# Clarification on position equation for simple harmonic motion

1. Apr 6, 2010

### darryw

I have 4 quick questions about the position wave equation, x = Acos(omega*t + delta), can someone please explain this (in very simple language) thanks!

1. Why do we have to use cos? why not sin? wouldnt sine be better so that amplitude starts at zero also? using cos must make things easier, but i dont see why?

2. Is "width" the distance between any two peaks or troughs of the wave? Is this the same as frequency of wave? (if its now, then what is the correct term for width?)

3. doesnt A gradually decrease over time ( this is obvious from intuition) and if so, then why are the waveforms always drawn with uniform amplitude? (is this just an idealized wave (no friction or damping)? AKA: SIMPLE harmonic motion?

4 how do we differentiate between small greek letters and big? for eg, delta looks totally different when small vs big, so what do we say "small delta" or "capital delta" or what? seems confusing..

thanks for any help

2. Apr 6, 2010

### James Leighe

1) I have no idea. One possibility is that Re(exp(i*theta)) is cos(theta) and when you are doing superpositions you can use the identity cosA+cosB=2cos(a+b/2)cos(a-b/2) to add your two waves like thus:

exp(i(A)) + exp(i(B)) = 2exp(i(A+B/2))exp(i(A-B/2)) = 2exp(i(A+B/2)+(A-B/2))
(or something like that)

Also, when you do superpositions, often the greatest amplitude is right at the zero point and gradually lessens as you go further.

(Could someone else chime in on this one?)

2) Yes, but it's called wavelength.

3) Not in simple harmonic motion, since the definition of which is motions with no dampening or 'driving' (thanks wiki).

4) No idea. I think most people would call a capitol delta a delta the same as a lowercase delta.

3. Apr 6, 2010

### bp_psy

You don't have to use the cos function ,The sin function can describe the motion also. The use of cos is more standard because cos(0)=1 which makes a lots of things cleaner.
It is called wavelength. Note that SHM is not the description of a wave just of an oscillator.Frequency is a measure of how many complete cycles the oscillator has made in a set unit of time .
SHM is an idealization. In reality there is always some damping. You study SHM first because it is easy and useful.if you take a more advanced class in waves you will study the effects of damping.
Small deltas are used for partial derivatives. Capital Deltas are used to express small nonzero changes or displacements. Nobody really says delta. you just write it down and expect everyone to Know what you want to say.

4. Apr 6, 2010

### darryw

Thanks for replies.. just have another question about the correct symbol for frequency:

textbook says frequency of a spring is omega = root k/m.
but then it says frequency is the inverse of the period, written like this: f = 1/T
please explain why they use both symbols to mean frequency?? thanks

5. Apr 6, 2010

### bp_psy

Omega is the angular frequency. it represents how many complete rotation cycles your vector has made per unit time. $$\omega=2\pi\cdot f$$
$$\ T=1/f$$
Look here for more details
http://hyperphysics.phy-astr.gsu.edu/HBASE/hframe.html

6. Apr 6, 2010

### Naty1

"doesnt A gradually decrease over time ( this is obvious from intuition) and if so, then why are the waveforms always drawn with uniform amplitude? (is this just an idealized wave (no friction or damping)? AKA: SIMPLE harmonic motion?'

The answer given above is correct...if you want a gradually decreasing amplitude, one uses a term like
e-t or 1/t as an example...

7. Apr 6, 2010

### darryw

Is all of this correct (this is in my own words):

In SHM graph of an oscillating spring, when the y-axis is positive the spring is stretching beyond its EQ point.
The EQ point is where the spring comes to rest with no forces acting on it .
When y-axis is negative, the spring is compressing.
Amplitude represents the spring's max and min displacement from the EQ point.
It is convention to set t=0 at the waveforms max amplitude, this is why we use cosine in the position equation.

8. Apr 7, 2010

### James Leighe

Yup, that's right.

One thing tho, it doesn't matter (mathematically, in this case) if you consider the spring stretching or compressing at the positive y-axis. But you have the idea.

9. Apr 7, 2010

### bp_psy

Yes.
The mass on the spring is not at rest at the equilibrium. Only the accelerations is 0.
Yes but this is due to the choice of coordinates.
Yes.