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 of finding the angular frequency \(\omega\) of a particle of mass \(m\) moving under the potential \(V(x) = \frac{kx^2}{2} + \lambda\), we will follow these steps:

### Step 1: Identify the Potential Energy
The potential energy of the particle is given as:
\[ V(x) = \frac{kx^2}{2} + \lambda \]
where \(k\) is a positive constant.

### Step 2: Determine the Force
The force \(F\) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dV}{dx} \]
We will differentiate \(V(x)\) with respect to \(x\).

### Step 3: Differentiate the Potential Energy
Calculating the derivative:
\[ \frac{dV}{dx} = \frac{d}{dx}\left(\frac{kx^2}{2} + \lambda\right) = kx \]
Thus, the force becomes:
\[ F = -kx \]

### Step 4: Apply Newton's Second Law
According to Newton's second law, the force can also be expressed as:
\[ F = ma \]
where \(a\) is the acceleration of the particle. Therefore, we have:
\[ ma = -kx \]
This can be rearranged to express acceleration:
\[ a = -\frac{k}{m}x \]

### Step 5: Compare with the SHM Equation
The equation \(a = -\frac{k}{m}x\) resembles the standard form of simple harmonic motion (SHM):
\[ a = -\omega^2 x \]
By comparing the two equations, we can identify:
\[ \omega^2 = \frac{k}{m} \]

### Step 6: Solve for Angular Frequency
Taking the square root gives us the angular frequency:
\[ \omega = \sqrt{\frac{k}{m}} \]

### Conclusion
The angular frequency \(\omega\) of the particle oscillating under the given potential is:
\[ \omega = \sqrt{\frac{k}{m}} \]