A particle performs simple harmonic motion wit amplitude A. its speed is double at the instant when it is at distance `(A)/(3)` from equilibrium position. The new amplitude of the motion is

Options

A. `sqrt(11)A`
B. `(sqrt(22)A)/(3)`
C. `(sqrt(33)A)/(2)`
D. none of these

Explanation

To solve the problem, we need to find the new amplitude of a particle performing simple harmonic motion (SHM) given that its speed is double when it is at a distance of \( \frac{A}{3} \) from the equilibrium position.

### Step-by-step Solution:

1. **Understanding the SHM Velocity Formula**:
The velocity \( V \) of a particle in SHM at a distance \( y \) from the equilibrium position is given by:
\[ V = \omega \sqrt{A^2 - y^2} \]
where \( A \) is the amplitude and \( \omega \) is the angular frequency.

2. **Calculate Initial Velocity at \( y = \frac{A}{3} \)**:
Substitute \( y = \frac{A}{3} \) into the velocity formula:
\[ V = \omega \sqrt{A^2 - \left(\frac{A}{3}\right)^2} \]
Simplifying this:
\[ V = \omega \sqrt{A^2 - \frac{A^2}{9}} = \omega \sqrt{\frac{9A^2 - A^2}{9}} = \omega \sqrt{\frac{8A^2}{9}} = \frac{2\sqrt{2}}{3} A \omega \]

3. **Determine the New Velocity**:
According to the problem, the new speed \( V' \) is double the initial speed:
\[ V' = 2V = 2 \left(\frac{2\sqrt{2}}{3} A \omega\right) = \frac{4\sqrt{2}}{3} A \omega \]

4. **Set Up the Equation for New Amplitude**:
The new velocity \( V' \) can also be expressed using the new amplitude \( A' \):
\[ V' = \omega \sqrt{A'^2 - \left(\frac{A}{3}\right)^2} \]
Squaring both sides gives:
\[ V'^2 = \omega^2 \left(A'^2 - \left(\frac{A}{3}\right)^2\right) \]

5. **Substituting Values**:
Substitute \( V' = \frac{4\sqrt{2}}{3} A \omega \):
\[ \left(\frac{4\sqrt{2}}{3} A \omega\right)^2 = \omega^2 \left(A'^2 - \left(\frac{A}{3}\right)^2\right) \]
This simplifies to:
\[ \frac{32}{9} A^2 \omega^2 = \omega^2 \left(A'^2 - \frac{A^2}{9}\right) \]

6. **Canceling \( \omega^2 \)**:
Dividing both sides by \( \omega^2 \):
\[ \frac{32}{9} A^2 = A'^2 - \frac{A^2}{9} \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ A'^2 = \frac{32}{9} A^2 + \frac{A^2}{9} = \frac{32A^2 + A^2}{9} = \frac{33A^2}{9} \]

8. **Finding the New Amplitude**:
Taking the square root:
\[ A' = \sqrt{\frac{33}{9}} A = \frac{\sqrt{33}}{3} A \]

### Final Answer:
The new amplitude of the motion is:
\[ A' = \frac{\sqrt{33}}{3} A \]