# Simple Harmonic Motion without damping

1. Nov 4, 2012

### danielu13

So, simple harmonic motion without damping is described generally by
$x(t) = Acos(\omega*t +\delta)$

Which is derived from the differential equation
$x''+\frac{k}{m}x = 0$

We know that
$A = \sqrt{c_1^2+c_2^2}$

and

$tan\delta = \frac{c_1}{c_2}$

With the differential equation, dealing with an initial condition is relatively easy, but it does not work as easily if using the generalized equation. Is there a way of making a relationship between $A$ and $\delta$? I've worked with the equations a bit and can't find anything, but thought someone on here might know something different.

2. Nov 4, 2012

### Hetware

I'm sorry, but you didn't define your variables $c_1,c_2$.

3. Nov 4, 2012

### danielu13

Those are constants in the solution to the differential equation that are determined by the initial conditions.

4. Nov 5, 2012

### nasu

Not sure what you are trying to do.
The solution expressed in terms of c1 and c2 is a general solution too.
The one you wrote is just a different way to express the general solution.
In both cases you have two independent parameters (c1,c2 or A, δ).

5. Nov 5, 2012

### danielu13

In the problems that I have dealt with, I've been given the A value and had to calculate $\delta$. I was trying to see if there is a simple way to relate it without solving every time, but the more I think about it, the less likely it seems.

6. Nov 5, 2012

### Hetware

I don't get the question either. Is it the inverse tangent that is the problem? You actually have shown a relationship between A and δ. In general you can (so far as I know) not get exact values for arctan δ. It is a transcendental function.

Someone please correct me if I am wrong. It's been years since I looked at this in depth.

I asked about the constants of integration becaus I wanted to know how you arrived at them. It will not always be the case that you constants of integration work as you have presented.

7. Nov 5, 2012

### nasu

For a given A, δ may have all kind of values, depending on the initial conditions.
Imagine that you can have a pendulum oscillating with the same amplitude but either starting from the maximum displacement (and zero v) or starting from equilibrium position but having initial velocity. A would be the same in both cases but δ will not.
So you cannot find δ as a function of A only.

8. Nov 5, 2012

### danielu13

Okay, that's what I was thinking but I saw that both A and $\delta$ were related to the constants, and thought there may be some way to relate the two.

9. Nov 6, 2012

### Hetware

Are you familiar with the use of complex numbers to arive at the expressions you presented?