A particle is placed at the lowest point of a smooth wire frame in the shape of a parabola, lying in the vertical xy-plane having equatioin `x^(2)`=5y(x,y are in meter). After slight displacement, the particle is set free. Find angular frequency of osciallation (in rad/sec) (Take g=10 m/`s^(2))`

Options

A. 2 rad/s
B. 4 rad/s
C. 6 rad/s
D. 8 rad/s

Answer

To solve the problem of finding the angular frequency of oscillation of a particle placed at the lowest point of a smooth wire frame in the shape of a parabola, we can follow these steps:

### Step 1: Understand the Parabola Equation
The equation of the parabola given is \( x^2 = 5y \). This represents a parabola that opens upwards with its vertex at the origin (0,0).

### Step 2: Identify the Lowest Point
The lowest point of the parabola is at the origin (0,0). When the particle is at this point and is slightly displaced, it can oscillate about this point.

### Step 3: Determine the Slope of the Parabola
To find the slope of the parabola at any point, we differentiate the equation \( y = \frac{x^2}{5} \):
\[ \frac{dy}{dx} = \frac{2x}{5} \]
At the lowest point (x = 0), the slope is \( \frac{dy}{dx} = 0 \).

### Step 4: Analyze Forces Acting on the Particle
When the particle is displaced by a small angle \( \theta \), the gravitational force \( mg \) acts downwards. The component of this force acting along the slope of the parabola can be expressed as:
\[ F = -mg \sin(\theta) \]
For small angles, we can approximate \( \sin(\theta) \approx \theta \).

### Step 5: Relate the Angle to Displacement
From the slope of the parabola, we have:
\[ \tan(\theta) = \frac{dy}{dx} = \frac{2x}{5} \]
For small angles, \( \tan(\theta) \approx \theta \), thus:
\[ \theta \approx \frac{2x}{5} \]

### Step 6: Set Up the Equation of Motion
The force acting on the particle can be written as:
\[ F = -mg \theta \]
Substituting \( \theta \) from the previous step:
\[ F = -mg \left(\frac{2x}{5}\right) \]
Using Newton's second law \( F = ma \), where \( a = \frac{d^2x}{dt^2} \):
\[ m \frac{d^2x}{dt^2} = -mg \left(\frac{2x}{5}\right) \]
Dividing through by \( m \):
\[ \frac{d^2x}{dt^2} = -g \left(\frac{2x}{5}\right) \]

### Step 7: Substitute the Value of g
Given \( g = 10 \, \text{m/s}^2 \):
\[ \frac{d^2x}{dt^2} = -10 \left(\frac{2x}{5}\right) = -4x \]

### Step 8: Identify the Angular Frequency
This equation can be compared to the standard form of simple harmonic motion:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
From our equation, we have:
\[ \omega^2 = 4 \]
Thus, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{4} = 2 \, \text{rad/s} \]

### Final Answer
The angular frequency of oscillation is \( \omega = 2 \, \text{rad/s} \).

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