A particle performs `SHM` on x- axis with amplitude `A ` and time period period `T` .The time taken by the particle to travel a distance `A//5` starting from rest is

Options

A. `(T)/(20)`
B. `(T)/(2pi)cos^-1((4)/(5))`
C. `(T)/(2pi)cos^-1((1)/(5))`
D. `(T)/(2pi)sin^-1((1)/(5))`

Explanation

To solve the problem of finding the time taken by a particle to travel a distance of \( \frac{A}{5} \) starting from rest in Simple Harmonic Motion (SHM), we can follow these steps:

### Step 1: Understanding the Motion
The particle is performing SHM along the x-axis with amplitude \( A \) and time period \( T \). At the extreme position (where the particle starts from rest), its velocity is zero.

### Step 2: Determine the Distance to be Covered
The total distance the particle needs to travel is \( \frac{A}{5} \). Starting from the extreme position (where the particle is at \( x = A \)), it will move towards the mean position (where \( x = 0 \)).

### Step 3: Calculate the Remaining Distance
If the particle travels \( \frac{A}{5} \), the remaining distance to the mean position is:
\[ \text{Remaining distance} = A - \frac{A}{5} = \frac{4A}{5} \]

### Step 4: Relate the Distance to the Angle in Circular Motion
In SHM, the motion can be related to uniform circular motion. The amplitude \( A \) corresponds to the radius of the circle. The distance covered in terms of angle \( \theta \) can be calculated using the cosine function:
\[ \cos(\theta) = \frac{\text{Remaining distance}}{A} = \frac{\frac{4A}{5}}{A} = \frac{4}{5} \]
Thus,
\[ \theta = \cos^{-1}\left(\frac{4}{5}\right) \]

### Step 5: Calculate the Angular Frequency
The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]

### Step 6: Relate Time to Angle
The time \( t \) taken to cover the angle \( \theta \) is given by:
\[ \theta = \omega t \implies t = \frac{\theta}{\omega} \]
Substituting for \( \omega \):
\[ t = \frac{\theta}{\frac{2\pi}{T}} = \frac{T}{2\pi} \cdot \theta \]

### Step 7: Substitute for \( \theta \)
Substituting \( \theta = \cos^{-1}\left(\frac{4}{5}\right) \):
\[ t = \frac{T}{2\pi} \cdot \cos^{-1}\left(\frac{4}{5}\right) \]

### Final Answer
Thus, the time taken by the particle to travel a distance of \( \frac{A}{5} \) starting from rest is:
\[ t = \frac{T}{2\pi} \cdot \cos^{-1}\left(\frac{4}{5}\right) \]