Two identical pendulums oscillate with a constant phase difference `(pi)/(4)` and same amplitude. If the maximum velocity of one is v, the maximum velocity of the other will be.

Options

A. v
B. `sqrt2v`
C. 2v
D. `(v)/(sqrt2)`

Answer

To solve the problem, we need to analyze the motion of two identical pendulums that oscillate with a constant phase difference of \(\frac{\pi}{4}\) and the same amplitude. We are given that the maximum velocity of one pendulum is \(v\), and we need to determine the maximum velocity of the other pendulum.

### Step-by-step Solution:

1. **Understanding the Motion of the Pendulums:**
Both pendulums oscillate in simple harmonic motion (SHM) with the same amplitude \(A\) and the same angular frequency \(\omega\). The phase difference between them is \(\frac{\pi}{4}\).

2. **Equations of Motion:**
The maximum velocity \(V_{\text{max}}\) of a pendulum in SHM is given by the formula:
\[ V_{\text{max}} = A \omega \]
For the first pendulum, we can write its velocity as:
\[ v_1(t) = A \omega \sin(\omega t) \]
For the second pendulum, which has a phase difference of \(\frac{\pi}{4}\), the velocity can be expressed as:
\[ v_2(t) = A \omega \sin\left(\omega t + \frac{\pi}{4}\right) \]

3. **Finding the Maximum Velocities:**
The maximum velocity occurs when the sine function equals 1. Therefore, for the first pendulum:
\[ V_{\text{max,1}} = A \omega \cdot 1 = A \omega \]
For the second pendulum:
\[ V_{\text{max,2}} = A \omega \cdot 1 = A \omega \]

4. **Conclusion:**
Since both pendulums have the same amplitude \(A\) and angular frequency \(\omega\), their maximum velocities are equal:
\[ V_{\text{max,1}} = V_{\text{max,2}} = A \omega \]
Given that the maximum velocity of one pendulum is \(v\), we conclude that the maximum velocity of the other pendulum is also \(v\).

### Final Answer:
The maximum velocity of the other pendulum will be \(v\).

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