In a certain oscillatory system (particle is performing SHM), the amplitude of motion is 5 m and the time period is 4 s. the minimum time taken by the particle for passing betweens points, which are at distances of 4 m and 3 m from the centre and on the same side of it will approximately be

Options

A. `(16)/(45)s`
B. `(7)/(45)s`
C. `(8)/(45)s`
D. `(13)/(45)s`

Explanation

To solve the problem, we need to find the minimum time taken by a particle performing Simple Harmonic Motion (SHM) to travel between two points at distances of 4 m and 3 m from the center of motion. Here’s a step-by-step solution:

### Step 1: Understand the Parameters
Given:
- Amplitude (A) = 5 m
- Time period (T) = 4 s

### Step 2: Calculate the Angular Frequency
The angular frequency (ω) can be calculated using the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the given time period:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ rad/s} \]

### Step 3: Set Up the Equations for the Distances
The displacement in SHM is given by:
\[ x = A \sin(\omega t) \]
For the first position (x1 = 3 m):
\[ 3 = 5 \sin(\omega t_1) \implies \sin(\omega t_1) = \frac{3}{5} \]
For the second position (x2 = 4 m):
\[ 4 = 5 \sin(\omega t_2) \implies \sin(\omega t_2) = \frac{4}{5} \]

### Step 4: Calculate Time for Each Position
Using the inverse sine function:
1. For \( t_1 \):
\[ \omega t_1 = \sin^{-1}\left(\frac{3}{5}\right) \]
2. For \( t_2 \):
\[ \omega t_2 = \sin^{-1}\left(\frac{4}{5}\right) \]

### Step 5: Find the Time Difference
The time difference \( t_2 - t_1 \) can be expressed as:
\[ t_2 - t_1 = \frac{1}{\omega} \left( \sin^{-1}\left(\frac{4}{5}\right) - \sin^{-1}\left(\frac{3}{5}\right) \right) \]

### Step 6: Calculate \( \sin^{-1}\left(\frac{3}{5}\right) \) and \( \sin^{-1}\left(\frac{4}{5}\right) \)
Using a calculator:
- \( \sin^{-1}\left(\frac{3}{5}\right) \approx 0.6435 \text{ radians} \)
- \( \sin^{-1}\left(\frac{4}{5}\right) \approx 0.9273 \text{ radians} \)

### Step 7: Substitute Values
Now substitute these values into the time difference equation:
\[ t_2 - t_1 = \frac{1}{\frac{\pi}{2}} \left( 0.9273 - 0.6435 \right) \]
Calculating the difference:
\[ 0.9273 - 0.6435 = 0.2838 \text{ radians} \]
Thus,
\[ t_2 - t_1 = \frac{2}{\pi} \times 0.2838 \]

### Step 8: Final Calculation
Calculating the time difference:
\[ t_2 - t_1 \approx \frac{2 \times 0.2838}{3.1416} \approx 0.180 \text{ seconds} \]

### Conclusion
The minimum time taken by the particle to pass between the points at distances of 4 m and 3 m from the center is approximately **0.180 seconds**.