If a simple pendulum oscillates with an amplitude of 50 mm and time period of 2 sec, then its maximum velocity is

Options

A. (a)`0.10 ms ^(-1)`
B. (b)`0.15 ms ^(-1)`
C. (c)`0.8 ms ^(-1)`
D. (d)`0.26 ms ^(-1)`

Answer

To find the maximum velocity of a simple pendulum, we can follow these steps:

### Step 1: Understand the formula for maximum velocity
The maximum velocity \( V_{max} \) of a simple harmonic oscillator (like a pendulum) is given by the formula:
\[ V_{max} = A \omega \]
where:
- \( A \) is the amplitude of the oscillation,
- \( \omega \) is the angular frequency.

### Step 2: Convert amplitude to meters
The amplitude is given in millimeters, so we need to convert it to meters:
\[ A = 50 \text{ mm} = 50 \times 10^{-3} \text{ m} = 0.050 \text{ m} \]

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]
Given that the time period \( T = 2 \text{ s} \), we can substitute this value:
\[ \omega = \frac{2\pi}{2} = \pi \text{ rad/s} \]

### Step 4: Substitute values into the maximum velocity formula
Now we can substitute the values of \( A \) and \( \omega \) into the maximum velocity formula:
\[ V_{max} = A \omega = (0.050 \text{ m}) \times (\pi \text{ rad/s}) \]

### Step 5: Calculate the maximum velocity
Using \( \pi \approx 3.14 \):
\[ V_{max} = 0.050 \times 3.14 \approx 0.157 \text{ m/s} \]

### Final Answer
Thus, the maximum velocity of the pendulum is approximately:
\[ V_{max} \approx 0.157 \text{ m/s} \]