How Does Amplitude Affect Total Distance Traveled in SHM?

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SUMMARY

The discussion focuses on calculating the total distance traveled by a particle undergoing Simple Harmonic Motion (SHM) with an amplitude of 0.25 meters. The key formula for determining the total distance in one period of SHM is derived from the relationship between amplitude and distance, specifically noting that the total distance is four times the amplitude. Therefore, the total distance traveled in one complete cycle is 1 meter (4 * 0.25m). The participant also referenced various formulas related to SHM, but the essential calculation is straightforward and based on the amplitude alone.

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  • Understanding of Simple Harmonic Motion (SHM)
  • Familiarity with the concept of amplitude in physics
  • Basic knowledge of periodic motion and cycles
  • Ability to manipulate mathematical formulas related to motion
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Vibrations and Waves - please help!

I'm having trouble with my physics homework. I can't seem to figure out which formula to use to solve a particular problem.

"If a particle undergoes SHM with amplitude 0.25m, what is the total distance it travels in one period?"

The formulas I have that deal with amplitude are:
[tex]V[/tex]= [tex]\pm[/tex][tex]V{o}[/tex][tex]\sqrt{1 - x^{2}/A^{2}}[/tex]

[tex]V[/tex]=[tex]A\sqrt{k/m}[/tex]

x=[tex]Acos2[/tex]* [tex]\pi[/tex](frequency)t​
I don't know which formula to use because it seems like I don't have enough information for any of them. Could someone please tell me which formula to use and why?
 
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A=4(pi)r^2
so r=sqrt(A/4pi)

your A=0.25
so:
r=sqrt(0.25m/12.57)=.020m
 

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