Question

A 2.15 kg lightly damped harmonic oscillator has an angular oscillation frequency of 0.261 rad/s. If the maximum displacement of 2.9 m occurs when t = 0.00 s, and the damping constant b is 0.74 kg/s what is the object's displacement when t = 4.01 s?

Answer 1

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Below are the choices:

A) 0.50 m
B) 0.43 m
C) 0.58 m
D) 0.65 m
Answer: A
Var: 21

Answer 2

Answer:

x(4.01) = 0.72823 m

Explanation:

Given:

- Mass of the oscillator m = 2.15 kg

- Angular frequency w_d = 0.261 rad/s

- Amplitude of motion A = 2.9 m

- Damping constant b = 0.74 kg/s

Find:

- what is the object's displacement when t = 4.01 s?

Solution:

- The solution to the second order ODE of an harmonic damped oscillator is given by:

                             x(t) = A*e^(-p*t) *cos(w_d*t)

Where,

A: amplitude.

w_d: is the damped frequency

t: Time in seconds

p: is the damping factor

Where, p is the damping factor related by:

                               p = b / 2*m

- Plug values in:     p = 0.74/ (2*2.15)

                               p = 0.172

- Formulate the solution x(t):

                               x(t) = 2.9*e^(-0.172*t) *cos(0.261*t)

- Evaluate the displacement x(4.01) :

                               x(4.01) = 2.9*e^(-0.172*4.01) * cos (0.261*4.01)

                               x(4.01) = 2.9*0.50171653*0.50051

                               x(4.01) = 0.72823 m