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If $x_{1}(t) = \cos(\omega t - \frac{\pi}{6})$ and $x_{2}(t) = \sin(\omega t)$ are two simple harmonic oscillators in the same direction and with the same angular frequency $\omega$, how to calculate the resultant movement $x_{1}+x_{2}$?

What I tried to do First, I wrote $ x_{2}(t) = \cos(\omega t + \frac{\pi}{2})$.

I know superpositions like these might be evaluated using complex numbers: $$z_{1}(t) = e^{i(\omega t - \frac{\pi}{6})}$$ $$z_{2}(t) = e^{i (\omega t + \frac{\pi}{2})}$$

However, I don't know how to work on these complex numbers (probably because I have never seen a example - I just know the theory that harmonic oscillators are projections of uniform circular motions). Can someone help me on that? If someone knows another approach without using complex numbers, it would be also welcome;

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The use of the following identities may prove helpful:

$$\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B$$ $$\sin(A \pm B) = \sin A \cos B \pm \sin B \cos A$$

For example:

$$\cos(\omega t - \pi / 6) = \frac{\sqrt{3}}{2}\cos(\omega t) + \frac{1}{2}\sin(\omega t)$$

Then you have a sum of sines and cosines, which you can use the identities again to turn in a sine or cosine term with a phase offset and an amplitude other than one, usually.

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  • $\begingroup$ And an amplitude of the combined wave of possibly greater than one... $\endgroup$ – hft Mar 29 '15 at 22:38
  • $\begingroup$ I've made the corresponding change to my answer $\endgroup$ – Involutius Mar 29 '15 at 22:42
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Maybe the graphical solution for the Real-parts helps (f=1Hz): plot Realpart

Notice, your x2(t) conversion to cosine is wrong.

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