Question

A 0.52·kg mass vibrates according to the equation x = 0.49 cos (8.37t + 3.54), where x is in meters, and t is in seconds.

(a) Determine the amplitude.
m

(b) Determine the frequency.

Hz

(c) Determine the total energy.

J

(d) Determine the kinetic energy and potential energy when x = 0.30 m.
kinetic energy =

J
potential energy = J

Answer 1

Here

part A)

as the equation of the simple harmonic motion is

x = A * cos(w * t + phi)

x = 0.49 cos (8.37t + 3.54

comparing the equations

Amplitude = A = 0.49 m

part B)

frequency = w/(2pi)

frequency = 8.37/(2pi)

frequency = 1.33 Hz

c)

for the total energy

total energy = 0.50 * m * Vmax^2

total energy = 0.50 * 0.52 * (0.49 * 8.37)^2

total energy = 4.37 J

d)

at x = 0.30 m

potential energy = total energy * x^2/A^2

potential energy = 4.37 * 0.30^2/0.49^2

potential energy = 1.64 J

for the kinetic energy

kinetic energy = 4.37 - 1.64

kinetic energy = 2.73 J