Question

An object of mass of 0.2 kg rests on a frictionless surface and is connected to a horizontal spring of force constant k=20 N/m. The object is set into simple harmonic oscillation with an initial potential energy of 0.5 J and an initial kinetic energy of 0.2 J. Take: x=Acos(ω)t+φ). a) What is the amplitude of oscillation? b) What is the potential energy when the displacement is one half the amplitude?

Answer 1

The amplitude of the oscillation is 0.2646 m, and the potential energy when the displacement is half the amplitude is 0.661 J.

Step-by-step explanation:

The student's question deals with simple harmonic motion (SHM) of a mass-spring system. Given the total mechanical energy and the spring constant, we can determine the amplitude of the oscillation. The initial potential energy (U) of a spring in SHM is given by U = 1/2 kx^2, where k is the spring constant and x is the displacement from equilibrium. The amplitude (A) of the oscillation is the maximum displacement where all the mechanical energy is potential. As the total mechanical energy E is the sum of kinetic energy (K) and potential energy (U), and is conserved, E = K + U. Thus, the amplitude A can be found from the initial total energy.

Amplitude Calculation:
Given: U_initial = 0.5 J, K_initial = 0.2 J, k = 20 N/m
Total Energy E = U_initial + K_initial = 0.5 J + 0.2 J = 0.7 J
A = √(2E/k) = √(2*0.7 J/20 N/m) = √(0.07 m^2) = 0.2646 m

Potential Energy at Half Amplitude:
When the displacement is one-half the amplitude (A/2), we use U = 1/2 kx^2 to find the potential energy.
U = 1/2 * k * (A/2)^2 = 1/2 * 20 N/m * (0.2646 m / 2)^2 = 0.5 * 20 * 0.0661 m^2 = 0.661 J