Forced SHM with damping problem

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Homework Statement


A mass, [itex]m[/itex], attached to a spring hangs vertically downwards under gravity. It is subject to a viscous damping force proportional to its velocity. The spring constant is [itex]k > 0[/itex].
A periodic force is applied to the spring and the equation of motion for the mass is
[tex]\ddot x+ \frac{b}{m}\dot x+\omega_0^2x=fe^{i\omega t}[/tex] with [itex]b>0[/itex], [itex]m[/itex] where [itex]f = |f|e^{i\phi_f}[/itex] with [itex]|f|[/itex] and f real constants. The applied force per unit mass should be taken to be the real part of the term on the right hand side.
(i) Explain the origin of the terms in the equation of motion of the mass.
(ii) For the case f = 0, assume a solution of the form [tex]x(t)=Ce^{i\alpha t} \text{ with } C=|C|e^{i\phi_C}[/tex] where [itex]|C|[/itex] and [itex]\phi_C[/itex] are real constants. Find the real and imaginary parts of [itex]\alpha[/itex] for the case [itex]b/2m<\omega_0[/itex].
(iii) Now consider the case with [itex]f \neq 0[/itex]. Assume a solution of the form [tex]x(t) = Ae^{i\omega t} [/tex] and find an expression for the complex constant [itex]A = |A|e^{i\phi_A}[/itex] .
(iv) Show that the sum of the solutions from parts (ii) and (iii) above is also a solution to the equation of motion. The initial conditions are [itex]x(0) = 0[/itex] and [itex]\dot x(0) = 0[/itex]. Use these to find two relations between the constants [itex]|C|, C, |A|[/itex] and [itex]A[/itex]. Hence show that [tex]\tan{\phi_c} = \frac{\omega}{\omega '} \tan{\phi_A} - \frac{b}{2m\omega '},[/tex] where [itex]\omega '[/itex] is the real part of [itex]\alpha[/itex] from part (ii).


Homework Equations





The Attempt at a Solution


Part (i) easy

Part (ii) get [itex]\alpha = \frac{ib}{2m}\pm \sqrt{\omega_0^2 - \frac{b^2}{4m^2}}[/itex]

Part (iii) get [tex]A=\frac{f}{[(\omega_0^2 - \omega^2)^2 + b^2/m^2]^{1/2}}\left(\frac{\omega_0^2 - \omega^2}{[(\omega_0^2 - \omega^2)^2 + b^2/m^2]^{1/2}} - i\frac{b\omega/m}{[(\omega_0^2 - \omega^2)^2 + b^2/m^2]^{1/2}}\right)[/tex] giving [itex]\tan\phi_A = \frac{-b\omega/m}{\omega_0^2 - \omega^2}[/itex]

Part (iv) Obviously a combination of the two is a solution. Using the given initial conditions [itex]x(0) = A+C = 0 \Rightarrow A=-C[/itex] and [itex]\dot x(0) = i\alpha Ce^{i\alpha t} + i\omega Ae^{i\omega t} \Rightarrow \alpha e^{i\alpha t} - \omega e^{i\omega t} = 0[/itex] (not going anywhere with this part)

Since [itex] A = -C[/itex] i.e. [itex]|A|e^{i\phi_A} = -|C|e^{i\phi_C}[/itex] we also have [itex] |A||e^{i\phi_A}| = |C||e^{i\phi_C}| \Rightarrow |A|=|C| [/itex] and [itex]\phi_A = \phi_C + \pi[/itex].

Taking [itex]\tan[/itex] of both sides, [itex]\tan\phi_A = \tan\phi_B[/itex] which is wrong...

This is from a past exam paper, only 1 person was able to do it in the exam. Anyone have a flash of brilliance?
 
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Anyone?
 

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