A particle performs simple harmonic motion with amplitude A. Its speed is tripled at the instant that it is at a distance
`(2A)/3` from equilibrium position. The new amplitude of the motion is:

Options

A. 3A
B. `A sqrt(3)`
C. `(7A)/(3)`
D. `(A)/(3) sqrt(41)`

Answer

To solve the problem, we need to find the new amplitude of a particle performing simple harmonic motion (SHM) when its speed is tripled at a distance of \( \frac{2A}{3} \) from the equilibrium position.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The particle performs SHM with an amplitude \( A \).
- At a distance \( x = \frac{2A}{3} \), the speed of the particle is tripled.

2. **Determine the Velocity at \( x = \frac{2A}{3} \)**:
- The velocity \( v \) in SHM is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
- Substituting \( x = \frac{2A}{3} \):
\[ v = \omega \sqrt{A^2 - \left(\frac{2A}{3}\right)^2} = \omega \sqrt{A^2 - \frac{4A^2}{9}} = \omega \sqrt{\frac{5A^2}{9}} = \frac{\omega A \sqrt{5}}{3} \]

3. **Calculate the New Velocity**:
- The new velocity \( v' \) when the speed is tripled:
\[ v' = 3v = 3 \left(\frac{\omega A \sqrt{5}}{3}\right) = \omega A \sqrt{5} \]

4. **Conservation of Mechanical Energy**:
- The total mechanical energy in SHM is given by:
\[ E = \frac{1}{2} k A^2 \]
- At the position \( x = \frac{2A}{3} \), the total mechanical energy is the sum of kinetic and potential energy:
\[ E = \frac{1}{2} m v'^2 + \frac{1}{2} k x^2 \]
- Substituting \( v' \) and \( x \):
\[ E = \frac{1}{2} m (\omega A \sqrt{5})^2 + \frac{1}{2} k \left(\frac{2A}{3}\right)^2 \]
- This simplifies to:
\[ E = \frac{1}{2} m \cdot 5 \omega^2 A^2 + \frac{1}{2} k \cdot \frac{4A^2}{9} \]

5. **Expressing \( k \) in terms of \( \omega \)**:
- We know \( k = m \omega^2 \), so substituting \( k \):
\[ E = \frac{5}{2} m \omega^2 A^2 + \frac{1}{2} m \omega^2 \cdot \frac{4A^2}{9} \]
- Factor out \( \frac{1}{2} m \omega^2 \):
\[ E = \frac{1}{2} m \omega^2 \left(5A^2 + \frac{4A^2}{9}\right) \]

6. **Finding the New Amplitude \( A' \)**:
- The new total energy can also be expressed as:
\[ E = \frac{1}{2} k A'^2 \]
- Setting the two expressions for energy equal gives:
\[ \frac{1}{2} m \omega^2 \left(5A^2 + \frac{4A^2}{9}\right) = \frac{1}{2} k A'^2 \]
- Substituting \( k = m \omega^2 \):
\[ \frac{1}{2} m \omega^2 \left(5A^2 + \frac{4A^2}{9}\right) = \frac{1}{2} m \omega^2 A'^2 \]
- Canceling \( \frac{1}{2} m \omega^2 \) gives:
\[ 5A^2 + \frac{4A^2}{9} = A'^2 \]
- Finding a common denominator (9):
\[ \frac{45A^2}{9} + \frac{4A^2}{9} = A'^2 \implies \frac{49A^2}{9} = A'^2 \]
- Taking the square root:
\[ A' = \frac{7A}{3} \]

### Final Answer:
The new amplitude of the motion is \( \frac{7A}{3} \).