A particle performs SHM on a straight line with time period `T` and amplitude`A`. The average speed of the particle between two successive instants, when potential energy and kinetic energy become same is

Options

A. `(A)/(T)`
B. `(4sqrt(2)A)/(T)`
C. `(2A)/(T)`
D. `(2sqrt(2)A)/(T)`

Explanation

To solve the problem, we need to find the average speed of a particle performing Simple Harmonic Motion (SHM) between two successive instants when its potential energy (PE) and kinetic energy (KE) are equal.

### Step-by-Step Solution:

1. **Understanding the Condition for PE and KE**:
In SHM, the total mechanical energy (E) is constant and is given by:
\[ E = KE + PE \]
When the potential energy equals the kinetic energy, we have:
\[ PE = KE \]
Therefore, we can express this as:
\[ PE = \frac{E}{2} \quad \text{and} \quad KE = \frac{E}{2} \]

2. **Expression for Potential Energy and Kinetic Energy**:
The potential energy at a displacement \( x \) from the mean position is given by:
\[ PE = \frac{1}{2} k x^2 \]
The kinetic energy is given by:
\[ KE = \frac{1}{2} k (A^2 - x^2) \]
Setting \( PE = KE \):
\[ \frac{1}{2} k x^2 = \frac{1}{2} k (A^2 - x^2) \]
Simplifying this, we get:
\[ x^2 = A^2 - x^2 \]
\[ 2x^2 = A^2 \quad \Rightarrow \quad x = \pm \frac{A}{\sqrt{2}} \]

3. **Finding the Time Taken to Reach These Points**:
The position of the particle in SHM can be described by:
\[ x(t) = A \cos(\omega t) \]
where \( \omega = \frac{2\pi}{T} \). Setting \( x = \frac{A}{\sqrt{2}} \):
\[ \frac{A}{\sqrt{2}} = A \cos(\omega t) \]
Dividing both sides by \( A \):
\[ \frac{1}{\sqrt{2}} = \cos(\omega t) \]
The angle whose cosine is \( \frac{1}{\sqrt{2}} \) is \( \frac{\pi}{4} \). Thus:
\[ \omega t = \frac{\pi}{4} \quad \Rightarrow \quad t = \frac{\pi}{4\omega} = \frac{\pi}{4 \cdot \frac{2\pi}{T}} = \frac{T}{8} \]

4. **Calculating the Average Speed**:
The average speed \( v_{avg} \) is defined as the total distance traveled divided by the total time taken. The total distance traveled between the two points \( x = \frac{A}{\sqrt{2}} \) and \( x = -\frac{A}{\sqrt{2}} \) is:
\[ \text{Distance} = \left(\frac{A}{\sqrt{2}} - (-\frac{A}{\sqrt{2}})\right) = \frac{A}{\sqrt{2}} + \frac{A}{\sqrt{2}} = \frac{2A}{\sqrt{2}} = A\sqrt{2} \]
The total time taken to travel this distance is:
\[ \Delta t = t_2 - t_1 = \frac{T}{8} - 0 = \frac{T}{8} \]
Therefore, the average speed is:
\[ v_{avg} = \frac{\text{Distance}}{\Delta t} = \frac{A\sqrt{2}}{\frac{T}{8}} = \frac{8A\sqrt{2}}{T} \]

### Final Answer:
The average speed of the particle between the two successive instants when the potential energy and kinetic energy become equal is:
\[ v_{avg} = \frac{8A\sqrt{2}}{T} \]