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 is smaller than A
B. Amplitude of B is greater than A
C. Amplitude will be same
D. None of the above

Answer

To solve the problem step by step, we will analyze the given information and apply the relevant formulas.

### Step 1: Understand the given information
We have two simple pendulums:
- Pendulum 1: Bob mass \( M_1 \) and length \( L_1 \)
- Pendulum 2: Bob mass \( M_2 \) and length \( L_2 \)

Given:
- \( M_1 = M_2 \)
- \( L_1 = 2L_2 \)

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

### Step 3: Find the frequencies of both pendulums
For Pendulum 1:
\[ f_1 = \frac{1}{2\pi} \sqrt{\frac{g}{L_1}} \]

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

### Step 4: Establish the relationship between the frequencies
Using the relationship between the lengths:
\[ L_1 = 2L_2 \]
We can express the frequency ratio:
\[ \frac{f_1}{f_2} = \sqrt{\frac{L_2}{L_1}} = \sqrt{\frac{L_2}{2L_2}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Relate the frequencies
From the frequency ratio, we can express:
\[ f_1 = \frac{1}{\sqrt{2}} f_2 \]
This implies:
\[ f_2 = \sqrt{2} f_1 \]

### Step 6: Write the expression for vibrational energy
The vibrational energy \( E \) of a pendulum is given by:
\[ E = \frac{1}{2} m A^2 \omega^2 \]
where \( A \) is the amplitude and \( \omega = 2\pi f \).

### Step 7: Set the energies equal
Since the vibrational energies of both pendulums are the same:
\[ E_1 = E_2 \]
This leads to:
\[ \frac{1}{2} M_1 A_1^2 (2\pi f_1)^2 = \frac{1}{2} M_2 A_2^2 (2\pi f_2)^2 \]

### Step 8: Substitute the known values
Since \( M_1 = M_2 \):
\[ A_1^2 (2\pi f_1)^2 = A_2^2 (2\pi f_2)^2 \]
This simplifies to:
\[ A_1^2 f_1^2 = A_2^2 f_2^2 \]

### Step 9: Express amplitudes in terms of frequencies
Rearranging gives:
\[ \frac{A_1^2}{A_2^2} = \frac{f_2^2}{f_1^2} \]
Substituting \( f_2 = \sqrt{2} f_1 \):
\[ \frac{A_1^2}{A_2^2} = \frac{(\sqrt{2} f_1)^2}{f_1^2} = 2 \]
Thus:
\[ A_1^2 = 2 A_2^2 \implies A_1 = \sqrt{2} A_2 \]

### Conclusion
Since \( A_1 > A_2 \), we can conclude that the amplitude of Pendulum 1 is greater than that of Pendulum 2.