A simple harmonic oscillator
(A) Always has maximum KE at the equilibrium position
(B) Always has zero KE at the extreme position
(C) Always has maximum PE at extreme position
(D) always has zero PE at the equilibrium position

Options

A. A, B, C, D are correct
B. Only A, B, C are correct
C. Only A, B, D are correct
D. Only B, C, D are correct

Answer

To solve the question about the properties of a simple harmonic oscillator (SHO), we need to analyze each option one by one based on the principles of SHOs.

### Step 1: Analyze Option A
**Statement**: "Always has maximum KE at the equilibrium position."

- In a simple harmonic oscillator, the kinetic energy (KE) is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]
- At the equilibrium position (x = 0), the velocity (v) is at its maximum because the restoring force is zero, and the particle is moving through the equilibrium point.
- Therefore, since KE is dependent on the square of the velocity, the KE is maximum at the equilibrium position.

**Conclusion**: Option A is **correct**.

### Step 2: Analyze Option B
**Statement**: "Always has zero KE at the extreme position."

- At the extreme positions (x = +A or x = -A), the particle momentarily comes to rest before changing direction. Hence, the velocity (v) is zero at these points.
- Since KE is directly related to velocity, when v = 0, KE must also be zero.

**Conclusion**: Option B is **correct**.

### Step 3: Analyze Option C
**Statement**: "Always has maximum PE at extreme position."

- The potential energy (PE) in a simple harmonic oscillator is given by:
\[ PE = \frac{1}{2} kx^2 \]
- At the extreme positions (x = +A or x = -A), the displacement (x) is at its maximum (±A), which means that the potential energy is also at its maximum.
- Since the kinetic energy is zero at these points, all the energy is in the form of potential energy.

**Conclusion**: Option C is **correct**.

### Step 4: Analyze Option D
**Statement**: "Always has zero PE at the equilibrium position."

- At the equilibrium position (x = 0), the displacement is zero, which means:
\[ PE = \frac{1}{2} k(0)^2 = 0 \]
- However, while the potential energy is at a minimum at this point, it is not necessarily zero for every system. The minimum potential energy can vary based on the system's configuration.

**Conclusion**: Option D is **incorrect**.

### Final Answer
The correct options are:
- A: Correct
- B: Correct
- C: Correct
- D: Incorrect

Thus, the answer to the question is that options A, B, and C are correct.

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