A particle performs simple harmonic motion with 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 `(sqrt(33)A)/(beta)`. Find `beta`

Explanation

To solve the problem step-by-step, we will use the principles of simple harmonic motion (SHM) and the given conditions.

### Step 1: Understand the velocity in SHM
In simple harmonic motion, the velocity \( v \) of a particle at a distance \( x \) from the equilibrium position is given by the formula:
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( \omega \) is the angular frequency and \( A \) is the amplitude.

### Step 2: Substitute the given values
We are given that the particle is at a distance \( x = \frac{A}{3} \) from the equilibrium position. We can substitute this value into the velocity equation:
\[ v = \omega \sqrt{A^2 - \left(\frac{A}{3}\right)^2} \]
Calculating \( \left(\frac{A}{3}\right)^2 \):
\[ \left(\frac{A}{3}\right)^2 = \frac{A^2}{9} \]
Thus, we have:
\[ v = \omega \sqrt{A^2 - \frac{A^2}{9}} = \omega \sqrt{\frac{9A^2}{9} - \frac{A^2}{9}} = \omega \sqrt{\frac{8A^2}{9}} = \omega \frac{2\sqrt{2}A}{3} \]

### Step 3: Determine the new speed
According to the problem, the new speed \( v' \) is double the original speed \( v \):
\[ v' = 2v = 2 \left(\omega \frac{2\sqrt{2}A}{3}\right) = \frac{4\sqrt{2}A\omega}{3} \]

### Step 4: Relate new speed to new amplitude
The new amplitude is denoted as \( A' \). The velocity at the same distance \( x = \frac{A}{3} \) for the new amplitude can also be expressed as:
\[ v' = \omega \sqrt{A'^2 - \left(\frac{A}{3}\right)^2} \]
Substituting \( v' \):
\[ \frac{4\sqrt{2}A\omega}{3} = \omega \sqrt{A'^2 - \frac{A^2}{9}} \]
Dividing both sides by \( \omega \):
\[ \frac{4\sqrt{2}A}{3} = \sqrt{A'^2 - \frac{A^2}{9}} \]

### Step 5: Square both sides
Squaring both sides gives:
\[ \left(\frac{4\sqrt{2}A}{3}\right)^2 = A'^2 - \frac{A^2}{9} \]
Calculating the left side:
\[ \frac{32A^2}{9} = A'^2 - \frac{A^2}{9} \]

### Step 6: Rearranging the equation
Now, rearranging the equation gives:
\[ A'^2 = \frac{32A^2}{9} + \frac{A^2}{9} = \frac{33A^2}{9} \]

### Step 7: Solve for \( A' \)
Taking the square root of both sides:
\[ A' = \sqrt{\frac{33A^2}{9}} = \frac{\sqrt{33}A}{3} \]

### Step 8: Find \( \beta \)
According to the problem, the new amplitude is given as:
\[ A' = \frac{\sqrt{33}A}{\beta} \]
Comparing both expressions for \( A' \):
\[ \frac{\sqrt{33}A}{3} = \frac{\sqrt{33}A}{\beta} \]
Thus, we find:
\[ \beta = 3 \]

### Final Answer
The value of \( \beta \) is:
\[ \boxed{3} \]