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` string 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 performing Simple Harmonic Motion (SHM) to travel a distance of \( \frac{A}{5} \) from rest, we can follow these steps:

### Step 1: Understand the SHM Equation
The position of a particle in SHM can be described by the equation:
\[ x(t) = A \cos(\omega t) \]
where:
- \( x(t) \) is the position at time \( t \),
- \( A \) is the amplitude,
- \( \omega \) is the angular frequency.

### Step 2: Define the Initial Condition
Since the particle starts from rest at the maximum amplitude, at \( t = 0 \):
\[ x(0) = A \cos(0) = A \]

### Step 3: Set Up the Equation for the Given Distance
We need to find the time taken to travel a distance of \( \frac{A}{5} \). Thus, the new position \( x \) can be expressed as:
\[ x = A - \frac{A}{5} = \frac{4A}{5} \]

### Step 4: Substitute into the SHM Equation
Now, we can set up the equation:
\[ \frac{4A}{5} = A \cos(\omega t) \]

### Step 5: Simplify the Equation
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{4}{5} = \cos(\omega t) \]

### Step 6: Solve for \( \omega t \)
Taking the inverse cosine:
\[ \omega t = \cos^{-1}\left(\frac{4}{5}\right) \]

### Step 7: Relate \( \omega \) to the Time Period \( T \)
The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

### Step 8: Substitute \( \omega \) into the Equation
Thus, we can express \( t \) as:
\[ t = \frac{T}{2\pi} \cos^{-1}\left(\frac{4}{5}\right) \]

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