Two simple pendulum first of bob mass `M_(1)` and length `L_(1)` second of bob mass `M_(2)` and length `L_(2) M_(1) = M_(2)` and `L_(1)` = 2L_(2)`. if the vibrational energy of both is same which is correct?

Options

A. Amplitude of B greater than A
B. Amplitude of B smaller than A
C. Amplitudes will be same
D. none of these

Answer

To solve the problem step by step, we need to analyze the given information about the two pendulums and their vibrational energies.

### Given:
1. Mass of the first pendulum bob: \( M_1 \)
2. Length of the first pendulum: \( L_1 \)
3. Mass of the second pendulum bob: \( M_2 \)
4. Length of the second pendulum: \( L_2 \)
5. Conditions: \( M_1 = M_2 \) and \( L_1 = 2L_2 \)
6. The vibrational energy of both pendulums is the same.

### Step 1: Write the formula for the frequency of a simple pendulum
The frequency \( f \) of a simple pendulum is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{g}{L}} \]
where \( g \) is the acceleration due to gravity and \( L \) is the length of the pendulum.

### Step 2: Determine the frequencies of both pendulums
For the first pendulum:
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{g}{L_1}} \]

For the second pendulum:
\[ f_2 = \frac{1}{2\pi} \sqrt{\frac{g}{L_2}} \]

### Step 3: Substitute \( L_1 \) in terms of \( L_2 \)
Since \( L_1 = 2L_2 \), we can substitute this into the frequency formula for \( f_1 \):
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{g}{2L_2}} = \frac{1}{2\pi} \frac{1}{\sqrt{2}} \sqrt{\frac{g}{L_2}} = \frac{f_2}{\sqrt{2}} \]

### Step 4: Calculate the ratio of frequencies
From the above, we have:
\[ \frac{f_1}{f_2} = \frac{1}{\sqrt{2}} \quad \text{or} \quad f_2 = \sqrt{2} f_1 \]

### Step 5: Write the expression for vibrational energy
The vibrational energy \( E \) of a pendulum is given by:
\[ E = \frac{1}{2} m \omega^2 A^2 \]
where \( \omega = 2\pi f \) and \( A \) is the amplitude.

### Step 6: Set the energies equal
Since the energies are the same:
\[ E_1 = E_2 \implies \frac{1}{2} M_1 \omega_1^2 A_1^2 = \frac{1}{2} M_2 \omega_2^2 A_2^2 \]
Given \( M_1 = M_2 \), we can cancel the mass terms:
\[ \omega_1^2 A_1^2 = \omega_2^2 A_2^2 \]

### Step 7: Substitute \( \omega \) in terms of frequency
We know that \( \omega = 2\pi f \):
\[ (2\pi f_1)^2 A_1^2 = (2\pi f_2)^2 A_2^2 \]
This simplifies to:
\[ f_1^2 A_1^2 = f_2^2 A_2^2 \]

### Step 8: Substitute the frequency ratio
Using \( f_2 = \sqrt{2} f_1 \):
\[ f_1^2 A_1^2 = ( \sqrt{2} f_1 )^2 A_2^2 \]
This gives:
\[ f_1^2 A_1^2 = 2 f_1^2 A_2^2 \]
Cancelling \( f_1^2 \) (assuming \( f_1 \neq 0 \)):
\[ A_1^2 = 2 A_2^2 \]
Taking the square root:
\[ A_1 = \sqrt{2} A_2 \]

### Conclusion
Thus, the amplitude of the first pendulum \( A_1 \) is greater than the amplitude of the second pendulum \( A_2 \).

### Final Answer
The correct statement is that \( A_1 > A_2 \).

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