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I am not understanding what I am doing wrong here. Please see the question below:

Two harmonic oscillators are made using identical springs with identical masses and are set up side by side. You notice that they cross (i.e., have the same position) when they are moving in opposite directions and both are at half of their maximum amplitude. What is the phase difference between the two oscillators?

Note: $\phi \in [\pi,2\pi].$

Now, I defined the phase difference to simply be the difference in phase offset between the two oscillators.

I made a picture where they were both at half of their maximum amplitude, and I was quite convinced that if the first oscillator (call it oscillator A) is moving down (say) from the top of the motion, it would have a phase offset of $\phi_1 = \frac{\pi}{4}$. Similarly, the second oscillator (call it oscillator B) moving upwards towards the top of the motion would have a phase offset of $\phi_2 = \frac{7\pi}{4}$ and so the phase difference between the two would simply be $\phi_2 - \phi_1 = \frac{3\pi}{2}$, which is wrong.

Any hints for this? I really don't understand what I'm doing wrong.

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  • $\begingroup$ The simplest way to visualise the problem is to draw a phasor diagram and then remember that a phase lag of $x$ can be considered as a phase lead of $2\pi-x$. $\endgroup$ – Farcher Apr 3 '19 at 10:41
  • $\begingroup$ @Farcher Thanks! Could you show me a phasor diagram that would be applicable to this situation if you can find one? $\endgroup$ – psa Apr 3 '19 at 16:35
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Consider two values of theta that make sin(theta) = 0.5.

Consider the difference between those two values.

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  • $\begingroup$ I require $\phi \in [\pi,2\pi]$ only. $\endgroup$ – psa Apr 3 '19 at 5:43
  • $\begingroup$ Does taking $\cos(\theta)=0.5$ suffice, since $x(t)=A\cos({\omega}t)$? $\endgroup$ – psa Apr 3 '19 at 5:49
  • $\begingroup$ There are two solutions in that interval and yes, cos is fine too, as is any phase shift as you are interested in the phase difference. $\endgroup$ – Paul Childs Apr 3 '19 at 6:02
  • $\begingroup$ @PaulChilds But $\sin(x)=+0.5$ only for $x_1 = \frac{\pi}{6}$ and $x_2 = \frac{5\pi}{6}$ when $x \in [0,2\pi]$, whereas $x_1 \not\in [0,2\pi]$. $\endgroup$ – psa Apr 3 '19 at 6:18
  • $\begingroup$ $0<\frac{\pi}{6}<\frac{5\pi}{6}<2\pi$ i.e. they are both on that interval and you thence have your answer then of $x_2 - x_1$. $\endgroup$ – Paul Childs Apr 3 '19 at 23:38

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