Question 63: Oscillations

Question

A simple pendulum doing small oscillations at a place $$R$$ height above earth surface has time period of $$T_1=4 \mathrm{~s}$$. $$\mathrm{T}_2$$ would be it's time period if it is brought to a point which is at a height $$2 \mathrm{R}$$ from earth surface. Choose the correct relation [$$\mathrm{R}=$$ radius of earth] :

Accepted Answer

The time period of a simple pendulum is given by the formula:
$$T = 2\pi \sqrt{\frac{l}{g}}$$
where $$T$$ is the time period, $$l$$ is the length of the pendulum, and $$g$$ is the acceleration due to gravity at the location of the pendulum.
The acceleration due to gravity changes with height above the Earth's surface. The acceleration due to gravity at a height $$h$$ above the Earth's surface can be expressed as:
$$g' = g \left(\frac{R}{R + h}\right)^2$$
where $$g$$ is the acceleration due to gravity at the surface of the Earth, $$R$$ is the radius of the Earth, and $$h$$ is the height above the EarthÃ¢ÂÂs surface. Since the time period of the pendulum depends on the square root of the inverse of the acceleration due to gravity, any change in $$g$$ due to a change in height will affect the time period.
Given that the time period of the pendulum at a height $$R$$ above Earth's surface is $$T_1$$, and we're to find the time period $$T_2$$ at a height of $$2R$$, we can use the formula for acceleration due to gravity at different heights to express the relationship between $$T_1$$ and $$T_2$$.
For the initial case at height $$R$$:
$$g_1 = g \left(\frac{R}{R + R}\right)^2 = g \left(\frac{R}{2R}\right)^2 = \frac{g}{4}$$
For the new case at height $$2R$$:
$$g_2 = g \left(\frac{R}{R + 2R}\right)^2 = g \left(\frac{R}{3R}\right)^2 = \frac{g}{9}$$
The time period is proportional to the square root of the inverse of $$g$$, so:
$$\frac{T_1}{T_2} = \sqrt{\frac{g_2}{g_1}} = \sqrt{\frac{\frac{g}{9}}{\frac{g}{4}}} = \sqrt{\frac{4}{9}} = \frac{2}{3}$$
Therefore:
$$T_1 = \frac{2}{3}T_2$$
Rearranging this equation:
$$3T_1 = 2T_2$$
This corresponds to Option A.

Answer

$$3 \mathrm{~T}_1=2 \mathrm{~T}_2$$

Answer

$$\mathrm{T}_1=\mathrm{T}_2$$

Answer

$$2 \mathrm{~T}_1=3 \mathrm{~T}_2$$

Answer

$$2 \mathrm{~T}_1=\mathrm{T}_2$$

Oscillations

Choose the correct answer: