Question

The vibration of a 0.3-kg mass on a spring can be described by the equation 0.7cos(1.2t+4.3), where tt is in seconds and x is in meters. Determine the following for this system:

a. The period of the oscillation (in seconds)

b. The total energy in the system (in Joules)

c. The potential energy (in Joules) when the spring is stretched 0.253 m.

d. The kinetic energy (in J) when the spring is stretched 0.253 m.

Answer 1

Here ,

m = 0.30 Kg

x = 0.70 * cos(1.2t + 4.3)

as x = A * cos(wt + phi)

hence , A = 0.70 m , w = 1.2 rad/s

a) period of oscillation = 2pi/w

period of oscillation = 2pi/1.2

period of oscillation = 5.24 s

b) total energy of system = maximum kinetic energy

total energy of system = 0.50 * m * (A * w)^2

total energy of system = 0.50 * 0.30 * (0.70 * 1.2)^2

total energy of system = 0.106 J

c) let the spring constant is k

w = sqrt(k/m)

1.2 = sqrt(k/0.30)

k = 0.432 N/m

potential energy = 0.50 * 0.432 * 0.253^2

potential energy = 0.014 J

d)

kinetic energy = total energy - potential energy

kinetic energy = 0.106 - 0.014

kinetic energy = 0.092 J