Why doesn't amplitude change period?

What an odd and interesting question. You are clearly thinking outside the box, and you've given me something to think about. If you'd been in one of my classes, what would I have said?

No, radius does not affect (with an "a" when it's the verb) the period. The period is about the timing, Remember playground merry-go-rounds? Or amusement park merry-go-rounds? The thing turns as fast as it turns, and if you lean way out you definitely are going faster in terms of covering more distance in the same time, measured as distance along the circumference, but the ANGULAR rate of change stays the same.

As you rightly state, the amplitude is only dependent on the radius of the circle. The amplitude is the maximum/minimum value of the graph's displacement from zero. Thus for a circle of radius R, the amplitude of the graph of sin(Ó¨) (resp. cos(Ó¨)) is +/- R.

And as you probably know, the amplitude is reached when Ө = (2k+/-1)π/2 for sin(Ө) (resp. Ө = +/- 2kπ for cos(Ө)).

However, consider the hour hand and minute hand moving around a clock of radius R. For the sake of argument, we'll assume in this instance that the length of each hand is equal to the radius of the clock's face.

Now if you were to graph the sine function of the movement of each hand as they travel around the clock (Start at 12 o'clock, say), the waveform of both graphs would have the same amplitude. However, since the minute hand will have completed 12 revolutions around the clock before the hour hand completes one, the period of the minute hand is one 12th that of the hour hand. In other words, it is the frequency or angular speed which determines the period. The slower the angular speed, the larger the wavelength or period.

Using the analogy of speed x time = distance

Let ωt = θ

where ω (Lower case Greek character "omega") is the angular speed of the minute hand. That is, travelling at angular speed ω, it takes t minutes to traverse an angle θ. Therefore in terms of minutes, the time it takes for the hand to complete one revolution ( θ = 2π) is

T = 2π/ω minutes

which is obviouly 60 minutes.

So the angular speed of the minute hand is

ω = 2π/60 = π/30 radians per minute.

Note that when dealing with angular speed (or really, velocity for that matter, which is speed and direction), we are not concerned about the radius of the clock - the minute hand on Big Ben has the same angular speed as the minute hand on your clock. The tips of the hands have different speeds which ARE determined by the radius, but their angular speeds are the same. In this case, Big Ben's minute hand would have to traverse a larger circumference than the minute hand of your clock, so it's speed in terms of distance around the clock is Rωt, which is what you might have been confused with.

Anyway.....

on the other hand (excuse the pun!) the hour hand takes 12 hours = 720 minutes to complete one revolution.

So, if we denote the angular speed of the hour hand by Ω (Upper case Greek character "omega"), then

Ω = 2π/720 = π/360 radians per minute

What this all means is that for every cycle (period) of the sine curve traced out by the hour hand, the minute hand would have traced out 12 cycles. That their amplitudes are the same has no bearing on the wavelength or period.

Since you used the radius analogy, let's take two wheels - one large, one small. Let's put a dot on each wheel's circumference, and then rotate each wheel vertically (axes horizontal) next to each other, at the same RPM.

If you plot the vertical position of the two dots, they would be going up and down the same number of times a second, but the amounts of vertical travel will be different for the two dots. There you have it - the same period, but different amplitude.

Got it ?

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