Question

A simple pendulum doing small oscillations at a place \( R \) height above the Earth's surface has a time period of \( T_1 = 4 \, {s} \). \( T_2 \) would be its time period if it is brought to a point which is at a height \( 2R \) from the Earth's surface. Choose the correct relation [\( R \) = radius of Earth]:

• A. \( T_1 = T_2 \)
• B. \( 2T_1 = 3T_2 \)
• C. \( 3T_1 = 2T_2 \)
• D. \( 2T_1 = T_2 \)

Hint:

The time period of a simple pendulum depends on the gravitational acceleration, which varies with height above the Earth's surface.

Answer 1

The Correct Option is C

Step 1: {Time period of a simple pendulum}
\[ T = 2\pi \sqrt{\frac{L}{g}} \quad {and} \quad g = \frac{GM}{(R + h)^2} \] Step 2: {Relating time periods at different heights}
\[ \frac{T_1}{T_2} = \frac{R + h_1}{R + h_2} = \frac{R + R}{R + 2R} = \frac{2}{3} \Rightarrow 3T_1 = 2T_2 \] Thus, the correct relation is \( 3T_1 = 2T_2 \).

Answer 2

Step-by-step Solution:

Step 1: Understand the formula for the time period of a simple pendulum
The time period \( T \) of a simple pendulum is given by:
\( T = 2\pi \sqrt{\dfrac{L}{g}} \)
Here, \( L \) is the length of the pendulum and \( g \) is the acceleration due to gravity at the location of the pendulum.

Step 2: Understand how gravity changes with height above Earth
The value of gravity \( g_h \) at a height \( h \) above the Earth’s surface is given by:
\( g_h = g \left( \dfrac{R}{R + h} \right)^2 \)
where \( R \) is the radius of the Earth and \( h \) is the height above the Earth's surface.

Step 3: Write time periods at two different heights
At height \( R \), time period is given as:
\( T_1 = 2\pi \sqrt{\dfrac{L}{g_1}} = 4 \, \text{s} \)
where \( g_1 = g \left( \dfrac{R}{2R} \right)^2 = \dfrac{g}{4} \)
So \( T_1 = 2\pi \sqrt{\dfrac{L}{g/4}} = 2\pi \sqrt{\dfrac{4L}{g}} \)

At height \( 2R \), gravity becomes:
\( g_2 = g \left( \dfrac{R}{3R} \right)^2 = \dfrac{g}{9} \)
So time period is:
\( T_2 = 2\pi \sqrt{\dfrac{L}{g/9}} = 2\pi \sqrt{\dfrac{9L}{g}} \)

Step 4: Take ratio of \( T_1 \) and \( T_2 \)
Let us compute the ratio \( \dfrac{T_1}{T_2} \):
From above:
\( T_1 = 2\pi \sqrt{\dfrac{4L}{g}} \),
\( T_2 = 2\pi \sqrt{\dfrac{9L}{g}} \)
So,
\( \dfrac{T_1}{T_2} = \dfrac{\sqrt{4L/g}}{\sqrt{9L/g}} = \dfrac{2}{3} \)

Step 5: Rearranging to match options
Multiplying both sides by 3:
\( 3T_1 = 2T_2 \)

Final Answer:
The correct relation is \( 3T_1 = 2T_2 \)