# Simple Harmonic Motion formula

1. Nov 22, 2005

### raven101

Hi. I've got a problem here about Simple Harmonic Motion. There is "formula" in our physics coursebook for the distance travelled by an object : s=4AN=4At/T
s-distance travelled
A-amplitude
N-number of cycles
t-time
T-period

Is it true for every case or just for N=1, 2 , 3 ?
I guess it is true for N=1, 2 , 3 but not sure, can u explain it please?

P.S Sorry for mistakes during translation

2. Nov 22, 2005

### mezarashi

The ultimate challenge to you here is to derive the formula!

What is the distance travelled by an object in 1 complete cycle or period (i.e. t = T)? Describe this in terms of the oscillation amplitude.

Then go for two, and three?

3. Nov 22, 2005

### El Hombre Invisible

In SHM the amplitude is constant, so yes it is true for all N and indeed any fractional value of N (i.e. a half cycle covers a distance of 2A), assuming it is actual distance travelled rather than displacement being measured and s = 0 and N = 0 at t = 0.

As for why it is true, consider an oscillator at equilibrium at t = 0. It will travel as far as it can from equilbrium (its amplitude, so s = A), then back down to equilibrium (A again, so total s = 2A), then its amplitude in the other direction (total s = 3A), then back to equilibrium (total s = 4A) making a cycle (N = 1). In SHM, each subsequent cycle will be the same as the 1st, so the total distance travelled will be 4A times the number of cycles.

The period T is the time taken to go through one cycle, so the total time taken t divided by T will tell you how many cycles have occurred, hence N = t/T.

Why did you think N = 1, 2, 3 would be special?

4. Nov 22, 2005

### raven101

In SHM "v" and "a" are not constant so i thought it is not true to find distance for example in t=T/6 T/5 and so on

5. Nov 23, 2005