A particle is in linear SHM of amplitude A and time period T. If v refers to its average speed during any interval of T/3 , then the maximum possible value of v is

Options

A. `(3sqrt(3))/(T)A`
B. `(sqrt(3)A)/(T)`
C. `(2sqrt(3))/(T)A`
D. `(3A)/(T)`

Explanation

To find the maximum possible average speed \( v \) of a particle in linear Simple Harmonic Motion (SHM) during any interval of \( \frac{T}{3} \), we can follow these steps:

### Step 1: Understand the Motion
In SHM, the particle moves back and forth between two extreme points with an amplitude \( A \) and a time period \( T \). The maximum displacement from the mean position is \( A \).

### Step 2: Determine the Angle for \( \frac{T}{3} \)
Since the time period \( T \) corresponds to a full cycle of \( 360^\circ \), the interval \( \frac{T}{3} \) corresponds to an angle of:
\[ \text{Angle} = \frac{360^\circ}{3} = 120^\circ \]

### Step 3: Calculate the Displacement
In SHM, the displacement \( x \) at any angle \( \theta \) can be expressed as:
\[ x = A \sin(\theta) \]
For \( \theta = 120^\circ \):
\[ x = A \sin(120^\circ) = A \cdot \frac{\sqrt{3}}{2} \]

### Step 4: Calculate the Distance Covered
The particle moves from one extreme position to another in the interval of \( \frac{T}{3} \). The maximum distance covered in this interval can be calculated as:
\[ \text{Distance} = A + A \sin(120^\circ) = A + A \cdot \frac{\sqrt{3}}{2} = A \left(1 + \frac{\sqrt{3}}{2}\right) \]

### Step 5: Calculate the Average Speed
The average speed \( v \) is defined as the total distance covered divided by the time taken. The time taken for the interval is \( \frac{T}{3} \):
\[ v = \frac{\text{Distance}}{\text{Time}} = \frac{A \left(1 + \frac{\sqrt{3}}{2}\right)}{\frac{T}{3}} = \frac{3A \left(1 + \frac{\sqrt{3}}{2}\right)}{T} \]

### Step 6: Simplify the Expression
To find the maximum possible average speed, we can simplify the expression:
\[ v = \frac{3A \left(1 + \frac{\sqrt{3}}{2}\right)}{T} = \frac{3A \left(\frac{2 + \sqrt{3}}{2}\right)}{T} = \frac{3A(2 + \sqrt{3})}{2T} \]

### Conclusion
Thus, the maximum possible value of the average speed \( v \) during the interval of \( \frac{T}{3} \) is:
\[ v = \frac{3A(2 + \sqrt{3})}{2T} \]