How Do You Solve for X(t) in an Underdamped Oscillator Problem?

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The discussion focuses on solving for x(t) in an underdamped oscillator problem with a mass of 0.05 kg, a spring constant of 5 N/m, and a damping constant of 0.4 Ns/m. The general solution for the position is given by x(t) = e^(-beta*t)[A1* e^(sqrt(Beta^2-w0^2)*t) + A2 *e^(-sqrt(Beta^2-w0^2)*t)]. The user correctly identifies that the initial conditions will lead to equations involving imaginary numbers for the coefficients A1 and A2, which is acceptable as the final expression can still be real. The key takeaway is that an expression can be real even if it contains imaginary components, as long as it equals its own complex conjugate.

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don_anon25
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Here's the problem:
A damped oscillator has a mass of .05 kg, a spring constant of 5 N/m, and a damping constant of .4 Ns/m. At t=0, the mass is moving at 3.0 m/s at x=.1m. Find x as a function of time.

What I have done:
I know the damping constant b = .4 and I have used this to find Beta. Also, I used k and m to find w0. I know the general solution for the equation of the harmonic oscillator -- please pardon my typing -- x(t) = e^(-beta*t)[A1* e^(sqrt (Beta^2-w0^2)) *t + A2 *e^(-sqrt(Beta^2-w0^2))].
I can use my initial condition for the position to get one equation with A1 and A2 in it. I can then take the derivative of x(t) and use the initial condition of the velocity to get the other. I now have two equations and two unknowns. The issue is that these unknowns will involve imaginary numbers because we have an underdamped case. Is this ok?
 
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Yes, it is OK. You should find that the imaginary parts of A_1 and A_2 are exactly canceled out by the imaginary parts of the complex exponentials to give you a real function.

Remember, to be real an expression doesn't have to have a complete absence of the imaginary unit i. It simply has to equal its own complex conjugate. For instance the expression (x-3i)(x+3i) is purely real, despite the fact that i appears.
 
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