A2.15 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?
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Ответ:
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
Ответ:
1.18*10^-6
Explanation: