The periodic time of a body executing SHM is 2s. After how much time interval from t=0, will its displacement be half of its amplitude?

Answer

To solve the problem, we need to find the time interval from \( t = 0 \) when the displacement of a body executing Simple Harmonic Motion (SHM) is half of its amplitude.

### Step-by-Step Solution:

1. **Understand the Equation of SHM**:
The displacement \( y \) in SHM can be expressed as:
\[ y = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Identify Given Values**:
- The periodic time \( T \) is given as \( 2 \) seconds.
- We need to find the time \( t \) when the displacement \( y \) is half of the amplitude \( A \). Therefore, we set:
\[ y = \frac{A}{2} \]

3. **Substituting into the SHM Equation**:
Substitute \( y = \frac{A}{2} \) into the SHM equation:
\[ \frac{A}{2} = A \sin(\omega t) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t) \]

4. **Calculate Angular Frequency \( \omega \)**:
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 2 \) seconds:
\[ \omega = \frac{2\pi}{2} = \pi \, \text{rad/s} \]

5. **Substituting \( \omega \) into the Equation**:
Now we substitute \( \omega \) back into our equation:
\[ \frac{1}{2} = \sin(\pi t) \]

6. **Finding the Angle**:
We know that:
\[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]
Therefore, we can set:
\[ \pi t = \frac{\pi}{6} \]

7. **Solving for \( t \)**:
Dividing both sides by \( \pi \):
\[ t = \frac{1}{6} \, \text{seconds} \]

### Final Answer:
The time interval from \( t = 0 \) when the displacement is half of the amplitude is:
\[ t = \frac{1}{6} \, \text{seconds} \]