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. Proportional to `1/sqrta`
B. Independent to a
C. Proportional to `sqrta`
D. Proportional to `a^(3/2)`

Answer

To find the time period \( T \) of a particle executing oscillations about the origin with a potential energy given by \( V(x) = k|x|^3 \), we can follow these steps:

### Step 1: Understand the Potential Energy
The potential energy of the particle is given as:
\[ V(x) = k |x|^3 \]
where \( k \) is a positive constant.

### Step 2: Determine the Dimensions of the Potential Energy
The dimension of potential energy \( V \) is:
\[ [V] = [M L^2 T^{-2}] \]
where \( M \) is mass, \( L \) is length, and \( T \) is time.

### Step 3: Find the Dimension of \( k \)
Since \( V(x) = k |x|^3 \), we can express the dimensions of \( k \) using the dimensions of \( |x|^3 \):
\[ |x|^3 \text{ has dimensions } [L^3] \]
Thus, we have:
\[ k [L^3] = [M L^2 T^{-2}] \]
From this, we can find the dimensions of \( k \):
\[ [k] = \frac{[M L^2 T^{-2}]}{[L^3]} = [M L^{-1} T^{-2}] \]

### Step 4: Set Up the Dimensional Analysis for Time Period
The time period \( T \) can depend on mass \( m \), the constant \( k \), and the amplitude \( a \). We can express the dimensions of \( T \) as:
\[ [T] = [M^x L^y T^z] \]
Equating the dimensions, we have:
\[ [M^0 L^0 T^1] = [M^x L^{y-1} T^{-2y}] \]

### Step 5: Create Equations from Dimension Analysis
From the dimensional analysis, we can set up the following equations:
1. For mass: \( 0 = x + 1 \) (from \( M \))
2. For length: \( 0 = -y + z \) (from \( L \))
3. For time: \( 1 = -2y \) (from \( T \))

### Step 6: Solve the Equations
From the third equation \( 1 = -2y \):
\[ y = -\frac{1}{2} \]
Substituting \( y \) into the second equation:
\[ 0 = -(-\frac{1}{2}) + z \implies z = -\frac{1}{2} \]
Substituting \( y \) into the first equation:
\[ 0 = x + 1 \implies x = -1 \]

### Step 7: Write the Expression for Time Period
Now we can express the time period \( T \):
\[ T \propto m^x k^y a^z = m^{-1} k^{-\frac{1}{2}} a^{-\frac{1}{2}} \]
Thus, we can write:
\[ T \propto \frac{1}{\sqrt{k}} \cdot \frac{1}{\sqrt{a}} \]

### Step 8: Final Expression for Time Period
The final expression for the time period \( T \) is:
\[ T = C \cdot \frac{1}{\sqrt{k}} \cdot \frac{1}{\sqrt{a}} \]
where \( C \) is a constant that may depend on the system.