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), we can follow these steps:

### Step 1: Understand the relationship between potential energy (PE), kinetic energy (KE), and total mechanical energy (TME) in SHM.
In SHM, the total mechanical energy (TME) is the sum of the potential energy (PE) and kinetic energy (KE) at any point in time. The formula for total mechanical energy is:
\[ \text{TME} = \text{PE} + \text{KE} \]

### Step 2: Write down the given information.
- Mass of the particle, \( m = 2 \, \text{kg} \)
- Potential energy, \( PE = 9x^2 \, \text{J} \)
- Total mechanical energy, \( TME = 36 \, \text{J} \)

### Step 3: Determine the maximum potential energy.
In SHM, the maximum potential energy occurs at the maximum displacement (amplitude, \( A \)). At this point, all the total mechanical energy is potential energy:
\[ TME = PE_{\text{max}} \]
Thus, we can set:
\[ 36 = 9A^2 \]

### Step 4: Solve for the amplitude \( A \).
Rearranging the equation gives:
\[ A^2 = \frac{36}{9} = 4 \]
Taking the square root, we find:
\[ A = 2 \, \text{m} \]

### Step 5: Use the total mechanical energy to find the maximum speed.
The total mechanical energy can also be expressed in terms of the maximum speed \( V_{\text{max}} \):
\[ TME = \frac{1}{2} m V_{\text{max}}^2 \]
Substituting the known values:
\[ 36 = \frac{1}{2} \times 2 \times V_{\text{max}}^2 \]

### Step 6: Simplify and solve for \( V_{\text{max}} \).
This simplifies to:
\[ 36 = V_{\text{max}}^2 \]
Taking the square root gives:
\[ V_{\text{max}} = \sqrt{36} = 6 \, \text{m/s} \]

### Final Answer:
The maximum speed of the particle is \( 6 \, \text{m/s} \).

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