Solving the SHM Calculus Twister

[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:

\begin{equation}<br /> x(t) = A \cdot sin( \omega t + \phi )\end{equation}

And that:

\begin{equation}<br /> dx(t)/dt = A \omega \cdot cos( \omega t + \phi )\end{equation}

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

 

[eJavier]

Ain't this same problem answered somewhere else on this same forum?

 

[Pi]

You need to write an equation relating displacement to velocity, which doesn't have any mention of time. Then, by plugging in the two pairs of displacement/velocity values, you get yourself two simultaneous equations in two unknowns (you can take the phase constant to be zero without loss of generality).

 

[nrqed]

The way to relate position to velocity without mention of time is through conservation of energy. Recall that

1/2 k x^2 + 1/2 m v^2 = 1/2 k A^2

where A is the amplitude and v, x are the velocity (or speed) and position at any time (I mean they must be taken at the same value of "t" but "t" can be anything).

Divinding by 1/2 and by m and using k/m = omega^2, you get

omega^2 x^2 + v^2 = omega^2 A^2

So knowing v and x at two different times allows you to find A and omega (they are positive values by definition so there is only one root allowed).

Pat

 

[noppakhuns]

The angular frequency is equal to omega, and the ammplitude is equal to A.