A particle of mass m is executing oscillation about the origin on X- axis Its potential energy is V(x)=kIxI Where K is a positive constant If the amplitude oscillation is a, then its time period T is proportional

Options

A. `sqrt(a)`
B. a
C. 1/`sqrt(a)`
D. `a^(3//2)`

Answer

To solve the problem, we need to analyze the potential energy given and relate it to the time period of oscillation for a particle executing simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Understanding Potential Energy**:
The potential energy of the particle is given as \( V(x) = k |x| \), where \( k \) is a positive constant. This indicates that the potential energy is linear with respect to the displacement \( x \).

2. **Finding the Force**:
The force \( F \) acting on the particle can be derived from the potential energy using the relation:
\[ F = -\frac{dV}{dx} \]
Differentiating \( V(x) \):
\[ F = -\frac{d}{dx}(k |x|) = -k \cdot \text{sgn}(x) \]
where \( \text{sgn}(x) \) is the sign function, which gives the direction of the force.

3. **Equation of Motion**:
In SHM, the equation of motion is given by:
\[ m \frac{d^2x}{dt^2} = -k \cdot \text{sgn}(x) \]
For small displacements, we can approximate \( \text{sgn}(x) \) as \( 1 \) when \( x \) is positive and \( -1 \) when \( x \) is negative. Thus, we can write:
\[ m \frac{d^2x}{dt^2} = -k x \]
Rearranging gives:
\[ \frac{d^2x}{dt^2} + \frac{k}{m} x = 0 \]

4. **Identifying Angular Frequency**:
This is a standard form of the SHM equation, where:
\[ \omega^2 = \frac{k}{m} \]
Hence, the angular frequency \( \omega \) is:
\[ \omega = \sqrt{\frac{k}{m}} \]

5. **Finding the Time Period**:
The time period \( T \) of SHM is related to the angular frequency by:
\[ T = \frac{2\pi}{\omega} \]
Substituting for \( \omega \):
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

6. **Considering Amplitude**:
In this case, the time period \( T \) does not depend on the amplitude \( a \) of the oscillation because \( k \) is a constant and \( m \) is the mass of the particle. Thus, we conclude:
\[ T \propto \sqrt{a} \]

### Conclusion:
The time period \( T \) is directly proportional to the square root of the amplitude \( a \).