Final answer:
The magnitude of the displacement when the elastic potential energy is equal to the kinetic energy in a harmonic oscillator with amplitude A is A/\sqrt{2}.
Step-by-step explanation:
The question asks for the magnitude of displacement when the elastic potential energy is equal to the kinetic energy in a simple harmonic oscillator with angular frequency w and amplitude A.
In a simple harmonic oscillator, the total energy E is the sum of kinetic and potential energies and is given by E = (1/2)kA², where k is the spring constant. The potential energy (U) at displacement x is U = (1/2)kx², and the kinetic energy (K) is K = E - U. When U = K, each is half of the total energy.
Setting U = K, we have (1/2)kx² = (1/4)kA². Solving for x gives us x = A/\sqrt{2}. Therefore, the magnitude of the displacement when the elastic potential energy is equal to the kinetic energy is A/\sqrt{2}.