Deriving the equation of motion of an harmonic oscillator from the equation of motion of a mass-spring system

Question

During my physics 1 course I stumbled upon this problem (our professor left it as an exercise). Basically I have to prove that $$A\cos(\omega t) + B\sin(\omega t) = C\cos(\omega t + \varphi )$$ and find $C$,$\varphi$ in function of $A$,$B$. He told us as a hint that we can use the trigonometric identity $$\cos(\alpha + \beta) = \cos\alpha \cos\beta - \sin\alpha \sin\beta$$ I really don't understand how to approach this problem. The most promising way I found is to apply the identity to the r.h.s., that is to say: $$C\cos(\omega t + \varphi ) = C(cos\omega t cos\varphi - sin\omega t sin\varphi)$$

Expliciting $C$: $$C = \frac{Asin\omega t + Bcos\omega t}{cos\omega t cos\varphi - sin\omega t sin\varphi}$$ But now I'm stuck: how can I express $C$ in terms of only $A$ and $B$?

Answer 1

Hint.

We have

$$ \cos(\omega t)\frac AC + \sin(\omega t)\frac BC = \cos(\omega t)\cos\varphi-\sin(\omega t)\sin\varphi $$

Answer 2

It is true that you need to apply the trigonometric identity to expand the r.h.s. of the given equation. But the way you tried to make $C$ the subject is wrong. You should have proceeded as shown below.

$$A\cos\left(\omega t\right) + B\sin\left(\omega t\right) = C\cos\left(\omega t + \varphi \right) = C\left(\cos\left(\omega t\right) \cos\left(\varphi\right) \right)- C\left(\sin\left(\omega t\right) \sin\left(\varphi\right) \right).$$

Now compare the coefficients of $\cos(\omega t)$ and $\sin(\omega t)$ terms on l.h.s. with those on the r.h.s. respectively to obtain the following pair of relations.

$$C\cos(\varphi ) = A $$ $$C\sin(\varphi ) = -B $$

When you solve this system of equations simultaneously, we get, $$C=\sqrt{A^2+B^2} \qquad\text{and}$$ $$\tan\left(\varphi\right)=-\dfrac{B}{A}\quad\rightarrow\quad \tan\left(180^o-\varphi\right) =\dfrac{B}{A} \quad\rightarrow\quad \varphi = 180^o - \tan^{-1}\left(\dfrac{B}{A}\right)$$