A particle oscillates as per the equation `x = (7 cos 0.5 pi t)m`, the time taken by the particle to move from the mean position to a point `3.5m` away is

Options

A. `1//3s`
B. `1//2s`
C. `1s`
D. `2//3s`

Explanation

To solve the problem, we need to determine the time taken by a particle oscillating according to the equation \( x = 7 \cos(0.5 \pi t) \) to move from the mean position (where \( x = 0 \)) to a point \( 3.5 \, m \) away.

### Step-by-Step Solution:

1. **Identify the Mean Position:**
The mean position in simple harmonic motion (SHM) is where the displacement \( x = 0 \).

2. **Set Up the Equation:**
We need to find the time \( t_1 \) when the particle is at \( x = 3.5 \, m \).
\[ x = 7 \cos(0.5 \pi t) \]
Setting \( x = 3.5 \):
\[ 3.5 = 7 \cos(0.5 \pi t_1) \]

3. **Rearranging the Equation:**
Divide both sides by 7:
\[ \frac{3.5}{7} = \cos(0.5 \pi t_1) \]
Simplifying gives:
\[ 0.5 = \cos(0.5 \pi t_1) \]

4. **Finding the Angle:**
We know that \( \cos(\frac{\pi}{3}) = 0.5 \). Therefore, we can set:
\[ 0.5 \pi t_1 = \frac{\pi}{3} \]

5. **Solving for \( t_1 \):**
To isolate \( t_1 \), divide both sides by \( 0.5 \pi \):
\[ t_1 = \frac{\frac{\pi}{3}}{0.5 \pi} = \frac{1}{3} \text{ seconds} \]

6. **Final Result:**
The time taken by the particle to move from the mean position to a point \( 3.5 \, m \) away is:
\[ t_1 = \frac{1}{3} \text{ seconds} \approx 0.333 \text{ seconds} \]