Finding amplitude of oscillation
So I know how to find the amplitude of oscillation. It's just the coefficient of the trig function. In general it would be $$ x = A\sin\left(\omega t +\phi_0\right) $$ and $A$ would be the amplitude. But my problem is asking for the amplitude of $$ x=-\cos(t)+3\sin\left(t-\frac{\pi}{6}\right) $$I don't exactly know what to do if there are 2 trig functions. The amplitude is $3.606$ but I don't know how to get that.
$\endgroup$ 21 Answer
$\begingroup$Expanding the explanations given in the comments by Artem and André Nicolas to the question.
We can always find constants $A,B$ and $C$ such that the equation
\begin{equation*} x(t)=A\cos \omega t+B\sin \omega t \end{equation*} is identical to the equation \begin{equation*} x(t)=C\sin \left( \omega t+\phi _{0}\right) =\left( C\sin \phi _{0}\right) \cos \omega t+\left( C\cos \phi _{0}\right) \sin \omega t. \end{equation*} To expand $C\sin \left( \omega t+\phi _{0}\right) $ we applied the trigonometric identity \begin{equation*} \sin (a+b)=\sin a\cos b+\cos a\sin b. \end{equation*} Equating the coefficients of $\cos \omega t$ and $\sin \omega t$ gives \begin{equation*} A=C\sin \phi _{0},\qquad B=C\cos \phi _{0}, \end{equation*} while squaring and adding these last equations gives \begin{equation*} C=\sqrt{A^{2}+B^{2}}\geq 0. \end{equation*} Dividing one by the other yields \begin{equation*} \tan \phi =\frac{A}{B}. \end{equation*}
In this case we have that \begin{equation*} x(t)=C\sin \left( \omega t+\phi _{0}\right) =-\cos t+3\sin (t-\frac{\pi }{6} )=-\frac{5}{2}\cos t+\frac{3\sqrt{3}}{2}\sin t. \end{equation*} We changed the notation of the amplitude of the wave $x(t)$ to $C$ instead of $A\ $as in the question. We see that $\omega =1$. The expansion of $\sin (t-\frac{\pi }{6})$ follows from the identity \begin{equation*} \sin (a-b)=\sin a\cos b-\cos a\sin b \end{equation*} and the trigonometric values \begin{equation*} \cos \frac{\pi }{6}=\frac{\sqrt{3}}{2},\quad \sin \frac{\pi }{6}=\frac{1}{2}. \end{equation*} From the numeric values \begin{equation*} A=-\frac{5}{2},\qquad B=\frac{3\sqrt{3}}{2}, \end{equation*} we find that the amplitude is \begin{equation*} C=\sqrt{A^{2}+B^{2}}=\sqrt{13}\approx 3. 6056, \end{equation*} the same value of yours. If we wanted to compute the phase angle $\phi _{0}\ $we would get \begin{eqnarray*} \tan \phi _{0} &=&\frac{A}{B}=\frac{-5\sqrt{3}}{9}, \\ \phi _{0} &=&\arctan \frac{-5\sqrt{3}}{9}\approx -0.766\,16\text{ }\mathrm{ rad}\approx -43.898{{}^\circ}. \end{eqnarray*} Graph of \begin{equation*} x(t)=\sqrt{13}\sin \left( t-0.766\,16\right) \end{equation*}
