Simple Harmonic Motion formula

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The formula for the distance traveled in Simple Harmonic Motion (SHM) is s = 4AN, where s is the distance, A is the amplitude, N is the number of cycles, t is time, and T is the period. This formula holds true for all integer values of N, as well as fractional values, since the amplitude remains constant throughout the motion. Each complete cycle results in a distance of 4A, and the total distance can be calculated by multiplying this by the number of cycles. However, the formula is only applicable for specific time intervals that are multiples of T/4. Understanding the relationship between distance and oscillation in SHM clarifies the application of this formula.
raven101
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Hi. I've got a problem here about Simple Harmonic Motion. There is "formula" in our physics coursebook for the distance traveled 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
 
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The ultimate challenge to you here is to derive the formula!

What is the distance traveled 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?
 
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 traveled 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 traveled 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?
 
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
 
Do you know any site or other source hat i could get more information about this?
 
raven101 said:
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
Actually, yes you're right. The right-hand side is only true for values of t that are multiples of T/4. However, this will always be the case for integer values of N, or even multiples of N/4, assuming N = 0 and s = 0 when t = 0. Sorry, I was focusing more on s = 4AN.
 
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