A particle starts SHM from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed, its displacement y is

Options

A. `A/2`
B. `A/sqrt2`
C. `(Asqrt3)/(2)`
D. `(2A)/(sqrt3)`

Explanation

To solve the problem step by step, we need to find the displacement \( y \) of a particle in Simple Harmonic Motion (SHM) when its speed is half of the maximum speed.

### Step-by-Step Solution:

1. **Identify Maximum Speed**:
The maximum speed \( V_{\text{max}} \) of a particle in SHM is given by the formula:
\[ V_{\text{max}} = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency. The angular frequency \( \omega \) can be expressed in terms of the time period \( T \) as:
\[ \omega = \frac{2\pi}{T} \]

2. **Calculate Half of Maximum Speed**:
Since we need the speed when it is half of the maximum speed:
\[ V = \frac{V_{\text{max}}}{2} = \frac{A \omega}{2} \]

3. **Use the Velocity Formula in SHM**:
The velocity \( V \) in SHM at displacement \( y \) is given by:
\[ V = \omega \sqrt{A^2 - y^2} \]
Setting this equal to \( \frac{A \omega}{2} \):
\[ \frac{A \omega}{2} = \omega \sqrt{A^2 - y^2} \]

4. **Cancel \( \omega \) from Both Sides**:
Assuming \( \omega \neq 0 \), we can divide both sides by \( \omega \):
\[ \frac{A}{2} = \sqrt{A^2 - y^2} \]

5. **Square Both Sides**:
To eliminate the square root, we square both sides:
\[ \left(\frac{A}{2}\right)^2 = A^2 - y^2 \]
\[ \frac{A^2}{4} = A^2 - y^2 \]

6. **Rearrange the Equation**:
Rearranging gives:
\[ y^2 = A^2 - \frac{A^2}{4} \]
\[ y^2 = A^2 \left(1 - \frac{1}{4}\right) = A^2 \left(\frac{3}{4}\right) \]

7. **Solve for \( y \)**:
Taking the square root of both sides:
\[ y = \sqrt{\frac{3}{4}} A = \frac{\sqrt{3}}{2} A \]

### Final Answer:
Thus, the displacement \( y \) when the speed is half of the maximum speed is:
\[ y = \frac{\sqrt{3}}{2} A \]