A particle of mass m is executing oscillations about the origin on the x-axis, it's potential energy is `U(x)=k|x|^(2)`, where k is a positive constant, if the amplitude of oscillation is a, then it's time period T is

Options

A. Proportional `o(1)/(sqrt(a))`
B. Independent of a
C. Proportional to `sqrt(a)`
D. Proportional to `a^(3//2)`

Explanation

To find the time period \( T \) of a particle of mass \( m \) executing oscillations about the origin with potential energy given by \( U(x) = k|x|^2 \), we can follow these steps:

### Step 1: Understand the potential energy function
The potential energy \( U(x) = k|x|^2 \) indicates that the force acting on the particle is derived from this potential energy. The force \( F \) can be found using the relation:
\[ F = -\frac{dU}{dx} \]

### Step 2: Calculate the force
Taking the derivative of the potential energy:
\[ F = -\frac{d}{dx}(k|x|^2) = -2k|x| \cdot \text{sgn}(x) \]
This shows that the force is proportional to the displacement \( x \) and acts towards the origin, indicating simple harmonic motion.

### Step 3: Identify the form of simple harmonic motion
For simple harmonic motion, the restoring force can be expressed as:
\[ F = -kx \]
where \( k \) is the force constant. Here, we see that the effective spring constant \( k' \) is \( 2k \) because of the factor of 2 in the force expression.

### Step 4: Use the formula for the time period of SHM
The time period \( T \) of a simple harmonic oscillator is given by:
\[ T = 2\pi \sqrt{\frac{m}{k'}} \]
Substituting \( k' = 2k \):
\[ T = 2\pi \sqrt{\frac{m}{2k}} \]

### Step 5: Analyze the dependence on amplitude
In simple harmonic motion, the time period \( T \) is independent of the amplitude \( a \). Thus, we can conclude that:
\[ T = 2\pi \sqrt{\frac{m}{2k}} \]
This shows that the time period does not depend on the amplitude \( a \).

### Final Answer
The time period \( T \) of the oscillation is:
\[ T = 2\pi \sqrt{\frac{m}{2k}} \]