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 requires determining the value of \( x \) based on the time period of a simple pendulum. This pendulum has a specific length and is positioned at an altitude equivalent to the Earth's radius (\( R \)) above the Earth's surface.

Concept Used:

The solution employs two fundamental physics principles:

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

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

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

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

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

Here, \( R \) is the Earth's radius.

Step-by-Step Solution:

Step 1: Enumerate the provided parameters.

Pendulum length, \( L = 4 \, \text{m} \).

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

Pendulum's time period, \( T = 2\pi \sqrt{\frac{8}{x}} \, \text{seconds} \).

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

Applying the formula for gravitational variation with altitude:

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

With \( h = R \) substituted:

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

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

Step 3: Determine the pendulum's time period at the specified altitude using its standard formula.

The time period \( T \) is defined as:

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

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

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

Step 4: Align the derived time period expression with the given time period expression.

The problem states \( T = 2\pi \sqrt{\frac{8}{x}} \). Equating this to the derived expression:

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

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

By canceling \( 2\pi \) from both sides and squaring the equation:

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

Rearranging to solve for \( x \):

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

Final Computation & Result:

To obtain the numerical value of \( x \), substitute the standard value of \( g \approx 9.8 \, \text{m/s}^2 \).

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

Therefore, the value of x is 5 (Nearest Integer).