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 step by step, we will follow the process outlined in the video transcript.

### Step 1: Understand the Problem
We have a particle performing Simple Harmonic Motion (SHM) with:
- Amplitude (A) = 5 m
- Time period (T) = 4 s

We need to find the minimum time taken by the particle to move between two points:
- Point X at a distance of 3 m from the center
- Point Y at a distance of 4 m from the center

### Step 2: Determine Angular Frequency (ω)
The angular frequency (ω) is given by the formula:
\[ \omega = \frac{2\pi}{T} \]
Substituting the value of T:
\[ \omega = \frac{2\pi}{4} = \frac{\pi}{2} \text{ radians/second} \]

### Step 3: Write the Equation of Motion
The position of the particle in SHM can be described by the equation:
\[ x(t) = A \sin(\omega t) \]
Substituting the values of A and ω:
\[ x(t) = 5 \sin\left(\frac{\pi}{2} t\right) \]

### Step 4: Find Time to Reach Point X (3 m)
Set \(x = 3\) m:
\[ 3 = 5 \sin\left(\frac{\pi}{2} t_3\right) \]
Rearranging gives:
\[ \sin\left(\frac{\pi}{2} t_3\right) = \frac{3}{5} \]
Taking the inverse sine:
\[ \frac{\pi}{2} t_3 = \sin^{-1}\left(\frac{3}{5}\right) \]
Calculating \(\sin^{-1}\left(\frac{3}{5}\right)\) gives approximately \(0.6435\) radians. Thus:
\[ t_3 = \frac{2}{\pi} \cdot 0.6435 \approx \frac{2 \cdot 0.6435}{3.1416} \approx 0.409 \text{ seconds} \]

### Step 5: Find Time to Reach Point Y (4 m)
Set \(x = 4\) m:
\[ 4 = 5 \sin\left(\frac{\pi}{2} t_4\right) \]
Rearranging gives:
\[ \sin\left(\frac{\pi}{2} t_4\right) = \frac{4}{5} \]
Taking the inverse sine:
\[ \frac{\pi}{2} t_4 = \sin^{-1}\left(\frac{4}{5}\right) \]
Calculating \(\sin^{-1}\left(\frac{4}{5}\right)\) gives approximately \(0.9273\) radians. Thus:
\[ t_4 = \frac{2}{\pi} \cdot 0.9273 \approx \frac{2 \cdot 0.9273}{3.1416} \approx 0.590 \text{ seconds} \]

### Step 6: Calculate Time Interval
The time taken to move from point X to point Y is:
\[ \Delta t = t_4 - t_3 \]
Substituting the values:
\[ \Delta t = 0.590 - 0.409 \approx 0.181 \text{ seconds} \]

### Final Answer
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 **0.181 seconds**.