A linear harmonic oscillator of force constant `2 xx 10^6 N//m` and amplitude (0.01 m) has a total mechanical energy of (160 J). Its.

Options

A. maximum potential enregy is `160 J`
B. maximum potential energy is `100 J`
C. maximum potential energy is zero
D. maximum potential energy is 100J`

Answer

To solve the problem regarding the linear harmonic oscillator, we will follow these steps:

### Step 1: Understand the System
We are given:
- Force constant (k) = \(2 \times 10^6 \, \text{N/m}\)
- Amplitude (A) = \(0.01 \, \text{m}\)
- Total mechanical energy (E) = \(160 \, \text{J}\)

### Step 2: Recall the Energy in a Simple Harmonic Oscillator
In a simple harmonic oscillator, the total mechanical energy (E) is given by the sum of kinetic energy (KE) and potential energy (PE). At the maximum displacement (amplitude), all the energy is potential, and at the mean position, all the energy is kinetic.

### Step 3: Maximum Potential Energy
At the maximum displacement (which is the amplitude), the potential energy is maximum, and the kinetic energy is zero. Therefore, the maximum potential energy (PE_max) is equal to the total mechanical energy (E).

\[ PE_{\text{max}} = E \]

### Step 4: Substitute the Values
Since we know the total mechanical energy is given as \(160 \, \text{J}\):

\[ PE_{\text{max}} = 160 \, \text{J} \]

### Step 5: Conclusion
Thus, the maximum potential energy of the linear harmonic oscillator is:

\[ \text{Maximum Potential Energy} = 160 \, \text{J} \]

### Final Answer
The maximum potential energy is \(160 \, \text{J}\).

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