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 solve the problem of finding the minimum time interval for a particle performing Simple Harmonic Motion (SHM) to go from position A to position B, we can 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. This means:
\[ KE_A = PE_A \]

Since the total mechanical energy in SHM is constant, we can express this as:
\[ KE_A + PE_A = KE_{max} \]
Given that \( KE_A = PE_A \), we can write:
\[ 2KE_A = KE_{max} \]
Thus, we can express \( KE_A \) as:
\[ KE_A = \frac{KE_{max}}{2} \]

### Step 2: Relate kinetic energy to velocity
The kinetic energy is given by:
\[ KE = \frac{1}{2} m v^2 \]
At position A, we have:
\[ \frac{1}{2} m v_A^2 = \frac{KE_{max}}{2} \]
This leads to:
\[ v_A^2 = \frac{KE_{max}}{m} \]
Since \( KE_{max} = \frac{1}{2} m v_{max}^2 \), we can substitute:
\[ v_A^2 = \frac{v_{max}^2}{2} \]
Thus, we find:
\[ v_A = \frac{v_{max}}{\sqrt{2}} \]

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

### Step 4: Use the velocity equation for SHM
The velocity in SHM can be expressed as:
\[ v = \omega \sqrt{A^2 - x^2} \]
Where \( \omega = \frac{2\pi}{T} \) is the angular frequency, \( A \) is the amplitude, and \( x \) is the displacement from the mean position.

### Step 5: Calculate the displacement at positions A and B
1. For position A:
\[ v_A = \omega \sqrt{A^2 - x_A^2} \]
Substituting \( v_A = \frac{v_{max}}{\sqrt{2}} \):
\[ \frac{v_{max}}{\sqrt{2}} = \omega \sqrt{A^2 - x_A^2} \]
This gives:
\[ \sqrt{A^2 - x_A^2} = \frac{v_{max}}{\sqrt{2\omega}} \]
Squaring both sides:
\[ A^2 - x_A^2 = \frac{v_{max}^2}{2\omega^2} \]
Since \( \omega = \frac{v_{max}}{A} \):
\[ A^2 - x_A^2 = \frac{A^2}{2} \]
Thus:
\[ x_A = \frac{A}{\sqrt{2}} \]

2. For position B:
\[ v_B = \omega \sqrt{A^2 - x_B^2} \]
Substituting \( v_B = \frac{v_{max}}{2} \):
\[ \frac{v_{max}}{2} = \omega \sqrt{A^2 - x_B^2} \]
This gives:
\[ \sqrt{A^2 - x_B^2} = \frac{v_{max}}{2\omega} \]
Squaring both sides:
\[ A^2 - x_B^2 = \frac{v_{max}^2}{4\omega^2} \]
Substituting \( \omega = \frac{v_{max}}{A} \):
\[ A^2 - x_B^2 = \frac{A^2}{4} \]
Thus:
\[ x_B = \frac{\sqrt{3}A}{2} \]

### Step 6: Calculate the time intervals
Using the displacement equation \( x = A \sin(\omega t) \):
1. For position A:
\[ \sin(\omega t_A) = \frac{1}{\sqrt{2}} \Rightarrow \omega t_A = \frac{\pi}{4} \Rightarrow t_A = \frac{\pi}{4\omega} \]

2. For position B:
\[ \sin(\omega t_B) = \frac{\sqrt{3}}{2} \Rightarrow \omega t_B = \frac{\pi}{3} \Rightarrow t_B = \frac{\pi}{3\omega} \]

### Step 7: Find the time interval \( t_B - t_A \)
\[ t_B - t_A = \frac{\pi}{3\omega} - \frac{\pi}{4\omega} = \frac{\pi}{\omega} \left(\frac{4 - 3}{12}\right) = \frac{\pi}{12\omega} \]
Substituting \( \omega = \frac{2\pi}{T} \):
\[ t_B - t_A = \frac{\pi}{12 \cdot \frac{2\pi}{T}} = \frac{T}{24} \]

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