The displacement of a particle executing simple harmonic motion is given by $\mathrm{y} = {\mathrm{A}}_0+\mathrm{Asin} \mathrm{\omega t}+\mathrm{B} \cos \mathrm{\omega t}$. Then the amplitude of its oscillation is given by

$\mathrm{y} = {\mathrm{A}}_0+(\mathrm{A} \sin \mathrm{\omega t} + \mathrm{B} \cos \mathrm{\omega t})$

    = $A_0+A\sin \omega t+B \sin (\omega t+\frac{\mathrm{\pi }}{2})$

Hence 2 SHMs are superimposed with phase difference of $\frac{\mathrm{\pi }}{2}$

Amplitude a = $\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos}∆\varnothing }$

As, $∆\varnothing = \frac{\mathrm{\pi }}{2} \mathrm{Hence} \mathrm{a} = \sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2}$

$\mathrm{y} = {\mathrm{A}}_0+\sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2} \sin \left(\mathrm{\omega t}6\mathrm{\varphi }\right)$

${\mathrm{A}}_0 \mathrm{is} \mathrm{mean} \mathrm{position}, \mathrm{and} \sqrt{{\mathrm{A}}^2+{\mathrm{B}}^2} \mathrm{is} \mathrm{amplitude}$