A particle is executing simple harmonic motion of amplitude 5 cm and period 6 s. How long will it take to move from one end of its path on one side of mean position to a position 2.5 cm on the same side of the mean position ?

Options

A. 1 s
B. 1.5 s
C. 3 s
D. 3.5 s

Answer

To solve the problem, we need to determine how long it takes for a particle executing simple harmonic motion (SHM) to move from one end of its path to a position that is 2.5 cm from the mean position on the same side.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Amplitude (A) = 5 cm
- Period (T) = 6 s
- Position to reach (x) = 2.5 cm (which is half the amplitude)

2. **Understand the Motion:**
- The particle starts at one extreme position (let's say at +5 cm) and moves towards the mean position (0 cm).
- The distance from the extreme position to the mean position is 5 cm, and we want to find the time taken to reach 2.5 cm from the mean position.

3. **Use the SHM Displacement Equation:**
- The displacement equation for SHM when starting from the extreme position is given by:
\[ x = A \cos(\omega t) \]
- Here, \( \omega \) is the angular frequency, which can be calculated as:
\[ \omega = \frac{2\pi}{T} \]
- Substituting the period:
\[ \omega = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad/s} \]

4. **Set Up the Equation:**
- We want to find the time \( t \) when the displacement \( x \) is 2.5 cm:
\[ 2.5 = 5 \cos(\omega t) \]

5. **Simplify the Equation:**
- Dividing both sides by 5:
\[ \frac{2.5}{5} = \cos(\omega t) \]
\[ 0.5 = \cos(\omega t) \]

6. **Find \( \omega t \):**
- The cosine of \( \frac{\pi}{3} \) is 0.5, so:
\[ \omega t = \frac{\pi}{3} \]

7. **Solve for Time \( t \):**
- Substitute \( \omega = \frac{\pi}{3} \):
\[ \frac{\pi}{3} t = \frac{\pi}{3} \]
- Therefore, solving for \( t \):
\[ t = 1 \text{ second} \]

### Final Answer:
The time taken for the particle to move from one end of its path to a position 2.5 cm on the same side of the mean position is **1 second**.