Help me do some algebra!

[Matt R.]

OK, so I'm trying to find the angular frequency of a particle underdoing simple harmonic motion. It has a velocity x1' when the displacement is x1 and a velocity x2' when the displacement is x2.

I'm not 100% sure the method that I used to get this equality is correct (but I think it is). Basically, I'm more concerned with the algebra at this point and whether or not it is solvable.... If it's not solvable I know I need to go a different route.

So, I need to solve for angular frequency, w, in terms of x1 x1' x2 and x2'. Here is the equality:

x1 cos (wt) + (x1'/w)sin wt = -(x2'/w)cos wt + x2 sin wt

Again, I need to solve this equation for w in terms of the displacements and velocities. I cannot properly get rid of the sine and cosine terms. I'm always "off" by a minus sign somewhere. I thought I solved it once, and it was close, but the signs were slightly off from the correct answer.

Any help would be appreciated!

[Borodog]

I think your problem must be overspecified. It only takes 3 quantities to describe SHM (for example, amplitude, frequency, and phase). You have 4. Something seems amiss.

Edit: Nevermind. You're not trying to describe SHM.

I'm still not sure I agree with that expression, though.

[Matt R.]

me neither [img]/images/graemlins/frown.gif[/img]

Here was my method: I used the general solution of the differential equation describing SHM of the form:

x(t) = A cos wt + B sin wt

Where A and B are real. I've played around with the expressions using the complex exponential and a few other 'forms'... but I can't get those to work either.

Anyway, I took the particle at x1 to be its initial displacement at t=0. Using the expression for x(t) gives x1 = A. Differentiating gives its velocity to be x1' = Bw.

Here's where I'm iffy... I took the position of x2 to be the equilibrium position... so t = pi/(2*w) [edit]. Thus x2 = B and x2' = -Aw.

I just substituted these values for A and B back into the original equation for x(t). Since they describe the same motion, I set them equal to one another (as you can see from my equality). There was a small section in my analytical mechanics text that did something very similar, but only with one displacement... x0 and x0'. I think this method should work but I'm not sure....

[Borodog]

[ QUOTE ]
me neither [img]/images/graemlins/frown.gif[/img]

Here was my method: I used the general solution of the differential equation describing SHM of the form:

x(t) = A cos wt + B sin wt

Where A and B are real. I've played around with the expressions using the complex exponential and a few other 'forms'... but I can't get those to work either.

Anyway, I took the particle at x1 to be its initial displacement at t=0. Using the expression for x(t) gives x1 = A. Differentiating gives its velocity to be x1' = Bw.

Here's where I'm iffy... I took the position of x2 to be the equilibrium position... so t = pi/(2*w) [edit]. Thus x2 = B and x2' = -Aw.

I just substituted these values for A and B back into the original equation for x(t). Since they describe the same motion, I set them equal to one another (as you can see from my equality). There was a small section in my analytical mechanics text that did something very similar, but only with one displacement... x0 and x0'. I think this method should work but I'm not sure....

[/ QUOTE ]
Ok, there's no way this is right. [img]/images/graemlins/wink.gif[/img]

First of all, your equation for x(t) does not describe simple harmonic motion, because you have two different amplitudes (A & B). What you want is something of the form:

x(t) = Acos(w(t-t0))

The A is the amplitude, the w (should be an omega, as I'm sure you know) is the angular frequency, and the t0 is the initial time (where the oscillator is maximally displaced and has zero velocity).

But you don't need any of that. You should use energy conservation:

Etotal = Ti + Ui = Tf + Uf

Where T is the kinetic energy and U is the potential energy.

T = 1/2 mv^2, U = 1/2 kx^2

The factors of 1/2 all cancel out. You can divide out by m, and get:

v2^2 + w^2 x2^2 = v1^2 + w^2 x1^2.

Just solve this for omega and you're done:

w^2 = (v1^2 - v2^2) / (x2^2 - x1^2)

Edit: recall that k/m = w^2

[Matt R.]

egad

Thanks Boro. Yep, that's definitely much easier. I kept glossing over conservation of energy because I was so sure the problem called for the displacement expressions (since we were given a displacement/velocity combo).