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 find the frequency of oscillation of a particle executing simple harmonic motion (SHM) given the relationship between its velocity \( v \) and position \( x \) as:

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

where \( a \) and \( b \) are positive constants.

### Step-by-Step Solution:

1. **Rearranging the Equation**:
Start with the given equation:
\[ v^2 = b - ax^2 \]

2. **Expressing Velocity**:
Taking the square root of both sides gives:
\[ v = \sqrt{b - ax^2} \]

3. **Identifying the Form of SHM**:
In SHM, the velocity \( v \) can also be expressed in terms of angular frequency \( \omega \) and displacement \( x \):
\[ v = \omega \sqrt{A^2 - x^2} \]
where \( A \) is the amplitude of the motion.

4. **Comparing the Two Expressions**:
From our expression for \( v \), we can compare:
\[ \sqrt{b - ax^2} = \omega \sqrt{A^2 - x^2} \]

5. **Identifying Amplitude**:
To find the amplitude \( A \), we can rewrite the expression \( b - ax^2 \) in a form that resembles the SHM equation:
- Set \( A^2 = \frac{b}{a} \) (since when \( x = 0 \), \( v = \sqrt{b} \)).
- Thus, the amplitude \( A \) is:
\[ A = \sqrt{\frac{b}{a}} \]

6. **Finding Angular Frequency**:
From the comparison, we can also identify \( \omega \):
\[ \omega = \sqrt{a} \]

7. **Calculating Frequency**:
The frequency \( f \) is related to angular frequency \( \omega \) by:
\[ \omega = 2\pi f \]
Therefore, we can express frequency as:
\[ f = \frac{\omega}{2\pi} = \frac{\sqrt{a}}{2\pi} \]

### Final Result:
The frequency of oscillation of the particle is:
\[ f = \frac{\sqrt{a}}{2\pi} \]