A particle of mass `10gm` is placed in a potential field given by `V = (50x^(2) + 100)J//kg`. The frequency of oscilltion in `cycle//sec` is

Options

A. `(10)/(pi)`
B. (5)/(pi)`
C. `(100)/(pi)`
D. `(50)/(pi)`

Explanation

To solve the problem, we need to find the frequency of oscillation of a particle in a given potential field. Let's go through the solution step by step.

### Step 1: Understand the Potential Energy
The potential energy \( V \) is given as:
\[ V = 50x^2 + 100 \text{ J/kg} \]
This represents the potential energy per unit mass, so the total potential energy \( U \) of the particle can be expressed as:
\[ U = m \cdot V = m \cdot (50x^2 + 100) \]
where \( m \) is the mass of the particle.

### Step 2: Convert Mass to Kilograms
The mass of the particle is given as \( 10 \text{ gm} \). We need to convert this to kilograms:
\[ m = 10 \text{ gm} = 10 \times 10^{-3} \text{ kg} = 0.01 \text{ kg} \]

### Step 3: Calculate the Total Potential Energy
Substituting the value of \( m \) into the potential energy equation:
\[ U = 0.01 \cdot (50x^2 + 100) = 0.5x^2 + 1 \text{ J} \]

### Step 4: Find the Force
The force \( F \) acting on the particle is given by the negative gradient of the potential energy:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ \frac{dU}{dx} = \frac{d}{dx}(0.5x^2 + 1) = 0.5 \cdot 2x = x \]
Thus, the force is:
\[ F = -x \]

### Step 5: Relate Force to Simple Harmonic Motion
In simple harmonic motion (SHM), the force can also be expressed as:
\[ F = -m\omega^2 x \]
Setting the two expressions for force equal to each other:
\[ -x = -m\omega^2 x \]
This simplifies to:
\[ 1 = m\omega^2 \]

### Step 6: Solve for Omega
Substituting the value of \( m \):
\[ 1 = 0.01 \cdot \omega^2 \]
\[ \omega^2 = \frac{1}{0.01} = 100 \]
\[ \omega = \sqrt{100} = 10 \text{ rad/s} \]

### Step 7: Calculate Frequency
The frequency \( f \) in cycles per second (Hz) is related to \( \omega \) by:
\[ f = \frac{\omega}{2\pi} \]
Substituting the value of \( \omega \):
\[ f = \frac{10}{2\pi} = \frac{5}{\pi} \text{ cycles/sec} \]

### Final Answer
The frequency of oscillation is:
\[ f = \frac{5}{\pi} \text{ cycles/sec} \]