Total distance in simple harmonic motion

[chococho]

Homework Statement

A mass oscillates along the x-axis in simple harmonic motion. It goes through 200 cycles in 10 seconds and its vibrational amplitude is 0.020 m. What is the frequency, in hertz, and the total distance traveled, in meters, by the mass in the 200 cycles?

Homework Equations

The Attempt at a Solution

I got the frequency but cannot understand how to get the total distance.
The explanation just says "one cycle goes through distance 4A. 200 * 4A = 800A = 16m". Can someone explain this to me please?

 

[Pythagorean]

A is the Amplitude.

 

[chococho]

Yes I know that, but I just don't understand how 800A equals 16.

 

[Pythagorean]

800*.02 = (8*100) * (2/100) = (100/100)*(8*2) = (1)*(16) = 16

 

[HalfLostAndSearching]

In simple harmonic motion, the total distance traveled by the mass in one cycle is equal to 4 times the amplitude (A) of the motion. This is because the mass moves from its equilibrium position to its maximum displacement (A), then back to its equilibrium position, and finally to its maximum displacement in the opposite direction (-A), before returning to its equilibrium position. This constitutes one full cycle, and the total distance traveled is equal to 4A.

In this case, the amplitude is given as 0.020 m, so the total distance traveled in one cycle would be 4 * 0.020 m = 0.080 m. Since the mass goes through 200 cycles in 10 seconds, the total distance traveled by the mass in 200 cycles would be 200 * 0.080 m = 16 m.

Therefore, the total distance traveled by the mass in 200 cycles is 16 meters. I hope this explanation helps to clarify the concept.