How Does Amplitude Affect Total Distance Traveled in SHM?

[Shiina-kun]

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:

$$V$$= $$\pm$$$$V{o}$$$$\sqrt{1 - x^{2}/A^{2}}$$

$$V$$=$$A\sqrt{k/m}$$

x=$$Acos2$$* $$\pi$$(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?

 

[la_diabla]

A=4(pi)r^2
so r=sqrt(A/4pi)

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

 

[baba89]

Hi there,

Based on the information given, it seems like you are dealing with simple harmonic motion (SHM), where the particle moves back and forth in a periodic manner. In this case, the total distance traveled in one period is equal to the amplitude of the motion, which in this case is 0.25m. This is because in SHM, the particle moves from its equilibrium position (where x=0) to its maximum displacement (amplitude) and back to the equilibrium position in one period.

Therefore, the formula you need to use is x=Acos(2πft), where x is the displacement, A is the amplitude, f is the frequency, and t is time. In this case, you are given the amplitude (0.25m) and you can assume a value for the frequency (e.g. 1Hz or 2π rad/s) to solve for the total distance traveled.

I hope this helps! If you are still having trouble, I suggest discussing it with your teacher or seeking help from a tutor or classmate. Good luck with your homework!