A particle of mass `m` moving along x-axis has a potential energy `U(x)=a+bx^2` where a and b are positive constant. It will execute simple harmonic motion with a frequency determined by the value of

Options

A. b and m alone
B. b, a and m alone
C. b alone
D. b and a alone

Answer

To solve the problem, we need to determine the frequency of the simple harmonic motion (SHM) for a particle of mass \( m \) moving along the x-axis with a given potential energy \( U(x) = a + b x^2 \), where \( a \) and \( b \) are positive constants.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function:**
The potential energy is given as:
\[ U(x) = a + b x^2 \]

2. **Calculate the Force Acting on the Particle:**
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dU}{dx} \]
First, we differentiate \( U(x) \) with respect to \( x \):
\[ \frac{dU}{dx} = \frac{d}{dx}(a + b x^2) = 0 + 2b x = 2b x \]
Therefore, the force is:
\[ F = -2b x \]

3. **Relate Force to Acceleration:**
According to Newton's second law, the force can also be expressed as:
\[ F = m a \]
where \( a \) is the acceleration of the particle. Since \( a = \frac{d^2x}{dt^2} \), we can write:
\[ m \frac{d^2x}{dt^2} = -2b x \]

4. **Formulate the Equation of Motion:**
Rearranging the equation gives:
\[ \frac{d^2x}{dt^2} + \frac{2b}{m} x = 0 \]
This is the standard form of the equation for simple harmonic motion, which can be written as:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
where \( \omega^2 = \frac{2b}{m} \).

5. **Determine the Angular Frequency:**
From the comparison, we find:
\[ \omega = \sqrt{\frac{2b}{m}} \]

6. **Find the Frequency of SHM:**
The frequency \( f \) of the simple harmonic motion is related to the angular frequency \( \omega \) by the formula:
\[ f = \frac{\omega}{2\pi} \]
Substituting for \( \omega \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{2b}{m}} \]

### Conclusion:
The frequency of the simple harmonic motion is determined by the values of \( b \) and \( m \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{2b}{m}} \]