \) and \( g = 9 \, \text{m/s}^2 \), we get: \[ T = 2\pi \sqrt{\frac{4 \times 4}{9}} = 2\pi \sqrt{\frac{16}{9}} = 2\pi \times \frac{4}{3} \] Thus, the time period is: \[ T = \frac{8\pi}{3} \approx 8 \, \text{s} \] Step 4: Conclusion The time period of oscillation is approximately 8 seconds. Thus, the correct answer is: \[ \boxed{(C)} \, 8 \, \text{s} \]