Consider a simple harmonic motion of time period T. Calculate the time taken for the displacement of change value from half the amplitude to the amplitude.

Explanation

To solve the problem of finding the time taken for a particle in simple harmonic motion (SHM) to move from half the amplitude (A/2) to the amplitude (A), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the SHM Equation**:
The displacement \( x \) of a particle in SHM can be described by the equation:
\[ x = A \sin(\omega t) \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Identify the Time Period**:
The time period \( T \) of the SHM is related to the angular frequency \( \omega \) by the formula:
\[ T = \frac{2\pi}{\omega} \]

3. **Find the Time at \( x = \frac{A}{2} \)**:
Set \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t_1) \]
Dividing both sides by \( A \) (assuming \( A \neq 0 \)):
\[ \frac{1}{2} = \sin(\omega t_1) \]
The sine function equals \( \frac{1}{2} \) at \( \omega t_1 = \frac{\pi}{6} \) (30 degrees). Thus:
\[ t_1 = \frac{\pi}{6\omega} \]

4. **Substituting for \( \omega \)**:
Substitute \( \omega = \frac{2\pi}{T} \):
\[ t_1 = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12} \]

5. **Find the Time at \( x = A \)**:
Set \( x = A \):
\[ A = A \sin(\omega t_2) \]
Dividing both sides by \( A \):
\[ 1 = \sin(\omega t_2) \]
The sine function equals 1 at \( \omega t_2 = \frac{\pi}{2} \) (90 degrees). Thus:
\[ t_2 = \frac{\pi}{2\omega} \]

6. **Substituting for \( \omega \)**:
Substitute \( \omega = \frac{2\pi}{T} \):
\[ t_2 = \frac{\pi}{2} \cdot \frac{T}{2\pi} = \frac{T}{4} \]

7. **Calculate the Time Difference**:
The time taken to go from \( A/2 \) to \( A \) is given by:
\[ \Delta t = t_2 - t_1 \]
Substituting the values:
\[ \Delta t = \frac{T}{4} - \frac{T}{12} \]
To subtract these fractions, find a common denominator (which is 12):
\[ \Delta t = \frac{3T}{12} - \frac{T}{12} = \frac{2T}{12} = \frac{T}{6} \]

### Final Answer:
The time taken for the displacement to change from half the amplitude to the amplitude is:
\[ \Delta t = \frac{T}{6} \]