How long after the beginning of motion is the displacment of a harmonically oscillation particle equal to one half its amplitude if the period is `24s` and perticle starts from rest.

Options

A. `12s`
B. `2s`
C. `4s`
D. `6s`

Explanation

To solve the problem step by step, we will follow the principles of simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understand the Problem**:
We need to find the time \( t \) when the displacement \( y \) of a particle in SHM is equal to half of its amplitude \( A \). The period \( T \) of the motion is given as 24 seconds, and the particle starts from rest.

2. **Displacement Equation**:
The displacement \( y \) in SHM can be expressed as:
\[ y = A \cos(\omega t) \]
where \( \omega \) is the angular frequency.

3. **Setting Up the Equation**:
We want to find \( t \) when \( y = \frac{A}{2} \):
\[ \frac{A}{2} = A \cos(\omega t) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \cos(\omega t) \]

4. **Finding the Angle**:
The cosine function equals \( \frac{1}{2} \) at specific angles. The principal angle is:
\[ \omega t = \frac{\pi}{3} \quad \text{(and also at } \frac{5\pi}{3} \text{, but we will focus on the first quadrant)} \]

5. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is related to the period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]
Substituting \( T = 24 \) seconds:
\[ \omega = \frac{2\pi}{24} = \frac{\pi}{12} \]

6. **Substituting Back to Find Time**:
Now substituting \( \omega \) back into the equation:
\[ \frac{\pi}{3} = \frac{\pi}{12} t \]
To solve for \( t \), multiply both sides by \( 12 \):
\[ 12 \cdot \frac{\pi}{3} = \pi t \]
Simplifying gives:
\[ 4\pi = \pi t \]
Dividing both sides by \( \pi \):
\[ t = 4 \text{ seconds} \]

### Final Answer:
The time after the beginning of motion when the displacement of the harmonically oscillating particle is equal to one half its amplitude is **4 seconds**.