Two S.H.Ms. ${\mathrm{s}}_1 = \mathrm{a} \sin \mathrm{\omega t} \mathrm{and} {\mathrm{s}}_2 = \mathrm{b} \sin \mathrm{\omega t}$ are superimposed on a particle. The ${\text{s}}_1 \mathrm{and} {\mathrm{s}}_2$ are along the directions which make angle 37° with each other.

Resultant displacement will be the vector sum of two displacements:

$\mathrm{S} = \sqrt{{{\mathrm{S}}_1}^2+{{\mathrm{S}}_2}^2+2{\mathrm{S}}_1{\mathrm{S}}_2\cos 37°}$

    = $\sqrt{a^{2}si{n}^2\omega t+b^{2}si{n}^2\omega t+2ab si{n}^2\omega t\times \frac{4}{5}}$

S = $\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+\frac{8\mathrm{ab}}{5}}\sin \mathrm{\omega t}$

Which shows that the particle will perform SHM.