What is the displacement equation for simple harmonic motion without damping?

[imagemania]

(This isn't homework)
I've been crating some notes, and noticed that we haven't been told one equation.
I know for:
$$\frac{d^2 x}{dt^2}$$ $$= \frac{-kx}{m} -\frac{b}{m} \frac{dx}{dt}$$
I know the displacement is:
$$x = A \omega e^{-bt/2m} cos(\omega t+ \phi)$$

[Ie damping]

I know for:
$$\frac{d^2 x}{dt^2}$$ $$= \frac{-kx}{m} -\frac{b}{m} \frac{dx}{dt} + \frac{F}{m}$$
I know the displacement for this is:
$$x = Asin(\omega t + \phi)$$ (omega is for driving force).

But what about:
$$\frac{d^2 x}{dt^2}$$ $$= \frac{-kx}{m} + \frac{F}{m}$$
I.e. No damping, what would x be for this?

Thank you!

[This is out of pure interest]

 

[AlephZero]

When $\omega^2 \ne k/m$, it makes no difference if b = 0, except that $\phi$ will always be 0 or $\pi$

When $\omega^2 = k/m$ the solution is of the form

$$x = A t \cos (\omega t)$$

In words, the response grows without limit, and it is 90 degrees out of phase with the applied force.

 

[mathman]

The solution for your second equation should have a b in it, but it does not appear. For the third equation use the second equation solution with b=0.

 

[imagemania]

Gah first off i made a mistake on the first one, should have no omega next to A (demensions wouldn't equate)

Ok it would seem i have some equations written down wrong (well two).

$$\frac{d^2 x}{dt^2}$$ $$= \frac{-kx}{m} + \frac{F}{m}$$
Would
$$x = Asin(\omega t + \phi)$$ (i.e. the cos shifts by pi/2 to make sine?)

And
$$\frac{d^2 x}{dt^2}$$ $$= \frac{-kx}{m} -\frac{b}{m} \frac{dx}{dt} + \frac{F}{m}$$

$$x = A e^{-bt/2m} sin(\omega t + \phi)$$

Let me know if these are right :)

 

[mathman]

Have you tried verifying them yourself. Simply plug your answers back into the equations.