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`

Answer

To solve the problem step by step, we need to find the minimum time taken by a particle performing simple harmonic motion (SHM) to travel between two points, 3 m and 4 m from the center of oscillation.

### Step 1: Understand the parameters of SHM
- Amplitude (A) = 5 m
- Time period (T) = 4 s

### Step 2: Calculate the angular frequency (ω)
The angular frequency (ω) is given by 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: Write the equation of motion
The displacement (x) in SHM can be expressed as:
\[ x = A \sin(\omega t) \]
Substituting the values of A and ω:
\[ x = 5 \sin\left(\frac{\pi}{2} t\right) \]

### Step 4: Find the time when the particle is at x = 3 m
Set x = 3 m in the equation:
\[ 3 = 5 \sin\left(\frac{\pi}{2} T_3\right) \]
Rearranging gives:
\[ \sin\left(\frac{\pi}{2} T_3\right) = \frac{3}{5} \]
Now, take 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:
\[ T_3 = \frac{2 \cdot 0.6435}{\pi} \approx \frac{37}{90} \text{ seconds} \]

### Step 5: Find the time when the particle is at x = 4 m
Set x = 4 m in the equation:
\[ 4 = 5 \sin\left(\frac{\pi}{2} T_4\right) \]
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:
\[ T_4 = \frac{2 \cdot 0.9273}{\pi} \approx \frac{53}{90} \text{ seconds} \]

### Step 6: Calculate the time taken to travel from 3 m to 4 m
The time taken (ΔT) to go from 3 m to 4 m is:
\[ \Delta T = T_4 - T_3 = \frac{53}{90} - \frac{37}{90} = \frac{16}{90} \text{ seconds} \]
Simplifying gives:
\[ \Delta T = \frac{8}{45} \text{ seconds} \]

### 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 \(\frac{8}{45}\) seconds.