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

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:

$T = 8 \, s.$