Question

Harmonic oscillator potential is:

• A. \(V = \frac{m\omega^2 x^2}{2}\)
• B. \(V = mgh\)
• C. \(V = kx\)
• D. \(V = 0\)

Hint:

A harmonic oscillator potential is always a symmetric, quadratic, parabolic well centered at the origin: \(V(x) \propto x^2\). - Option (2) represents a linear uniform gravitational potential. - Option (3) is a linear potential which gives a constant force. - Option (4) represents a completely free particle system.

Answer 1

The Correct Option is A

Concept: A classical or quantum harmonic oscillator describes a particle subject to a conservative restoring force that is directly proportional to its displacement from an equilibrium position. This relationship is defined by Hooke's Law: \[ F = -kx \] where \(k\) is the positive force constant (or spring constant) and \(x\) represents the displacement from the equilibrium position (\(x=0\)).

Step 1: Finding potential energy from the restoring force.
The potential energy function \(V(x)\) associated with a conservative force field is defined by the negative spatial integral of that force: \[ F = -\frac{dV}{dx} \quad \Rightarrow \quad dV = -F \, dx \] Substituting Hooke's linear restoring force \(F = -kx\) into this fundamental integration relation: \[ dV = -(-kx) \, dx = kx \, dx \] Integrating both sides with respect to position, assuming that the potential energy is zero at the equilibrium origin (\(V(0) = 0\)): \[ V(x) = \int_{0}^{x} kx' \, dx' = k \left[ \frac{(x')^2}{2} \right]_{0}^{x} = \frac{1}{2}kx^2 \]

Step 2: Expressing the spring constant in terms of angular frequency.
For a physical body of mass \(m\) undergoing periodic oscillations, its characteristic angular frequency \(\omega\) is dynamically linked to the spring stiffness parameter \(k\) by the formula: \[ \omega = \sqrt{\frac{k}{m}} \] Squaring both sides of this relationship allows us to solve for \(k\): \[ \omega^2 = \frac{k}{m} \quad \Rightarrow \quad k = m\omega^2 \]

Step 3: Final substitution into the potential formula.
Now substitute this expression for \(k = m\omega^2\) directly back into the potential energy equation derived in
Step 1: \[ V(x) = \frac{1}{2}(m\omega^2)x^2 = \frac{m\omega^2 x^2}{2} \] This matches option (1) exactly.