A particle of mass (m) is executing oscillations about the origin on the (x) axis. Its potential energy is `V(x) = k|x|^3` where (k) is a positive constant. If the amplitude of oscillation is a, then its time period (T) is.

Options

A. (a)Propertional to `(1)/(sqrta)`
B. (b)Independent of a.
C. (c)Propertional to `sqrta`
D. (d)Propertional to `a^(3//2)`

Answer

To find the time period \( T \) of a particle executing oscillations with a potential energy given by \( V(x) = k|x|^3 \), we can use dimensional analysis and the properties of simple harmonic motion.

### Step-by-Step Solution:

1. **Identify the Potential Energy Function**:
The potential energy is given as \( V(x) = k|x|^3 \), where \( k \) is a positive constant.

2. **Determine the Dimensions of Each Variable**:
- The dimensions of mass \( m \) are \( [M] \).
- The dimensions of length \( x \) are \( [L] \).
- The dimensions of time \( t \) are \( [T] \).
- The constant \( k \) has dimensions that we need to determine.

3. **Find the Dimensions of the Potential Energy**:
The potential energy \( V \) has dimensions of energy, which is \( [ML^2T^{-2}] \).

4. **Set Up the Dimensional Equation**:
From the potential energy function, we can express the dimensions of \( k|x|^3 \):
\[ [V] = [k][L^3] \]
Therefore,
\[ [k] = \frac{[V]}{[L^3]} = \frac{[ML^2T^{-2}]}{[L^3]} = [ML^{-1}T^{-2}] \]

5. **Express the Time Period in Terms of Dimensions**:
The time period \( T \) can be expressed in terms of \( m \), \( k \), and the amplitude \( a \):
\[ T = m^x k^y a^z \]

6. **Substituting the Dimensions**:
Substituting the dimensions into the equation:
\[ [T] = [M^x][M^{1}L^{-1}T^{-2}]^y[L^z] \]
This simplifies to:
\[ [T] = [M^{x+y}L^{z-y}T^{-2y}] \]

7. **Equating Dimensions**:
For the dimensions to be consistent with time \( [T] \), we have:
- \( x + y = 0 \) (for mass)
- \( z - y = 0 \) (for length)
- \( -2y = 1 \) (for time)

8. **Solving the Equations**:
From \( -2y = 1 \), we find \( y = -\frac{1}{2} \).
Substituting \( y \) into \( x + y = 0 \) gives \( x = \frac{1}{2} \).
Substituting \( y \) into \( z - y = 0 \) gives \( z = -\frac{1}{2} \).

9. **Final Expression for Time Period**:
Thus, we can express the time period as:
\[ T \propto a^{-\frac{1}{2}} \implies T \propto \frac{1}{\sqrt{a}} \]

### Conclusion:
The time period \( T \) of the oscillation is inversely proportional to the square root of the amplitude \( a \):
\[ T \propto \frac{1}{\sqrt{a}} \]