Power in driven oscillations. A driven oscillator with mass m, spring constant k

Question

Power in driven oscillations. A driven oscillator with mass m, spring constant k, and damping coefficient b is driven by a force F_0 cos omega t. The resulting steady-state oscillations are described by x(t) = A cos (omega t + phi). The instantaneous power delivered to the oscillator by the driving force is P_drive(t) = F(t) v(t) (since F and v are collinear.) (a) Show that the average power delivered by the driving force during one complete cycle is 1/2 omega F_0 A sin (-phi). (b) The instantaneous power dissipated by the damping force is P_damp(t) = F_d v = -bv^2. Show that the average power dissipated during one cycle is -1/2 b omega^2 A^2. (c) In steady state, your answers to (a) and (b) must be equal in magnitude: power in must equal power dissipated so that the amplitude and mechanical energy of the oscillations remain constant. Using your answer to (b), find the driving frequency at which the power dissipated by damping (and thus the power drawn from the driving force) is a maximum. At this maximum power, what is phi and what is the phase difference between the driving force and the velocity? Assume that Q_0 for the system is much larger than 2 so that the system is strongly underdamped.

Explanation / Answer

a. given instantaneous power, P = F(t)v(t)

F(t) = Focos(wt)

x(t) = Acos(wt + phi)

v(t) = d(x(t))/dt = -Awsin(wt + phi)

so dP = Fo*A*w*cos(wt)sin(wt + phi)

using sin(A)*cos(B) = {sin(A + B) + sin(A - B)}/2

dP = Fo*A*w*{sin(2wt + phi) + sin(phi)}/2

average power = integrate (dP * dt)/integrate dt

Pav = Fo*A*w*{[-cos(2wt + phi) + cos(phi)]/2w + sin(-phi)*t}/2*t

for t = 2*pi/w ( ome complete cycle)

Pav = Fo*A*w*{sin(-phi)}/2 = Fo*A*w*sin(-phi)/2

b. Pdamp(t) = Fdv = -bv^2

so Pdamp(avg) = integrate (Pdamp(t)*dt)/integrate dt

Pdamp(avg) = integrate(-bv^2 dt)/t

now v(t) = -Awsin(wt + phi)

v^2 = A^2*w^2*sin^2(wt + phi)

Pdamp(avg) = -b(integrate ( A^2*w^2*sin^2(wt + phi))dt)/t

Pdamp(avg) = -bA^2*w^2(integrate sin^2(wt + phi))/t

for one cyccle, t = 2*pi/w

Pdamp(avg) = -bA^2*w^2/2

c. so from steady state

Pdamp(avg) = Pavg

bA^2*w^2/2 = Fo*A*w*sin(phi)/2

bA*w = Fo*sin(phi)

w = Fosin(phi)/bA

d. for maximum power

Pdamp(avg) = -bA^2*w^2/2 = -bA^2*Fo^2*sin^2(phi)/2*b^2*A^2 = -Fo^2*sin^2(phi)/2*b

so phi = 90 deg

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