For a particle executing simple harmonic motion, the amplitude is A and time period is T. The maximum speed will be

Options

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

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 relationship between angular frequency and time period
The angular frequency \( \omega \) is related to the time period \( T \) by the formula:
\[ \omega = \frac{2\pi}{T} \]

### Step 2: Write the formula for velocity in SHM
The velocity \( v \) of a particle in SHM at a displacement \( x \) from the mean position is given by:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude and \( x \) is the displacement.

### Step 3: Determine the maximum speed
The maximum speed occurs when the displacement \( x \) is zero (i.e., at the mean position). Substituting \( x = 0 \) into the velocity formula gives:
\[ v_{\text{max}} = \omega \sqrt{A^2 - 0^2} = \omega A \]

### Step 4: Substitute the expression for \( \omega \)
Now, substitute \( \omega = \frac{2\pi}{T} \) into the equation for maximum speed:
\[ v_{\text{max}} = \left(\frac{2\pi}{T}\right) A \]

### Step 5: Final expression for maximum speed
Thus, the maximum speed \( v_{\text{max}} \) can be expressed as:
\[ v_{\text{max}} = \frac{2\pi A}{T} \]

### Conclusion
The maximum speed of a particle executing simple harmonic motion is given by:
\[ v_{\text{max}} = \frac{2\pi A}{T} \]

---