Amplitude of a particle executing SHM is a and its time period is T. Its maximum speed is

Options

A. `(2a)/(T)`
B. `2pi sqrt((a)/(T))`
C. `(2 pi a)/(T)`
D. `4aT`

Answer

To find the maximum speed of a particle executing Simple Harmonic Motion (SHM) with amplitude \( a \) and time period \( T \), we can follow these steps:

### Step 1: Understand the SHM Equation
The position of a particle in SHM can be described by the equation:
\[ x(t) = a \sin(\omega t + \phi) \]
where:
- \( x(t) \) is the position of the particle at time \( t \),
- \( a \) is the amplitude,
- \( \omega \) is the angular frequency,
- \( \phi \) is the phase constant.

### Step 2: Determine the Angular Frequency
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

### Step 3: Find the Velocity
The velocity \( v(t) \) of the particle is the derivative of the position with respect to time:
\[ v(t) = \frac{dx}{dt} = a \omega \cos(\omega t + \phi) \]

### Step 4: Find Maximum Speed
The maximum speed \( v_{\text{max}} \) occurs when \( \cos(\omega t + \phi) \) is at its maximum value, which is 1. Thus:
\[ v_{\text{max}} = a \omega \]

### Step 5: Substitute for Angular Frequency
Now, substituting \( \omega \) from Step 2 into the equation for maximum speed:
\[ v_{\text{max}} = a \left(\frac{2\pi}{T}\right) \]
This simplifies to:
\[ v_{\text{max}} = \frac{2\pi a}{T} \]

### Conclusion
The maximum speed of the particle executing SHM is:
\[ v_{\text{max}} = \frac{2\pi a}{T} \]