Free,Undamped Mechanical Vibrations

Question

In the case of free, undamped vibrations,the differential equation is $mu''+ku=0$ and solution to this differential equation is

\begin{align} \tag{1} u(t)=c_1\cos{(\omega_0 t)}+c_2\sin{(\omega_0 t)} \end{align}

Now this has been written by the author in the following form,

\begin{align} \tag{2} u(t)=R\cos{(\omega_0 t-\delta)} \end{align}

where he says R is the amplitude of displacement $u(t)$ and $\delta$ is the phase shift or phase angle of displacement $u(t)$

Now why did he assume that both the equations $(1)$ and $(2)$ are equivalent and on what ground?

Answer 1

One can use $$c_1=R\cos\delta\\c_2=R\sin\delta$$ or one can go the other way around $$R=\sqrt{c_1^2+c_2^2}\\ \delta=\arctan{c_2/c_1}$$ Then: $$c_1\cos\omega_0 t+c_2\sin\omega_0 t=R(\cos\omega_0t\cos\delta+\sin\omega_0t\sin\delta)=R\cos(\omega_0t-\delta)$$

Answer 2

If $c_1=c_2=0$ you may take any $\delta$ and $R=0$. Otherwise put $R:=\sqrt{c_1^2+c_2^2}$ and observe that $d_1^2+d_2^2=1$ where $d_i:=\frac{c_i}{R}$. Thus there ist some $\delta$ auch that $d_1=\cos\delta$ and $d_2=\sin\delta$. Finally note that $\cos\delta \cos(\omega_0 t)+\sin\delta \sin(\omega_0 t)=\cos(\omega_0 t-\delta)$ by the subtraction formula of $\cos$.