Simple Harmonic Motion Brain Teaser

1. Jul 22, 2004

cj

I saw this in an old, junior-level, classical mechanics
textbook and haven't been able to figure it out.

A particle undergoing simple harmonic motion has a velocity:

$$\frac{dx_1}{dt}$$

when the displacement is:

$$x_1$$

and a velocity

$$\frac{dx_2}{dt}$$

when the displacement is:

$$x_2$$

What is the angular frequency and the amplitude of the motion in terms of the given quantities?

I know the solution to the SHM wave equation is:

$$x(t) = A \cdot sin( \omega t + \phi )$$

And that:

$$dx(t)/dt = A \omega \cdot cos( \omega t + \phi )$$

But can't see how to express omega or A in these terms.

Last edited: Jul 22, 2004
2. Jul 22, 2004

rayjohn01

If x = A.sin(w.t) then

x1 = A sin(w.t1)
x2 = A.sin(w.t2)
dx/dt 1= A.w.cos(w.t1)
dx/dt 2 = A.w.cos(w.t2)

I am not going to do it but there appears to be enough information , to eliminate t1,t2 and get A and w.
For instance a) and c) can eliminate t1 , b) and d) eliminate t2 .

Last edited: Jul 22, 2004
3. Jul 22, 2004

Staff: Mentor

Apply what you know. I'll get you started. You are given: At time $t_1$ the displacement is $x_1$ and the speed is $v_1$. (I didn't like your notation, so I changed it. )

So... just plug into your SHM equations:
$x_1 = A sin(\omega t_1)$
$v_1 = A \omega cos(\omega t_1)$
Combine these equations to get a relationship between $\omega$ and A.

Now do the same for time $t_2$, and then you should be able to solve for $\omega$ and A.