
#1
Sep2307, 03:20 PM

P: 60

1. The problem statement, all variables and given/known data
PROBLEM STATEMENT: Under these conditions, the motion of the mass when displaced from equilibrium by A is simply that of a damped oscillator, x = A cos(ω_0t) e^(−γt/2) where ω_0 = K/M, K =2k,and γ = b/M. Later we will discuss your measurement of this phenomenon. Now suppose that the right hand end of the right hand spring is vibrated so instead of the end being fixed at x1 + x2 its position is given by x1 + x2 + s sin ωt. Then the sum of the forces includes the driving force, and the equation of motion becomes M = −Kx − bv + F_0 sin ωt (1) where F_0 = Ks. Equation 1 is the very famous damped, forced oscillator equation that reappears over and over in the physical sciences. There are many possible solutions to this equation, but only those that correspond to physical reality are sought. Experimentally it is clear that the mass will oscillate at the driving frequency that can be varied over a wide range. The motion of the mass differs in phase from the drive even though its frequency is the same, but this will be hard to see in this experiment. A difference in phase means that the mass does not always move in the direction of the applied force, but may sometimes move in the opposite direction. Try a solution of the equation in which the force is out of phase with the motion. For simplicity change the phase of the force instead of the phase of the motion (this clearly is the same as changing the phase of the motion by the negative of the force phase change) so x = A cos ωt and then Eq. 1 will become −MAω^2 cos (ωt) = −AK cos (ωt) + bAω sin (ωt) + F_0 sin (ωt − φ) (2) where φ is the phase difference. This equation must hold for all values of time, so we can choose any time to evaluate it. At t =0, and at t = π/2ω, we find MAω^2 = +AK + F_0 sin (φ) and 0= bAω + F_0 cos (φ) (3) since sin (π/2 − φ)=cos φ. Solve the second part of Eq. 3 for cos φ and calculate sin φ using sin2 φ =1−cos2 φ and substitute it into the first Eq. 3. Isolate the radical on one side of the equation and then square both sides. The result is A^2(K − Mω^2)^2 = _F0^2 − (bAω)^2 from which A = [F_0/M] / {[(ω_0^2  ω^2)^2 + (γω)^2 ]^(1/2)} (4) QUESTION: In this derivation, sines and cosines were used in place of exponential notation, and the consequence was considerable extra algebra. Replace the sin (ωt − φ) term by Im [e^i(ωt − φ)], replace the trial expression for x with Re [Ae^iωt], separate the real and imaginary parts of the result, and solve for A. (Here Re and Im refer to real and imaginary parts respectively.) Compare your expression for A with Eq. 4. 2. Relevant equations e^it = cos t + i sin t Re [e^it] = cos t Im [e^it] = sin t 3. The attempt at a solution I wasn't sure how to approach this. I plugged in x = Re [Ae^iωt] and sin (ωt − φ) = Im [e^i(ωt − φ)] into Eq (2), and after some algebra, was able to show: A = { (F_0/M) Im [e^i(ωt − φ)] } / {(ω_0^2  ω^2 + ωiγ) Re [Ae^iωt] } Which isn't equivalent to the original A...help? 



#2
Sep2407, 12:57 AM

P: 60

Nevermind, I got it :)



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