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

Options

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

Explanation

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

### Step 1: Understand the SHM Parameters
The particle is performing SHM with:
- Amplitude \( A \)
- Time period \( T \)

### Step 2: Identify the Position in SHM
When the particle is at a distance of \( \frac{A}{5} \) from the mean position (which is at \( x = 0 \)), we can denote this position as:
\[ x = \frac{A}{5} \]

### Step 3: Use the Cosine Relation
In SHM, the position \( x \) can be expressed as:
\[ x = A \cos(\omega t) \]
where \( \omega \) is the angular frequency given by:
\[ \omega = \frac{2\pi}{T} \]

### Step 4: Set Up the Equation
From the position equation, we can set up the following:
\[ \frac{A}{5} = A \cos(\omega t) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{5} = \cos(\omega t) \]

### Step 5: Solve for \( \omega t \)
Taking the inverse cosine:
\[ \omega t = \cos^{-1}\left(\frac{1}{5}\right) \]

### Step 6: Substitute for \( \omega \)
Substituting \( \omega \) into the equation:
\[ \frac{2\pi}{T} t = \cos^{-1}\left(\frac{1}{5}\right) \]

### Step 7: Solve for \( t \)
Now, solving for \( t \):
\[ t = \frac{T}{2\pi} \cos^{-1}\left(\frac{1}{5}\right) \]

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

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