The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position is

Options

A. `0.1s`
B. `0.2s`
C. `0.4s`
D. `0.6s`

Explanation

To solve the problem, we need to find the time \( t_1 \) at which the velocity of the particle executing simple harmonic motion (SHM) is half of its maximum velocity. The time period of the oscillation is given as \( T = 1.2 \, \text{s} \).

### Step-by-step Solution:

1. **Understand the Maximum Velocity**:
The maximum velocity \( V_0 \) of a particle in SHM is given by the formula:
\[ V_0 = A \omega \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Determine Angular Frequency**:
The angular frequency \( \omega \) is related to the time period \( T \) by the equation:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 1.2 \, \text{s} \):
\[ \omega = \frac{2\pi}{1.2} = \frac{5\pi}{6} \, \text{rad/s} \]

3. **Set Up the Velocity Equation**:
The velocity \( V \) of the particle at any time \( t \) is given by:
\[ V = V_0 \cos(\omega t) \]
We need to find the time \( t_1 \) when the velocity \( V \) is half of the maximum velocity \( V_0 \):
\[ V = \frac{V_0}{2} \]

4. **Substitute into the Velocity Equation**:
Setting the equation for velocity equal to half the maximum velocity:
\[ \frac{V_0}{2} = V_0 \cos(\omega t_1) \]
Dividing both sides by \( V_0 \) (assuming \( V_0 \neq 0 \)):
\[ \frac{1}{2} = \cos(\omega t_1) \]

5. **Solve for \( \omega t_1 \)**:
The cosine function yields:
\[ \cos(\omega t_1) = \frac{1}{2} \]
This implies:
\[ \omega t_1 = \frac{\pi}{3} \quad \text{(since } \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\text{)} \]

6. **Substitute \( \omega \)**:
Now substituting \( \omega = \frac{5\pi}{6} \):
\[ \frac{5\pi}{6} t_1 = \frac{\pi}{3} \]

7. **Solve for \( t_1 \)**:
To find \( t_1 \), rearranging gives:
\[ t_1 = \frac{\frac{\pi}{3}}{\frac{5\pi}{6}} = \frac{6}{15} = \frac{2}{5} \, \text{s} = 0.4 \, \text{s} \]

### Final Answer:
The time starting from the mean position at which its velocity will be half of its velocity at the mean position is \( t_1 = 0.4 \, \text{s} \).