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?


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

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  • $\begingroup$ How the $c_1$ and $c_2$ is calculated? $\endgroup$ – Dhamnekar Winod Oct 20 '18 at 4:34
  • $\begingroup$ You need to have some initial conditions (like position $u(t=0)$ and speed $u'(t=0)$) $\endgroup$ – Andrei Oct 20 '18 at 4:38
  • $\begingroup$ suppose $u(t)=-\frac12 \cos{(6t)}+\frac16 \sin{(6t)}$ and in another form $u(t)=0.52705 \cos{(6t-2.81984)}$ where $R=0.52705$ and $\delta=2.81984$ Now i want the displacement at 5 seconds, so both the equations must be equal. But it is not so. What is wrong? $\endgroup$ – Dhamnekar Winod Oct 20 '18 at 5:16
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    $\begingroup$ I get for both of them $-0.2418$. Make sure you use radians $\endgroup$ – Andrei Oct 20 '18 at 5:35
  • $\begingroup$ If i use $u(t)=R \sin{(\omega_0t+ \delta)} $ shall i get the correct answer? $\endgroup$ – Dhamnekar Winod Oct 20 '18 at 16:37

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 $\cos$.

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