In an experiment, a boy plots a graph between `(v_(max))^(2)and (a_(max))^(2)` for a simple pendulum for different values of (small) amplitudes, where `v_(max)`and `a_(max)` is the maximum velocity and the maximum acceleration respectively. He found the graph to be a straight line with a negative slope, making an angle of `30^(@)` when the experiment was conducted on the earth surface. When the same experiment was conducted at a height h above the surface, the line was at an angle of `60^(@)`. The value of h is [radius of the earth = R]

Options

A. 0.5R
B. 0.24R
C. 0.73R
D. R

Answer

To solve the problem, we need to analyze the relationship between the maximum velocity \( v_{\text{max}} \) and the maximum acceleration \( a_{\text{max}} \) of a simple pendulum, and how these quantities change with respect to gravitational acceleration at different heights.

### Step 1: Understand the relationship between \( v_{\text{max}} \) and \( a_{\text{max}} \)

For a simple pendulum undergoing simple harmonic motion (SHM):
- The maximum velocity \( v_{\text{max}} \) is given by:
\[ v_{\text{max}} = A \omega \]
- The maximum acceleration \( a_{\text{max}} \) is given by:
\[ a_{\text{max}} = A \omega^2 \]
Where \( A \) is the amplitude and \( \omega \) is the angular frequency.

### Step 2: Express \( v_{\text{max}}^2 \) and \( a_{\text{max}}^2 \)

Squaring both equations, we get:
- \( v_{\text{max}}^2 = A^2 \omega^2 \)
- \( a_{\text{max}}^2 = A^2 \omega^4 \)

### Step 3: Plotting the graph

When we plot \( v_{\text{max}}^2 \) against \( a_{\text{max}}^2 \), we can express the relationship as:
\[ \frac{v_{\text{max}}^2}{a_{\text{max}}^2} = \frac{A^2 \omega^2}{A^2 \omega^4} = \frac{1}{\omega^2} \]
This means the slope of the graph is:
\[ \text{slope} = \frac{1}{\omega^2} \]

### Step 4: Relate \( \omega \) to gravitational acceleration

For a simple pendulum, the angular frequency \( \omega \) is given by:
\[ \omega = \sqrt{\frac{g}{L}} \]
where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum. Thus:
\[ \omega^2 = \frac{g}{L} \]
This gives us the slope as:
\[ \text{slope} = \frac{L}{g} \]

### Step 5: Analyze the slopes at different conditions

1. **On the Earth's surface**:
- The slope is given as \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \).
- Therefore:
\[ \frac{L}{g} = \frac{1}{\sqrt{3}} \implies g = L \sqrt{3} \quad \text{(Equation 1)} \]

2. **At height \( h \)**:
- The slope is given as \( \tan(60^\circ) = \sqrt{3} \).
- Therefore:
\[ \frac{L}{g'} = \sqrt{3} \implies g' = \frac{L}{\sqrt{3}} \quad \text{(Equation 2)} \]

### Step 6: Relate \( g \) and \( g' \)

From the equations derived:
\[ g' = \frac{g}{3} \]
This indicates that at height \( h \), the gravitational acceleration is one-third of that on the surface.

### Step 7: Use the formula for gravitational acceleration at height \( h \)

The formula for gravitational acceleration at a height \( h \) above the Earth's surface is:
\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]
Setting this equal to \( \frac{g}{3} \):
\[ \frac{g}{(1 + \frac{h}{R})^2} = \frac{g}{3} \]
Cancelling \( g \) from both sides:
\[ (1 + \frac{h}{R})^2 = 3 \]
Taking the square root:
\[ 1 + \frac{h}{R} = \sqrt{3} \]
Thus:
\[ \frac{h}{R} = \sqrt{3} - 1 \]
So:
\[ h = R(\sqrt{3} - 1) \]

### Final Result

The value of \( h \) is:
\[ h = R(0.732) \quad \text{(approximately)} \]