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 need to analyze the motion of the particle executing Simple Harmonic Motion (SHM) with the given parameters.

### Step 1: Understand the parameters of SHM
- **Mean position (x = 0)**: This is the center of the motion.
- **Amplitude (A)**: The maximum displacement from the mean position, which is given as 5 cm.
- **Time period (T)**: The time taken to complete one full cycle, which is given as 20 seconds.

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{20} = \frac{\pi}{10} \text{ rad/s} \]

### Step 3: Write the equation of motion
The equation of motion for SHM can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
Where:
- \( A = 5 \, \text{cm} \)
- \( \omega = \frac{\pi}{10} \, \text{rad/s} \)
- \( \phi \) is the phase constant.

### Step 4: Find the phase constant (φ)
At \( t = 0 \), the particle is at \( x = 4 \, \text{cm} \):
\[ 4 = 5 \sin(\phi) \]
Solving for \( \sin(\phi) \):
\[ \sin(\phi) = \frac{4}{5} \]
Thus, we can find \( \phi \):
\[ \phi = \sin^{-1}\left(\frac{4}{5}\right) \]

### Step 5: Find the time to reach \( x = -3 \, \text{cm} \)
We need to find the time \( t_1 \) when the particle is at \( x = -3 \, \text{cm} \):
\[ -3 = 5 \sin\left(\frac{\pi}{10} t_1 + \phi\right) \]
This simplifies to:
\[ \sin\left(\frac{\pi}{10} t_1 + \phi\right) = -\frac{3}{5} \]

### Step 6: Solve for \( t_1 \)
Using the sine inverse function:
\[ \frac{\pi}{10} t_1 + \phi = \sin^{-1}\left(-\frac{3}{5}\right) \]
We know that \( \sin^{-1}(-x) = -\sin^{-1}(x) \), thus:
\[ \frac{\pi}{10} t_1 + \phi = -\sin^{-1}\left(\frac{3}{5}\right) \]
Rearranging gives:
\[ t_1 = \frac{10}{\pi}\left(-\sin^{-1}\left(\frac{3}{5}\right) - \phi\right) \]

### Step 7: Calculate the values
Using the values of \( \sin^{-1}\left(\frac{4}{5}\right) \) and \( \sin^{-1}\left(\frac{3}{5}\right) \):
1. Calculate \( \sin^{-1}\left(\frac{4}{5}\right) \approx 0.927 \) radians.
2. Calculate \( \sin^{-1}\left(\frac{3}{5}\right) \approx 0.644 \) radians.

Now substituting these values into the equation:
\[ t_1 = \frac{10}{\pi}\left(-0.644 - 0.927\right) \]
\[ t_1 = \frac{10}{\pi}(-1.571) \approx 5 \text{ seconds} \]

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