A particle of mass m in a unidirectional potential field have potential energy `U(x)=alpha+2betax^(2)`, where `alpha` and `beta` are positive constants. Find its time period of oscillations.

Options

A. `2pi sqrt((2beta)/(m))`
B. `2pi sqrt((m)/(2 beta))`
C. `pi sqrt((m)/(beta))`
D. `pi sqrt((beta)/(m))`

Answer

To find the time period of oscillations for a particle of mass \( m \) in a unidirectional potential field with potential energy given by \( U(x) = \alpha + 2\beta x^2 \), we can follow these steps:

### Step 1: Identify the Force Acting on the Particle
The force \( F \) acting on the particle can be derived from the potential energy function. The force is given by:
\[ F = -\frac{dU}{dx} \]
Calculating the derivative of the potential energy:
\[ U(x) = \alpha + 2\beta x^2 \]
Taking the derivative:
\[ \frac{dU}{dx} = 0 + 4\beta x = 4\beta x \]
Thus, the force is:
\[ F = -4\beta x \]

### Step 2: Relate Force to Acceleration
According to Newton's second law, the force can also be expressed in terms of mass and acceleration:
\[ F = m a \]
Where \( a \) is the acceleration. Since acceleration \( a \) can be expressed as \( a = \frac{d^2x}{dt^2} \), we can write:
\[ m \frac{d^2x}{dt^2} = -4\beta x \]

### Step 3: Identify the Form of Simple Harmonic Motion (SHM)
The equation \( m \frac{d^2x}{dt^2} = -4\beta x \) resembles the standard form of simple harmonic motion:
\[ \frac{d^2x}{dt^2} = -\omega^2 x \]
Where \( \omega^2 = \frac{4\beta}{m} \). Thus, we can identify:
\[ \omega = \sqrt{\frac{4\beta}{m}} \]

### Step 4: Calculate the Time Period of Oscillation
The time period \( T \) of oscillation for a simple harmonic oscillator is given by:
\[ T = \frac{2\pi}{\omega} \]
Substituting the expression for \( \omega \):
\[ T = \frac{2\pi}{\sqrt{\frac{4\beta}{m}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{m}{4\beta}} = \pi \sqrt{\frac{m}{\beta}} \]

### Final Result
Thus, the time period of oscillations is:
\[ T = \pi \sqrt{\frac{m}{\beta}} \]

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