A simple harmonic oscillation has an amplitude `A` and time period `T`. The time required to travel from `x = A` to ` x= (A)/(2)` is

Options

A. `(T)/(6)`
B. `(T)/(4)`
C. `(T)/(3)`
D. `(T)/(12)`

Explanation

To solve the problem of finding the time required for a simple harmonic oscillator to travel from \( x = A \) to \( x = \frac{A}{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Simple Harmonic Motion (SHM)**:
- In SHM, the position \( x \) of a particle at any time \( t \) can be described by the equation:
\[ x(t) = A \cos(\omega t + \phi) \]
- Here, \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant.

2. **Determine the Angular Frequency**:
- The angular frequency \( \omega \) is related to the time period \( T \) by:
\[ \omega = \frac{2\pi}{T} \]

3. **Initial and Final Positions**:
- We start at \( x = A \) and need to find the time taken to reach \( x = \frac{A}{2} \).

4. **Set Up the Equations**:
- For the initial position \( x = A \):
\[ A = A \cos(\omega t_1 + \phi) \implies \cos(\omega t_1 + \phi) = 1 \implies \omega t_1 + \phi = 0 \quad \text{(taking } t_1 = 0 \text{ for simplicity)} \]
- For the final position \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \cos(\omega t_2 + \phi) \implies \cos(\omega t_2 + \phi) = \frac{1}{2} \]
- This gives:
\[ \omega t_2 + \phi = \frac{\pi}{3} \quad \text{(since } \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\text{)} \]

5. **Calculate the Time Difference**:
- Since we set \( \phi = 0 \) and \( t_1 = 0 \):
\[ \omega t_2 = \frac{\pi}{3} \]
- Substituting for \( \omega \):
\[ \frac{2\pi}{T} t_2 = \frac{\pi}{3} \]
- Solving for \( t_2 \):
\[ t_2 = \frac{T}{6} \]

6. **Final Result**:
- The time required to travel from \( x = A \) to \( x = \frac{A}{2} \) is:
\[ t = \frac{T}{6} \]