A partical of mass m is moving along the x-axis under the potential `V (x) =(kx^2)/2+lamda` Where k and x are positive constants of appropriate dimensions . The particle is slightly displaced from its equilibrium position . The particle oscillates with the the angular frequency `(omega)` given by

Options

A. `3 k/m`
B. `2m/k`
C. `sqrt(k/m)`
D. `sqrt(3m/k)`

Answer

To solve the problem, we need to find the angular frequency (ω) of a particle of mass \( m \) moving under the potential \( V(x) = \frac{1}{2} kx^2 + \lambda \). Here are the steps to derive the angular frequency:

### Step 1: Identify the Potential Energy Function
The potential energy of the particle is given by:
\[ V(x) = \frac{1}{2} kx^2 + \lambda \]
where \( k \) is a positive constant and \( \lambda \) is a constant that does not affect the motion of the particle.

### Step 2: Determine the Equilibrium Position
To find the equilibrium position, we need to find the point where the force acting on the particle is zero. The force \( F \) is related to the potential energy by:
\[ F = -\frac{dV}{dx} \]
Calculating the derivative of \( V(x) \):
\[ \frac{dV}{dx} = \frac{d}{dx}\left(\frac{1}{2} kx^2 + \lambda\right) = kx \]
Thus, the force becomes:
\[ F = -kx \]
Setting \( F = 0 \) gives us the equilibrium position at \( x = 0 \).

### Step 3: Analyze Small Displacements from Equilibrium
When the particle is slightly displaced from the equilibrium position (let's say \( x \) is small), we can use the linear approximation of the force. The equation of motion can be expressed as:
\[ m \frac{d^2x}{dt^2} = -kx \]
This is a second-order linear differential equation.

### Step 4: Identify the Angular Frequency
The standard form of the equation for simple harmonic motion is:
\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]
Comparing this with our equation:
\[ \frac{d^2x}{dt^2} + \frac{k}{m} x = 0 \]
we identify:
\[ \omega^2 = \frac{k}{m} \]
Taking the square root gives us the angular frequency:
\[ \omega = \sqrt{\frac{k}{m}} \]

### Final Answer
Thus, the angular frequency \( \omega \) of the particle is:
\[ \omega = \sqrt{\frac{k}{m}} \]