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I was wondering if maybe someone could look at this excerpt from a textbook (Attached). It states that “the displacement of the two masses will be in opposite directions (out of phase by pi)” but I thought that the Definition of Phase was: Phase = theta in the following formula, cos(wt+theta).

However, the two equations in this book excerpt have the exact same expression in parenthesis, and both are nested in the same function, cosine - so how is it that they have different phases? Many thanksSee image of book example attached.

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2 Answers 2

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Here's a simple GIF to complement @garyp's succinct answer:

enter image description here

I've plotted the two functions $-\cos{(\sqrt{5}t)}$ (blue) and $\cos{(\sqrt{5}t + \phi)}$ (orange), and I've varied $\phi$ between 0 and $2\pi$. As you can see, they start off being exactly exactly reflected, as you'd expect, and now as I tune $\phi$, the orange curve moves "closer" to the blue one until it matches it exactly when $\phi = \pi/2$, as one would expect.

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  • $\begingroup$ I was expecting an animation when you mentioned "GIF". Oh well... $\endgroup$
    – wyphan
    Dec 4, 2020 at 20:07
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    $\begingroup$ @wyphan It's animated for me! Doesn't it work for you? :'( $\endgroup$
    – Philip
    Dec 4, 2020 at 20:17
  • $\begingroup$ Oh, it works now! Must be my ISP messing around with me again... $\endgroup$
    – wyphan
    Dec 4, 2020 at 20:18
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    $\begingroup$ @wyphan Ok it turns out that the GIF I made didn't loop, so you basically see it once and then you don't see it again for a while :P Just made one that loops, so it should be better! :) $\endgroup$
    – Philip
    Dec 4, 2020 at 20:22
  • $\begingroup$ Wow thank you so much - what an awesome animation! What programming did you use to do that? $\endgroup$
    – Yelena
    Dec 4, 2020 at 21:35
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$$-4\cos{(\sqrt{5}t)} = +4\cos{(\sqrt{5}t + \pi)}$$

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  • $\begingroup$ That’s interesting and very helpful! But I’m interested to know why this works? If we were to graph both equations (y=4cos(sqrt(5)t) and y=-4cps(sqrt(5)t), we’d find that the negative one was simply reflected over the t axis, yes? So then, why the phase shift of pi? It doesn’t look like a phase shift (as in one has y=0 at origin vs. another one at y=pi)? $\endgroup$
    – Yelena
    Dec 4, 2020 at 19:33
  • $\begingroup$ @Yelena being "reflected" is one way to see it, but how about you "shift" the second equation until it exactly overlaps the first: now, see how far you had to shift it in order for this to happen. $\endgroup$
    – Philip
    Dec 4, 2020 at 19:35
  • $\begingroup$ Thanks so much, that is so interesting! $\endgroup$
    – Yelena
    Dec 4, 2020 at 19:37

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