Question

A harmonic oscillator has angular frequency w and amplitude A.

Part A
What is the magnitude of the displacement when the elastic potential energy is equal to the kinetic energy? (Assume that U-0 at equilibrium.)
Express your answer in terms of the variables w and 4.

Answer 1

When a harmonic oscillator's potential energy equals its kinetic energy, the magnitude of the displacement is A/sqrt(2), where A is the amplitude of the oscillation.

Step-by-step explanation:

In a harmonic oscillator, the total mechanical energy is the sum of kinetic energy (K) and potential energy (U). At the position where potential energy is equal to the kinetic energy, each represents half of the total energy in the system. The total energy of a simple harmonic oscillator is given by the equation E = 1/2 kA², where k is the force constant and A is the amplitude. Thus, at the point where potential energy equals kinetic energy, we have:

E = K + U

1/2 kA² = 2K

K = 1/4 kA²

The elastic potential energy of a spring is also defined as U = 1/2 kx², where x is the displacement from the equilibrium. At the point where U = K:

1/2 kx² = 1/4 kA²

To find the displacement x, we solve this equation:

x² = (1/2)A²

x = A/sqrt(2) or x = -A/sqrt(2)

Therefore, the magnitude of the displacement x, when the potential energy equals the kinetic energy, is A/sqrt(2), with the final answer independent of the angular frequency ω.