The potential energy of a particle of mass 2 kg in SHM is `(9x^(2))`J. Here x is the displacement from mean position . If total mechanical energy of the particle is 36 J. The maximum speed of the particle is

Options

A. `4 ms^(-1)`
B. `2 ms^(-1)`
C. `6 ms^(-1)`
D. `10 ms^(-1)`

Explanation

To find the maximum speed of a particle undergoing simple harmonic motion (SHM) given its potential energy and total mechanical energy, we can follow these steps:

### Step 1: Understand the relationship between potential energy and total mechanical energy in SHM.
In SHM, the total mechanical energy (E) is the sum of kinetic energy (K) and potential energy (U). The total mechanical energy is constant and can be expressed as:
\[ E = K + U \]

### Step 2: Identify the potential energy function.
The potential energy given in the problem is:
\[ U = 9x^2 \, \text{J} \]
where \( x \) is the displacement from the mean position.

### Step 3: Determine the total mechanical energy.
The total mechanical energy is given as:
\[ E = 36 \, \text{J} \]

### Step 4: Find the potential energy at the mean position.
At the mean position (where \( x = 0 \)):
\[ U = 9(0)^2 = 0 \, \text{J} \]
Thus, at the mean position, all the total mechanical energy is converted into kinetic energy.

### Step 5: Set up the equation for kinetic energy.
At the mean position, the total energy is equal to the kinetic energy:
\[ E = K \]
\[ 36 \, \text{J} = \frac{1}{2} m V_{\text{max}}^2 \]
where \( m = 2 \, \text{kg} \).

### Step 6: Substitute the mass into the kinetic energy equation.
Substituting the mass into the equation:
\[ 36 = \frac{1}{2} \times 2 \times V_{\text{max}}^2 \]

### Step 7: Simplify the equation.
The \( \frac{1}{2} \times 2 \) simplifies to 1:
\[ 36 = V_{\text{max}}^2 \]

### Step 8: Solve for maximum speed.
Taking the square root of both sides to find \( V_{\text{max}} \):
\[ V_{\text{max}} = \sqrt{36} = 6 \, \text{m/s} \]

### Conclusion:
The maximum speed of the particle is:
\[ V_{\text{max}} = 6 \, \text{m/s} \]