A particle of mass `4g` is in a gravitational potential `V =(800x^(2) +150) erg//g`. Its frequency of oscillation is

Options

A. `(5)/(pi)Hz`
B. `(1)/(5pi)Hz`
C. `50pi Hz`
D. `(50)/(pi)Hz`

Explanation

To find the frequency of oscillation for a particle of mass \(4g\) in the given gravitational potential \(V = 800x^2 + 150\) erg/g, we can follow these steps:

### Step 1: Understand the potential energy
The gravitational potential \(V\) is given as:
\[ V = 800x^2 + 150 \text{ erg/g} \]
This represents the potential energy per unit mass.

### Step 2: Calculate the potential energy \(U\)
The total potential energy \(U\) of the particle can be calculated by multiplying the potential \(V\) by the mass \(m\):
\[ U = V \cdot m = (800x^2 + 150) \cdot 4 \text{ g} \]
\[ U = 3200x^2 + 600 \text{ erg} \]

### Step 3: Determine the force \(F\)
The force \(F\) acting on the particle is given by the negative gradient of the potential energy:
\[ F = -\frac{dU}{dx} \]
We differentiate \(U\) with respect to \(x\):
\[ \frac{dU}{dx} = \frac{d}{dx}(3200x^2 + 600) = 6400x \]
Thus, the force is:
\[ F = -6400x \]

### Step 4: Relate force to acceleration
According to Newton's second law, \(F = ma\), where \(m\) is the mass of the particle. Therefore:
\[ ma = -6400x \]
Since the mass \(m = 4g\) (or \(0.004 \text{ kg}\)), we can simplify:
\[ 4a = -6400x \implies a = -1600x \]

### Step 5: Identify the oscillation equation
The equation \(a = -1600x\) is in the form of simple harmonic motion (SHM), which can be expressed as:
\[ a = -\omega^2 x \]
From this, we can identify:
\[ \omega^2 = 1600 \implies \omega = \sqrt{1600} = 40 \text{ rad/s} \]

### Step 6: Calculate the frequency \(f\)
The relationship between angular frequency \(\omega\) and frequency \(f\) is given by:
\[ \omega = 2\pi f \]
Thus, we can solve for \(f\):
\[ f = \frac{\omega}{2\pi} = \frac{40}{2\pi} = \frac{20}{\pi} \text{ Hz} \]

### Final Answer
The frequency of oscillation is:
\[ f = \frac{20}{\pi} \text{ Hz} \]

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