A particle performs `SHM` of time period `T` , along a straight line. Find the minimum time interval to go from position `A` to position `B`. At `A` both potential energy and kinetic energy are same and at `B` the speed is half of the maximum speed.

Explanation

To find the minimum time interval for a particle performing Simple Harmonic Motion (SHM) to go from position A to position B, we will follow these steps:

### Step 1: Understand the conditions at positions A and B
At position A, the potential energy (PE) and kinetic energy (KE) are equal. Therefore, we can express this as:
\[ KE_A = PE_A \]
The total mechanical energy (E) in SHM is given by:
\[ E = KE + PE = KE_{max} \]
Since \( KE_A = PE_A \), we can write:
\[ KE_A + KE_A = KE_{max} \]
This implies:
\[ 2 KE_A = KE_{max} \]
Thus, we can find the kinetic energy at position A:
\[ KE_A = \frac{KE_{max}}{2} \]

### Step 2: Determine the velocity at position A
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
At position A:
\[ \frac{1}{2} m v_A^2 = \frac{1}{2} \cdot KE_{max} \]
From this, we can derive:
\[ v_A^2 = \frac{KE_{max}}{m} \]
Since \( KE_{max} = \frac{1}{2} m V_{max}^2 \), we have:
\[ v_A^2 = \frac{1}{2} V_{max}^2 \]
Thus:
\[ v_A = \frac{V_{max}}{\sqrt{2}} \]

### Step 3: Determine the velocity at position B
At position B, the speed is given as half of the maximum speed:
\[ v_B = \frac{V_{max}}{2} \]

### Step 4: Use the SHM velocity equation
The velocity in SHM is expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
Where \( \omega = \frac{2\pi}{T} \) and \( A \) is the amplitude.

For position A:
\[ v_A = \omega \sqrt{A^2 - x_A^2} \]
Substituting \( v_A \):
\[ \frac{V_{max}}{\sqrt{2}} = \omega \sqrt{A^2 - x_A^2} \]

For position B:
\[ v_B = \omega \sqrt{A^2 - x_B^2} \]
Substituting \( v_B \):
\[ \frac{V_{max}}{2} = \omega \sqrt{A^2 - x_B^2} \]

### Step 5: Solve for positions A and B
From the equations above, we can express \( x_A \) and \( x_B \):
1. For position A:
\[ \frac{V_{max}}{\sqrt{2}} = \frac{2\pi}{T} \sqrt{A^2 - x_A^2} \]
Rearranging gives:
\[ A^2 - x_A^2 = \frac{V_{max}^2 T^2}{8\pi^2} \]
Thus:
\[ x_A = A \cdot \frac{1}{\sqrt{2}} \]

2. For position B:
\[ \frac{V_{max}}{2} = \frac{2\pi}{T} \sqrt{A^2 - x_B^2} \]
Rearranging gives:
\[ A^2 - x_B^2 = \frac{V_{max}^2 T^2}{8\pi^2} \]
Thus:
\[ x_B = A \cdot \frac{\sqrt{3}}{2} \]

### Step 6: Calculate the time interval
The time taken to go from position A to position B can be calculated using the angles corresponding to the positions:
- For position A:
\[ \sin(\omega t_A) = \frac{1}{\sqrt{2}} \Rightarrow \omega t_A = \frac{\pi}{4} \]
- For position B:
\[ \sin(\omega t_B) = \frac{\sqrt{3}}{2} \Rightarrow \omega t_B = \frac{\pi}{3} \]

The time interval \( \Delta t \) is:
\[ \Delta t = t_B - t_A = \frac{\pi}{3} - \frac{\pi}{4} = \frac{4\pi - 3\pi}{12} = \frac{\pi}{12} \]

Finally, substituting \( \omega = \frac{2\pi}{T} \):
\[ \Delta t = \frac{\pi}{12} \cdot \frac{T}{2\pi} = \frac{T}{24} \]

### Final Answer
The minimum time interval to go from position A to position B is:
\[ \Delta t = \frac{T}{24} \]