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: \[\nT = 2\pi \sqrt{\frac{L}{g}}\n\] 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's height equals the Earth's radius, affecting the effective \( g \). The acceleration due to gravity at a height \( h \) above the Earth's surface (where \( h = R_{\text{earth}} \)) is: \[\ng' = \frac{g}{(1 + \frac{h}{R_{\text{earth}}})^2}\n\] Since \( h = R_{\text{earth}} \), this simplifies to: \[\ng' = \frac{g}{4}\n\] The effective gravity at this height is \( \frac{g}{4} \). Step 3: Calculating the Time Period Substitute the effective gravity \( g' = \frac{g}{4} \) into the time period formula: \[\nT = 2\pi \sqrt{\frac{L}{g'}} = 2\pi \sqrt{\frac{L}{\frac{g}{4}}} = 2\pi \sqrt{\frac{4L}{g}}\n\] Given \( L = 4 \, \text{m} \) and \( g = 9 \, \text{m/s}^2 \): \[\nT = 2\pi \sqrt{\frac{4 \times 4}{9}} = 2\pi \sqrt{\frac{16}{9}} = 2\pi \times \frac{4}{3}\n\] The time period is: \[\nT = \frac{8\pi}{3} \approx 8 \, \text{s}\n\] Step 4: Conclusion The oscillation's time period is approximately 8 seconds. Thus, the correct answer is: \[\n\boxed{(C)} \, 8 \, \text{s}\n\]