# Driven un-damped harmonic motion

1. Jan 30, 2010

### KaiserBrandon

1. The problem statement, all variables and given/known data

A particle of mass 'm' is subject to a spring force, '-kx', and also a driving force, Fd*cos(wdt). But there is no damping force. Find the particular solution for x(t) by guessing x(t) = A*cos(wdt) + B*sin(wdt). If you write this in the form C*cos(wdt - $$\phi$$), where C > 0, what are C and $$\phi$$? Be careful about the phase (there are two cases to consider).

2. Relevant equations

basic inhomogeneous second-order equation solving.

3. The attempt at a solution

So I got the equation F = Fd*cos(wdt) - k*x = m*x'', and put it in the form x'' + w2*x = F*cos(wdt), where w2$$\equiv$$ k/m, and F $$\equiv$$ Fd/m. I took the first and second derivative of the equation they told us to use for our guess, and subbed them into x'' + w2*x = F*cos(wdt). I ended up with A(w2 - wd2)*cos(wdt) + B(w2 - wd2)*sin(wdt) = F*cos(wdt). I therefore deduced that F = A(w2 - wd2). Therefore the particular solution would be xp(t) = A(w2 - wd2)*cos(wdt). Is there anyway that this can somehow simplify for the form they want us to put it in, C*cos(wdt - $$\phi$$), or did I do something wrong?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Jan 31, 2010

### ehild

You know F and have to express A and B in terms of F, wd and w.

ehild

3. Feb 1, 2010

### KaiserBrandon

ok, so then I get x(t) = $$\frac{F}{w^2-wd^2}$$*cos(wd2*t) for the particular solution. I still can't figure out how this can take the form that is asked for.

4. Feb 1, 2010

### ehild

It is $$x(t) = \frac{F}{w^2-wd^2}*\cos{(w_d*t)}$$

instead and has the desired form if w>wd.
$$C = {\frac{F}{|w^2-wd^2|}, \Phi =0$$

In the opposite case

$$x(t) =- \frac{F}{|w^2-wd^2|}*\cos{(w_d*t)}$$

you have to include a phase constant of pi:

$$C = {\frac{F}{|w^2-wd^2|}, \Phi =\pi$$

$$x(t) =\frac{F}{|w^2-wd^2|}*\cos{(w_d*t+\pi)}$$

ehild