A particle is executing SHM $x=3\cos \omega t+4\sin \omega t$. Find the phase shift and amplitude.

$x=3\cos \omega t+4\sin \omega t$
lets take
$x_0\cos \varphi =4$ ............$(I)$
and $x_0\sin \varphi =3$ ............$(II)$
Now the equation becomes
$$\begin{gathered} x=x_0\sin \varphi \cos \omega t+x_0\cos \varphi \sin \omega t \\ \Rightarrow x=x_0\left(\sin \varphi \cos \omega t+\cos \varphi \sin \omega t\right) \\ \Rightarrow x=x_0\sin \left(\omega t+\varphi \right) \end{gathered}$$
Squaring and adding $(I)$ and $(II)$ we get
$$\begin{gathered} \Rightarrow {x_0}^2({\cos }^2\varphi +{\sin }^2\varphi )=4^2+3^2=25 \\ \Rightarrow x_0=5 \end{gathered}$$
dividing $(II)$ by $(I)$, we get
$$\begin{gathered} \tan \varphi =\frac{3}{4} \\ \Rightarrow \varphi =37^{\circ } \end{gathered}$$
$\Rightarrow x=5\sin \left(\omega t+37^{\circ }\right)$ which represents a SHM of amplitude of 5 units and a phase of $37^{\circ }$