- #1

- 3

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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

- Thread starter raven101
- Start date

- #1

- 3

- 0

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

mezarashi

Homework Helper

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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

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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

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- #5

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Do you know any site or other source hat i could get more information about this?

- #6

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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.raven101 said:

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