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 calculating the time taken for the displacement to change from half the amplitude to the amplitude in simple harmonic motion (SHM), we can follow these steps:

### Step 1: Understand the parameters of SHM
In SHM, the displacement \( x \) can be expressed as:
\[ x(t) = A \sin(\omega t + \phi) \]
where:
- \( A \) is the amplitude,
- \( \omega = \frac{2\pi}{T} \) is the angular frequency,
- \( T \) is the time period,
- \( \phi \) is the phase constant.

### Step 2: Define the initial and final displacements
We need to find the time taken for the displacement to change from \( \frac{A}{2} \) (half the amplitude) to \( A \) (the full amplitude).

### Step 3: Set up the equations for the displacements
1. For \( x = \frac{A}{2} \):
\[ \frac{A}{2} = A \sin(\omega t_1 + \phi) \]
Simplifying gives:
\[ \sin(\omega t_1 + \phi) = \frac{1}{2} \]

2. For \( x = A \):
\[ A = A \sin(\omega t_2 + \phi) \]
Simplifying gives:
\[ \sin(\omega t_2 + \phi) = 1 \]

### Step 4: Solve for \( t_1 \) and \( t_2 \)
1. From \( \sin(\omega t_1 + \phi) = \frac{1}{2} \):
The general solutions for this equation are:
\[ \omega t_1 + \phi = \frac{\pi}{6} + 2n\pi \quad \text{or} \quad \omega t_1 + \phi = \frac{5\pi}{6} + 2n\pi \]
Solving for \( t_1 \):
\[ t_1 = \frac{1}{\omega} \left(\frac{\pi}{6} - \phi + 2n\pi\right) \quad \text{or} \quad t_1 = \frac{1}{\omega} \left(\frac{5\pi}{6} - \phi + 2n\pi\right) \]

2. From \( \sin(\omega t_2 + \phi) = 1 \):
The solution is:
\[ \omega t_2 + \phi = \frac{\pi}{2} + 2m\pi \]
Solving for \( t_2 \):
\[ t_2 = \frac{1}{\omega} \left(\frac{\pi}{2} - \phi + 2m\pi\right) \]

### Step 5: Calculate the time difference \( \Delta t \)
The time taken to go from \( \frac{A}{2} \) to \( A \) is:
\[ \Delta t = t_2 - t_1 \]

Substituting the values of \( t_1 \) and \( t_2 \):
\[ \Delta t = \frac{1}{\omega} \left(\frac{\pi}{2} - \phi + 2m\pi - \left(\frac{\pi}{6} - \phi + 2n\pi\right)\right) \]
This simplifies to:
\[ \Delta t = \frac{1}{\omega} \left(\frac{\pi}{2} - \frac{\pi}{6} + 2(m-n)\pi\right) \]
Calculating \( \frac{\pi}{2} - \frac{\pi}{6} \):
\[ \frac{\pi}{2} - \frac{\pi}{6} = \frac{3\pi}{6} - \frac{\pi}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \]
Thus:
\[ \Delta t = \frac{1}{\omega} \left(\frac{\pi}{3} + 2(m-n)\pi\right) \]

### Step 6: Substitute \( \omega \)
Since \( \omega = \frac{2\pi}{T} \):
\[ \Delta t = \frac{T}{2\pi} \left(\frac{\pi}{3} + 2(m-n)\pi\right) \]
This results in:
\[ \Delta t = \frac{T}{6} + \frac{T}{\pi}(m-n) \]

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