Question

A mass m on a spring of stiffness k undergoes horizontal simple harmonic motion with amplitude A, centered around x = 0. a) What is the total "mechanical" energy (kinetic plus potential) of the mass-spring system? b) What is the value of x when the mass-spring system has twice as much kinetic energy as potential energy? Your answers should be in terms of the quantities m, k, and A--or some subset thereof.

Answer 1

a) the total "mechanical" energy = (1/2)*k*A^2

b)

given

2*kinetic energy = potential energy

2*(1/2)*m*v^2 = (1/2)*k*x^2

(1/2)*m*v^2 = (1/2)*(1/2)*k*x^2

we know, kinetic energy + potential energy = total mechanical energy

(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*k*A^2

(1/2)*(1/2)*k*x^2 + (1/2)*k*x^2 = (1/2)*k*A^2

(1/2)*x^2*(1/2 + 1) = (1/2)*A^2

(3/2)*x^2 = A^2

x^2 = (3/2)*A^2

x = sqrt(3/2)*A