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