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The differential equation: $y''+\omega^2 y=0$ has as a general solution: $$y=A\cos{(\omega t)}+B\sin{(\omega t)}$$

By taking: $$A=R\cos{(\omega t_0)}$$ and $$B=R\sin{(\omega t_0)}$$ We can rewrite the general solution into: $$y=R\cos{(\omega(t-t_0))}$$

However, this is also a solution: $$y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$$

QUESTION:

I understand we can rewrite this last solution, $y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$ in the form of $y=A\cos{(\omega t)}+B\sin{(\omega t)}$ (using some trig identities)

But... Can we also rewrite $y=\alpha\cos{(\omega(t-t_0))+\beta\sin{(\omega(t-t_0))}}$ in the form of $y=R\cos{(\omega(t-t_0))}$? If so, how?

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  • $\begingroup$ Yes/no. Yes provided your final answer has a different $t_0$. Your two-stage process works, first translate it to $A\cos\omega t+B\sin\omega t$, then translate that into $C\cos\omega(t-t_1)$. $\endgroup$ – almagest Apr 21 '16 at 8:22
  • $\begingroup$ The point is that I noticed that $\alpha\cos{\omega(t-t_0)}$ looks like $R\cos{\omega(t-t_0)}$ so I was wondering whether $\beta\sin{\omega(t-t_0)}$ can be rewritten in the form of $\alpha\cos{\omega(t-t_0)}$ $\endgroup$ – GambitSquared Apr 21 '16 at 8:45
  • $\begingroup$ @almagest Can you show me how it's done? $\endgroup$ – GambitSquared Apr 21 '16 at 8:50
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Let $f(t)=\alpha\cos\omega(t-t_0)+\beta\sin\omega(t-t_0)$. Then as usual we have $f(t)=\alpha\cos\omega t_0\cos\omega t-\alpha\sin\omega t_0\sin\omega t+\beta\sin\omega t_0\cos\omega t+\beta\cos\omega t_0\sin\omega t$ $=(\alpha\cos\omega t_0+\beta\sin\omega t_0)\cos\omega t-(\alpha\sin\omega t_0-\beta\cos\omega t_0)\sin\omega t$ $=\gamma\cos\omega t-\delta\sin\omega t$, where $\gamma=\alpha\cos\omega t_0+\beta\sin\omega t_0,\delta=\alpha\sin\omega t_0-\beta\cos\omega t_0$ are constants.

Now we use the reverse procedure. Put $h=\frac{\gamma}{\sqrt{\gamma^2+\delta^2}},k=\frac{\delta}{\sqrt{\gamma^2+\delta^2}}$ and take $t_1$, so that $h=\cos\omega t_1,k=\sin\omega t_1$. Let $R=\sqrt{\gamma^2+\delta^2}$. Then we have $f(t)=R(\cos\omega t\cos\omega t_1-\sin\omega t\sin\omega t_1)=R\cos\omega(t-t_1)$.

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  • $\begingroup$ Thanks! And also $R=\sqrt{\gamma^2+\delta^2}=\sqrt{\alpha^2+\beta^2}$ $\endgroup$ – GambitSquared Apr 24 '16 at 22:52
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We can also do it directly:

$$y=\alpha\cos(\omega(t-t_0))+\beta\sin(\omega(t-t_0))$$ Take:

$$\alpha=R\cos{\omega t_k}$$

$$\beta=R\sin{\omega t_k}$$

Where:

$$R=\sqrt{\alpha^2+\beta^2}$$

$$\omega t_k=\arctan\frac{\beta}{\alpha}$$

Then:

$$y=R\cos(\omega(t-t_0-t_k))$$

Take:

$$t_1=t_0+t_k$$ Then: $$y=R\cos(\omega(t-t_1))$$

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