Simple Harmonic Motion formula

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SUMMARY

The formula for the distance traveled by an object in Simple Harmonic Motion (SHM) is given by s = 4AN, where s is the distance, A is the amplitude, N is the number of cycles, t is the time, and T is the period. This formula holds true for any integer value of N, including N = 1, 2, 3, and extends to fractional cycles, as the amplitude remains constant. The total distance traveled in one complete cycle is 4A, and this pattern repeats for each subsequent cycle. The relationship between time and cycles is defined as N = t/T.

PREREQUISITES
  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with the concepts of amplitude and period
  • Basic knowledge of oscillatory motion
  • Ability to interpret mathematical formulas in physics
NEXT STEPS
  • Study the derivation of the SHM distance formula in detail
  • Explore the relationship between velocity and acceleration in SHM
  • Learn about the implications of fractional cycles in oscillatory motion
  • Investigate resources on advanced SHM topics, such as damping and resonance
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Students of physics, educators teaching SHM concepts, and anyone interested in understanding the mathematical principles behind oscillatory motion.

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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