A simple harmonic motino has amplitude A and time period T. The maxmum velocity will be

Options

A. `4AT`
B. `(2A)/(T)`
C. `2pi sqrt((A)/(T)`
D. `(2piA)/(T)`

Answer

To find the maximum velocity of a simple harmonic motion (SHM) with amplitude \( A \) and time period \( T \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the relationship in SHM**: In simple harmonic motion, the maximum velocity (\( V_{max} \)) can be expressed in terms of the amplitude (\( A \)) and the angular frequency (\( \omega \)).

2. **Formula for maximum velocity**: The formula for maximum velocity in SHM is given by:
\[ V_{max} = A \cdot \omega \]
where:
- \( V_{max} \) is the maximum velocity,
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

3. **Calculate angular frequency**: The angular frequency (\( \omega \)) is related to the time period (\( T \)) by the formula:
\[ \omega = \frac{2\pi}{T} \]

4. **Substitute \( \omega \) into the maximum velocity formula**: Now, substitute the expression for \( \omega \) into the maximum velocity equation:
\[ V_{max} = A \cdot \left(\frac{2\pi}{T}\right) \]

5. **Final expression for maximum velocity**: Simplifying the equation gives us:
\[ V_{max} = \frac{2\pi A}{T} \]

Thus, the maximum velocity of the simple harmonic motion is:
\[ V_{max} = \frac{2\pi A}{T} \]

### Conclusion:
The maximum velocity in simple harmonic motion with amplitude \( A \) and time period \( T \) is \( \frac{2\pi A}{T} \).

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