A particle is performing SHM along `x-`axis with amplitude `4.0cm` and time period `1.2s` . What is the minimum time is deci`-` second taken by the particle to move from `x=+2cm `to `x=+4cm` and back again.

Explanation

To solve the problem, we need to determine the time taken by a particle performing Simple Harmonic Motion (SHM) to move from \( x = +2 \, \text{cm} \) to \( x = +4 \, \text{cm} \) and then back to \( x = +2 \, \text{cm} \).

### Step-by-step Solution:

1. **Identify Given Values**:
- Amplitude \( A = 4.0 \, \text{cm} \)
- Time period \( T = 1.2 \, \text{s} \)

2. **Write the Equation of Motion**:
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
where \( \omega \) is the angular frequency.

3. **Calculate Angular Frequency**:
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of \( T \):
\[ \omega = \frac{2\pi}{1.2} \approx 5.24 \, \text{rad/s} \]

4. **Calculate Time \( t_2 \) when \( x = 2 \, \text{cm} \)**:
Using the equation of motion:
\[ t_2 = \frac{T}{2\pi} \sin^{-1}\left(\frac{x}{A}\right) \]
Substituting \( x = 2 \, \text{cm} \):
\[ t_2 = \frac{1.2}{2\pi} \sin^{-1}\left(\frac{2}{4}\right) = \frac{1.2}{2\pi} \sin^{-1}\left(\frac{1}{2}\right) \]
Since \( \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \):
\[ t_2 = \frac{1.2}{2\pi} \cdot \frac{\pi}{6} = \frac{1.2}{12} = 0.1 \, \text{s} \]

5. **Calculate Time \( t_4 \) when \( x = 4 \, \text{cm} \)**:
Using the same formula:
\[ t_4 = \frac{T}{2\pi} \sin^{-1}\left(\frac{x}{A}\right) \]
Substituting \( x = 4 \, \text{cm} \):
\[ t_4 = \frac{1.2}{2\pi} \sin^{-1}\left(\frac{4}{4}\right) = \frac{1.2}{2\pi} \sin^{-1}(1) \]
Since \( \sin^{-1}(1) = \frac{\pi}{2} \):
\[ t_4 = \frac{1.2}{2\pi} \cdot \frac{\pi}{2} = \frac{1.2}{4} = 0.3 \, \text{s} \]

6. **Calculate Time Taken from \( x = 2 \, \text{cm} \) to \( x = 4 \, \text{cm} \)**:
The time taken to move from \( x = 2 \, \text{cm} \) to \( x = 4 \, \text{cm} \) is:
\[ \Delta t = t_4 - t_2 = 0.3 - 0.1 = 0.2 \, \text{s} \]

7. **Calculate Total Time for the Complete Motion**:
Since the particle moves from \( x = 2 \, \text{cm} \) to \( x = 4 \, \text{cm} \) and then back to \( x = 2 \, \text{cm} \), the total time taken is:
\[ \text{Total time} = 2 \times \Delta t = 2 \times 0.2 = 0.4 \, \text{s} \]

8. **Convert to Deci-seconds**:
Since \( 1 \, \text{s} = 10 \, \text{deci-seconds} \):
\[ 0.4 \, \text{s} = 4 \, \text{deci-seconds} \]

### Final Answer:
The minimum time taken by the particle to move from \( x = +2 \, \text{cm} \) to \( x = +4 \, \text{cm} \) and back again is \( 4 \, \text{deci-seconds} \).