Driven un-damped harmonic motion

[KaiserBrandon]

Homework Statement

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

Homework Equations

basic inhomogeneous second-order equation solving.

The Attempt at a Solution

So I got the equation F = F_d*cos(w_dt) - k*x = m*x'', and put it in the form x'' + w²*x = F*cos(w_dt), where w²\equiv k/m, and F \equiv F_d/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(w_dt). I ended up with A(w² - w_d²)*cos(w_dt) + B(w² - w_d²)*sin(w_dt) = F*cos(w_dt). I therefore deduced that F = A(w² - w_d²). Therefore the particular solution would be x_p(t) = A(w² - w_d²)*cos(w_dt). Is there anyway that this can somehow simplify for the form they want us to put it in, C*cos(w_dt - \phi), or did I do something wrong?

 

[ehild]

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

ehild

 

[KaiserBrandon]

ok, so then I get x(t) = \frac{F}{w^2-wd^2}*cos(w_d²*t) for the particular solution. I still can't figure out how this can take the form that is asked for.

 

[ehild]

ok, so then I get x(t) = \frac{F}{w^2-wd^2}*cos(w_d²*t) for the particular solution. I still can't figure out how this can take the form that is asked for.

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