The instantaneous displacement of a simple pendulum oscillator is given by `x= A cos (omegat+ pi / 4 )`Its speed will be maximum at time

Options

A. `pi/(4 omega)`
B. `pi/(2 omega)`
C. `(pi)/(omega)`
D. `(2pi)/(omega)`

Answer

To determine when the speed of the simple pendulum oscillator is maximum, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Displacement Function**:
The displacement of the simple harmonic oscillator is given by:
\[ x = A \cos(\omega t + \frac{\pi}{4}) \]
where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( t \) is time.

2. **Identify the Condition for Maximum Speed**:
The speed of the oscillator is maximum when the displacement \( x \) is zero. This is because, in simple harmonic motion (SHM), the speed is maximum at the mean position (equilibrium position).

3. **Set the Displacement to Zero**:
To find when the speed is maximum, we set the displacement \( x \) to zero:
\[ 0 = A \cos(\omega t + \frac{\pi}{4}) \]

4. **Solve for the Cosine Function**:
From the equation above, we can simplify it to:
\[ \cos(\omega t + \frac{\pi}{4}) = 0 \]
The cosine function is zero at odd multiples of \( \frac{\pi}{2} \):
\[ \omega t + \frac{\pi}{4} = \frac{\pi}{2} + n\pi \quad (n \in \mathbb{Z}) \]

5. **Isolate \( t \)**:
Rearranging the equation gives:
\[ \omega t = \frac{\pi}{2} - \frac{\pi}{4} + n\pi \]
\[ \omega t = \frac{\pi}{4} + n\pi \]
\[ t = \frac{\pi}{4\omega} + \frac{n\pi}{\omega} \]

6. **Determine the First Instance of Maximum Speed**:
The first instance (for \( n = 0 \)):
\[ t = \frac{\pi}{4\omega} \]

7. **Conclusion**:
The speed of the simple pendulum oscillator will be maximum at:
\[ t = \frac{\pi}{4\omega} \]