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 will follow these steps:

### Step 1: Convert the mass from grams to kilograms
Given mass \( m = 10 \, \text{g} \).

To convert grams to kilograms:
\[ m = 10 \, \text{g} = 10 \times 10^{-3} \, \text{kg} = 0.01 \, \text{kg} \]

### Step 2: Write the potential energy expression
The potential field is given as:
\[ V(x) = 50x^2 + 100 \, \text{J/kg} \]
To find the potential energy \( U \), we multiply the potential field by the mass:
\[ U(x) = m \cdot V(x) = 0.01 \cdot (50x^2 + 100) = 0.5x^2 + 1 \, \text{J} \]

### Step 3: Find the force from the potential energy
The force \( F \) can be found using the relation:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative:
\[ U(x) = 0.5x^2 + 1 \]
\[ \frac{dU}{dx} = 0.5 \cdot 2x = x \]
Thus,
\[ F = -x \]

### Step 4: Relate force to the simple harmonic motion
In simple harmonic motion, the force can also be expressed as:
\[ F = -kx \]
where \( k \) is the spring constant. From our previous result:
\[ -k = -1 \implies k = 1 \, \text{N/m} \]

### Step 5: Find the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the spring constant \( k \) and mass \( m \) by the formula:
\[ \omega^2 = \frac{k}{m} \]
Substituting the values:
\[ \omega^2 = \frac{1}{0.01} = 100 \implies \omega = 10 \, \text{rad/s} \]

### Step 6: Calculate the frequency \( f \)
The frequency \( f \) is related to the angular frequency \( \omega \) 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:
\[ f = \frac{5}{\pi} \, \text{cycles/sec} \]

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