# Forced SHM with damping problem

1. Apr 28, 2014

1. The problem statement, all variables and given/known data
A mass, $m$, 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 $k > 0$.
A periodic force is applied to the spring and the equation of motion for the mass is
$$\ddot x+ \frac{b}{m}\dot x+\omega_0^2x=fe^{i\omega t}$$ with $b>0$, $m$ where $f = |f|e^{i\phi_f}$ with $|f|$ 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 $$x(t)=Ce^{i\alpha t} \text{ with } C=|C|e^{i\phi_C}$$ where $|C|$ and $\phi_C$ are real constants. Find the real and imaginary parts of $\alpha$ for the case $b/2m<\omega_0$.
(iii) Now consider the case with $f \neq 0$. Assume a solution of the form $$x(t) = Ae^{i\omega t}$$ and find an expression for the complex constant $A = |A|e^{i\phi_A}$ .
(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 $x(0) = 0$ and $\dot x(0) = 0$. Use these to find two relations between the constants $|C|, C, |A|$ and $A$. Hence show that $$\tan{\phi_c} = \frac{\omega}{\omega '} \tan{\phi_A} - \frac{b}{2m\omega '},$$ where $\omega '$ is the real part of $\alpha$ from part (ii).

2. Relevant equations

3. The attempt at a solution
Part (i) easy

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

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

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

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

Taking $\tan$ of both sides, $\tan\phi_A = \tan\phi_B$ 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?

Last edited: Apr 28, 2014
2. Apr 29, 2014