A linear harmonic oscillator has a total mechanical energy of `200 J`. Potential energy of it at mean position is `50J`. Find
(i) the maximum kinetic energy,
(ii)the minimum potential energy,
(iii) the potential energy at extreme positions.

Explanation

To solve the problem step by step, we will analyze the given information and apply the principles of Simple Harmonic Motion (SHM).

### Given:
- Total mechanical energy (E) = 200 J
- Potential energy (PE) at mean position = 50 J

### Step 1: Finding Maximum Kinetic Energy (KE_max)
In SHM, the total mechanical energy is the sum of kinetic energy and potential energy at any position. At the mean position, the potential energy is at its minimum, and the kinetic energy is at its maximum.

The relationship can be expressed as:
\[ E = KE_{max} + PE_{min} \]

Where:
- \( PE_{min} \) at the mean position = 50 J

Substituting the values:
\[ 200 J = KE_{max} + 50 J \]

Now, solve for \( KE_{max} \):
\[ KE_{max} = 200 J - 50 J \]
\[ KE_{max} = 150 J \]

### Step 2: Finding Minimum Potential Energy (PE_min)
The minimum potential energy occurs at the mean position. From the given information:
\[ PE_{min} = 50 J \]

### Step 3: Finding Potential Energy at Extreme Positions (PE_max)
At extreme positions, all the mechanical energy is converted into potential energy, and the kinetic energy is zero. Therefore, the potential energy at extreme positions is equal to the total mechanical energy.

Thus:
\[ PE_{max} = E \]
\[ PE_{max} = 200 J \]

### Summary of Results:
(i) Maximum Kinetic Energy: \( KE_{max} = 150 J \)
(ii) Minimum Potential Energy: \( PE_{min} = 50 J \)
(iii) Potential Energy at Extreme Positions: \( PE_{max} = 200 J \)

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