A particle performs `SHM` of amplitude `A` along a straight line .When it is at a distance of `sqrt(3)/(2)`A from mean position its kinetic energy gets increased by an amount of `(1)/(2) m omega^(2)A^(2)` due to an impulsive force. Then its new amplitude becomes

Options

A. `sqrt(5)/2`A
B. `sqrt(3)/2`A
C. `sqrt(2)`A
D. `2/sqrt(5)`A

Answer

To solve the problem step by step, we will analyze the situation of the particle performing Simple Harmonic Motion (SHM) and how the kinetic energy changes with the application of an impulsive force.

### Step 1: Understand the Initial Conditions
The particle is performing SHM with an amplitude \( A \). When the particle is at a distance \( x = \frac{\sqrt{3}}{2} A \) from the mean position, we need to find the initial kinetic energy and potential energy at this position.

### Step 2: Calculate Initial Kinetic Energy and Potential Energy
The total mechanical energy \( E \) in SHM is given by:
\[ E = \text{Kinetic Energy} + \text{Potential Energy} \]

The potential energy \( U \) at position \( x \) is given by:
\[ U = \frac{1}{2} k x^2 \]
where \( k = m \omega^2 \).

At \( x = \frac{\sqrt{3}}{2} A \):
\[ U = \frac{1}{2} k \left(\frac{\sqrt{3}}{2} A\right)^2 = \frac{1}{2} m \omega^2 \left(\frac{3}{4} A^2\right) = \frac{3}{8} m \omega^2 A^2 \]

The total energy \( E \) is also given by:
\[ E = \frac{1}{2} k A^2 = \frac{1}{2} m \omega^2 A^2 \]

Thus, the initial kinetic energy \( K \) when the particle is at \( x = \frac{\sqrt{3}}{2} A \) is:
\[ K = E - U = \frac{1}{2} m \omega^2 A^2 - \frac{3}{8} m \omega^2 A^2 = \frac{1}{8} m \omega^2 A^2 \]

### Step 3: Change in Kinetic Energy
According to the problem, the kinetic energy increases by an amount of \( \frac{1}{2} m \omega^2 A^2 \). Therefore, the new kinetic energy \( K' \) becomes:
\[ K' = K + \frac{1}{2} m \omega^2 A^2 = \frac{1}{8} m \omega^2 A^2 + \frac{1}{2} m \omega^2 A^2 = \frac{1}{8} m \omega^2 A^2 + \frac{4}{8} m \omega^2 A^2 = \frac{5}{8} m \omega^2 A^2 \]

### Step 4: Calculate the New Total Energy
The potential energy \( U \) at position \( x = \frac{\sqrt{3}}{2} A \) remains unchanged:
\[ U = \frac{3}{8} m \omega^2 A^2 \]

The new total mechanical energy \( E' \) is:
\[ E' = K' + U = \frac{5}{8} m \omega^2 A^2 + \frac{3}{8} m \omega^2 A^2 = 1 m \omega^2 A^2 \]

### Step 5: Relate Total Energy to New Amplitude
The total mechanical energy in SHM is also given by:
\[ E' = \frac{1}{2} k A'^2 = \frac{1}{2} m \omega^2 A'^2 \]

Setting the two expressions for total energy equal gives:
\[ 1 m \omega^2 A^2 = \frac{1}{2} m \omega^2 A'^2 \]

### Step 6: Solve for the New Amplitude \( A' \)
Cancelling \( m \omega^2 \) from both sides:
\[ 1 = \frac{1}{2} A'^2 \implies A'^2 = 2 \implies A' = \sqrt{2} A \]

### Conclusion
Thus, the new amplitude of the particle after the impulsive force is applied is:
\[ A' = \sqrt{2} A \]