Position vs Time Graph: Simple Harmonic Motion

[Dorian]

Homework Statement

[see attached photo]

I seek specific help with (a) only. The answers to this question are provided in the back of the textbook, so I know the answers (I hope).

Homework Equations

$x(t)=Acos(\omega t+\phi _{0}),$

$v_{x}(t)=-A\omega sin(\omega t+\phi _{0})=-v_{max}sin(\omega t+\phi _{0}),$

$v_{max}=\frac{2\pi A}{T}$

The Attempt at a Solution

For (b), I got $v_{x}(0)=13.6 \frac{cm}{s}$

For (c), I got $v_{max}=15.7 \frac{cm}{s}$

Both of these answers are correct, according to the back of the textbook

For (a) (the phase constant), however, the back of the book says the correct answer is $\phi _{0}=-\frac{2\pi}{3}$

I got: $\frac{1}{2}A=Acos(\phi _{0})\Rightarrow cos^{-1}(\frac{1}{2})=\phi _{0}=\pm \frac{\pi}{3}$, for which I got $-\frac{\pi}{3}$ since it's moving to the right at $t=0 s$

With this answer, I was able to acquire the right answers for (b) and (c). Furthermore, I was able to accurately graph the same graph provided in the text using my answer, but not so with the answer given in the textbook. I'm lost, really. Can someone please help point something out that I'm missing?

 

[haruspex]

I agree with your solution. As a check, if we go 1/6 of a wavelength to the right of the origin we hit the peak. cos(π/3-π/3)=cos(0)=1.

 

[neilparker62]

Arguably one can show equivalence between $cos(ωt-π/3)$ and $-sin(ωt-2π/3)$

 

[haruspex]

Arguably one can show equivalence between $cos(ωt-π/3)$ and $-sin(ωt-2π/3)$

The standard seems to be the cos form. It certainly appears that this is what the OP has been taught.

 

[neilparker62]

The standard seems to be the cos form. It certainly appears that this is what the OP has been taught.

Yes. Anyway I think I've got that wrong - too used to working in degrees! $$cos(ωt-π/3) = sin(5π/6-ωt)=-sin(ωt-5π/6)$$

 

[Dorian]

Thank you both for your help! And yes, the cos is what's in the textbook, although I remember from a trigonometry textbook I used once upon a time that both cos and sin can work.

 

[haruspex]

both cos and sin can work.

Sure, but in order to define the phase you need to know which is to be used.

 

[Dorian]

Sure, but in order to define the phase you need to know which is to be used.

I'm aware :) Thanks!

I'm more concerned that the textbook had a wrong answer (if this is in fact the case), which made me question my understanding in an unproductive way.

 

[neilparker62]

In essence a cos graph is essentially a sin graph left shifted $π/2$ radians. Thus we may right the following equivalences:

$$ cos(wt + ∅) = sin(wt + ∅ + π/2) $$ and $$ sin(ωt + ∅)=cos(ωt+∅-π/2) $$