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 find the average speed of a particle performing Simple Harmonic Motion (SHM) between two successive instants when its potential energy and kinetic energy are equal, we can follow these steps:

### Step 1: Understand the Conditions
In SHM, the total mechanical energy (E) is constant and is given by:
\[ E = \frac{1}{2} k A^2 \]
where \( k \) is the spring constant and \( A \) is the amplitude. The potential energy (PE) and kinetic energy (KE) at any position \( x \) can be expressed as:
\[ PE = \frac{1}{2} k x^2 \]
\[ KE = E - PE \]

### Step 2: Set the Condition for Equal Energies
We need to find the position \( x \) where the potential energy equals the kinetic energy:
\[ PE = KE \]
Thus,
\[ \frac{1}{2} k x^2 = E - \frac{1}{2} k x^2 \]
This simplifies to:
\[ \frac{1}{2} k x^2 = \frac{1}{2} k A^2 - \frac{1}{2} k x^2 \]
Combining terms gives:
\[ k x^2 = \frac{1}{2} k A^2 \]
Cancelling \( k \) (assuming \( k \neq 0 \)):
\[ x^2 = \frac{1}{2} A^2 \]
Taking the square root:
\[ x = \pm \frac{A}{\sqrt{2}} \]

### Step 3: Calculate the Distance Between Two Positions
The two positions where the potential energy equals the kinetic energy are \( x = \frac{A}{\sqrt{2}} \) and \( x = -\frac{A}{\sqrt{2}} \). The distance between these two points is:
\[ \text{Distance} = \left( \frac{A}{\sqrt{2}} - \left(-\frac{A}{\sqrt{2}}\right) \right) = 2 \cdot \frac{A}{\sqrt{2}} = \frac{2A}{\sqrt{2}} = A\sqrt{2} \]

### Step 4: Determine the Time Taken
To find the time taken to move from \( x = -\frac{A}{\sqrt{2}} \) to \( x = \frac{A}{\sqrt{2}} \), we need to find the time at these positions. The angular frequency \( \omega \) is given by:
\[ \omega = \frac{2\pi}{T} \]
At \( x = \frac{A}{\sqrt{2}} \):
\[ \frac{A}{\sqrt{2}} = A \cos(\omega t) \]
Thus,
\[ \cos(\omega t) = \frac{1}{\sqrt{2}} \]
This gives:
\[ \omega t = \frac{\pi}{4} \]
So,
\[ t = \frac{\pi}{4\omega} = \frac{\pi}{4 \cdot \frac{2\pi}{T}} = \frac{T}{8} \]

The time taken to move from \( -\frac{A}{\sqrt{2}} \) to \( \frac{A}{\sqrt{2}} \) is:
\[ \text{Total time} = 2 \cdot \frac{T}{8} = \frac{T}{4} \]

### Step 5: Calculate the Average Speed
The average speed \( v_{avg} \) is given by:
\[ v_{avg} = \frac{\text{Total Distance}}{\text{Total Time}} \]
Substituting the values:
\[ v_{avg} = \frac{A\sqrt{2}}{\frac{T}{4}} = \frac{4A\sqrt{2}}{T} \]

### Final Answer
The average speed of the particle between two successive instants when the potential energy and kinetic energy are equal is:
\[ v_{avg} = \frac{4A\sqrt{2}}{T} \]

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