A particle is executing SHM and its velocity v is related to its position (x) as `v^(2) + ax^(2) =b`, where a and b are positive constant. The frequency of oscillation of particle is

Options

A. `(1)/(2pi) sqrt((b)/(a))`
B. `(sqrta)/(2pi)`
C. `(sqrtb)/(2pi)`
D. `(1)/(2pi) sqrt((a)/(b))`

Answer

To solve the problem, we need to analyze the given equation relating the velocity \( v \) to the position \( x \) of a particle executing simple harmonic motion (SHM). The equation provided is:

\[ v^2 + ax^2 = b \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express \( v^2 \) in terms of \( x \):

\[ v^2 = b - ax^2 \]

### Step 2: Expressing Velocity in Terms of Position
Taking the square root of both sides, we find:

\[ v = \sqrt{b - ax^2} \]

### Step 3: Relating Velocity to SHM
In SHM, the velocity \( v \) can also be expressed as:

\[ v = \omega \sqrt{A^2 - x^2} \]

where \( A \) is the amplitude of the motion and \( \omega \) is the angular frequency.

### Step 4: Equating the Two Expressions for Velocity
Now we can equate the two expressions for \( v \):

\[ \sqrt{b - ax^2} = \omega \sqrt{A^2 - x^2} \]

### Step 5: Finding the Maximum Velocity
To find the maximum velocity, we can set \( x = 0 \):

\[ v_{\text{max}} = \sqrt{b} \]

### Step 6: Finding the Angular Frequency
From the expression \( v^2 = b - ax^2 \), we can see that the maximum value of \( v^2 \) occurs when \( x = 0 \):

\[ v_{\text{max}}^2 = b \]

In SHM, the maximum velocity is also given by:

\[ v_{\text{max}} = \omega A \]

Equating the two expressions for maximum velocity, we have:

\[ \sqrt{b} = \omega A \]

### Step 7: Finding the Frequency
The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:

\[ \omega = 2\pi f \]

Substituting this into the equation gives:

\[ \sqrt{b} = (2\pi f) A \]

### Step 8: Solving for Frequency
Rearranging the equation to solve for frequency \( f \):

\[ f = \frac{\sqrt{b}}{2\pi A} \]

### Conclusion
The frequency of oscillation of the particle is given by:

\[ f = \frac{\sqrt{b}}{2\pi A} \]