Question

A simple pendulum is placed at a place where its distance from the earth's surface is equal to the radius of the earth. If the length of the string is 4 m, then the time period of small oscillations will be _____ s. [take \( g = \pi^2 \, m/s^2 \)]

Answer 1

Correct Answer: 8

The time period of a pendulum is given by:

\[ T = 2\pi \sqrt{\frac{L}{g}}, \] where \( L \) is the length of the pendulum, and \( g \) is the acceleration due to gravity.

For the given scenario, the effective acceleration due to gravity is:

\[ g' = \frac{g}{4}. \]

Substituting this 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 \( L = 4 \, m \), the formula becomes:

\[ T = 2\pi \sqrt{\frac{4 \times 4}{g}}. \]

Simplifying further:

\[ T = 2\pi \sqrt{\frac{16}{g}}. \]

Given \( g = \pi^2 \, m/s^2 \), substitute into the equation:

\[ T = 2\pi \sqrt{\frac{16}{\pi^2}} = 2\pi \times \frac{4}{\pi}. \]

Simplify:

\[ T = 2 \times 4 = 8 \, s. \]

Final Answer:

\[ T = 8 \, s. \]