A particle exectes `S.H.M.` along a straight line with mean position `x = 0`, period `20 s` amplitude `5 cm`. The shortest time taken by the particle to go form `x = 4 cm` to `x = -3 cm` is

Options

A. `4s`
B. `7s`
C. `5s`
D. `6s`

Explanation

To solve the problem step by step, we will use the properties of Simple Harmonic Motion (SHM) and the relationship between position, time, and angular frequency.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Mean position \( x = 0 \)
- Amplitude \( A = 5 \, \text{cm} \)
- Period \( T = 20 \, \text{s} \)

2. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{20} = \frac{\pi}{10} \, \text{rad/s} \]

3. **Determine Phase Angles**:
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \cos(\omega t + \phi) \]
To find the phase angle for the positions \( x = 4 \, \text{cm} \) and \( x = -3 \, \text{cm} \):
- For \( x = 4 \):
\[ 4 = 5 \cos(\theta_1) \Rightarrow \cos(\theta_1) = \frac{4}{5} \]
\[ \theta_1 = \cos^{-1}\left(\frac{4}{5}\right) \approx 36.87^\circ \text{ or } 0.6435 \, \text{rad} \]
- For \( x = -3 \):
\[ -3 = 5 \cos(\theta_2) \Rightarrow \cos(\theta_2) = -\frac{3}{5} \]
\[ \theta_2 = \cos^{-1}\left(-\frac{3}{5}\right) \approx 126.87^\circ \text{ or } 2.219 \, \text{rad} \]

4. **Calculate Phase Difference**:
The phase difference \( \Delta \theta \) between the two positions is:
\[ \Delta \theta = \theta_2 - \theta_1 = 2.219 - 0.6435 \approx 1.5755 \, \text{rad} \]

5. **Convert Phase Difference to Time**:
The time taken to cover a phase difference \( \Delta \theta \) is given by:
\[ t = \frac{\Delta \theta}{\omega} \]
Substituting the values:
\[ t = \frac{1.5755}{\frac{\pi}{10}} = \frac{1.5755 \times 10}{\pi} \approx 5 \, \text{s} \]

### Final Answer:
The shortest time taken by the particle to go from \( x = 4 \, \text{cm} \) to \( x = -3 \, \text{cm} \) is approximately **5 seconds**.

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