Question

A pendulum of length 4 m is located at a height R above the earth's surface.The period of the simple pendulum is \(2\pi\sqrt{\frac{8}{x}}\) seconds. Find x.

Answer 1

Correct Answer: 5

The problem asks to find the value of \( x \) given the time period of a simple pendulum of a specific length located at a height equal to the Earth's radius (\( R \)) above the Earth's surface.

Concept Used:

The solution involves two key concepts from physics:

1. Time Period of a Simple Pendulum: The time period \( T \) of a simple pendulum is given by the formula:

\[ T = 2\pi\sqrt{\frac{L}{g_{\text{eff}}}} \]

where \( L \) is the length of the pendulum and \( g_{\text{eff}} \) is the effective acceleration due to gravity at the pendulum's location.

2. Variation of Gravity with Altitude: The acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface is related to the acceleration due to gravity on the surface (\( g \)) by:

\[ g' = g \left( \frac{R}{R+h} \right)^2 \]

where \( R \) is the radius of the Earth.

Step-by-Step Solution:

Step 1: Identify the given values.

Length of the pendulum, \( L = 4 \, \text{m} \).

Height above the Earth's surface, \( h = R \).

Time period of the pendulum, \( T = 2\pi \sqrt{\frac{8}{x}} \, \text{seconds} \).

Step 2: Calculate the acceleration due to gravity (\( g' \)) at the height \( h = R \).

Using the formula for variation of gravity with altitude:

\[ g' = g \left( \frac{R}{R+h} \right)^2 \]

Substitute \( h = R \):

\[ g' = g \left( \frac{R}{R+R} \right)^2 = g \left( \frac{R}{2R} \right)^2 = g \left( \frac{1}{2} \right)^2 \] \[ g' = \frac{g}{4} \]

Here, \( g \) is the acceleration due to gravity at the Earth's surface, which is approximately \( 9.8 \, \text{m/s}^2 \).

Step 3: Determine the time period of the pendulum at this height using its formula.

The time period \( T \) is given by:

\[ T = 2\pi\sqrt{\frac{L}{g'}} \]

Substitute the values \( L = 4 \, \text{m} \) and \( g' = g/4 \):

\[ T = 2\pi\sqrt{\frac{4}{g/4}} = 2\pi\sqrt{\frac{16}{g}} \]

Step 4: Equate the calculated time period with the given expression for the time period.

We are given \( T = 2\pi \sqrt{\frac{8}{x}} \). Equating this with our derived expression:

\[ 2\pi \sqrt{\frac{8}{x}} = 2\pi \sqrt{\frac{16}{g}} \]

Step 5: Solve the equation for \( x \).

Cancel \( 2\pi \) from both sides and square the equation:

\[ \left(\sqrt{\frac{8}{x}}\right)^2 = \left(\sqrt{\frac{16}{g}}\right)^2 \] \[ \frac{8}{x} = \frac{16}{g} \]

Now, solve for \( x \):

\[ 8g = 16x \] \[ x = \frac{8g}{16} = \frac{g}{2} \]

Final Computation & Result:

Substitute the standard value of \( g \approx 9.8 \, \text{m/s}^2 \) to find the numerical value of \( x \).

\[ x = \frac{9.8}{2} \] \[ x = 4.9 \]

Thus, the value of x is 5 (Nearest Integer)