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?
Best Answer
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: $$c_1\cos\omega_0 t+c_2\sin\omega_0 t=R(\cos\omega_0t\cos\delta+\sin\omega_0t\sin\delta)=R\cos(\omega_0t-\delta)$$