A particle is subjected two simple harmonic motions as :
\(\mathrm{x}_1=\sqrt{7} \sin 5 \mathrm{tcm}\)
and \(x_2=2 \sqrt{7} \sin \left(5 t+\frac{\pi}{3}\right) \mathrm{cm}\)
where x is displacement and \(t\) is time in seconds. The maximum acceleration of the particle is \(\mathrm{x} \times 10^{-2} \mathrm{~ms}^{-2}\). The value of x is :

\(\begin{aligned} & \mathrm{x}_1=\sqrt{7} \sin 5 \mathrm{t} \\ & \mathrm{x}_2=2 \sqrt{7} \sin \left(5 \mathrm{t}+\frac{\pi}{3}\right)\end{aligned}\)
From phasor,

\(\therefore\) Amplitude of resultant \(\mathrm{SHM}=7\)
\(\begin{aligned} & \phi=\tan ^{-1} \frac{2 \sqrt{7} \times \sqrt{3} / 2}{\sqrt{7}+2 \sqrt{7} \times \frac{1}{2}}=\tan ^{-1} \frac{\sqrt{21}}{2 \sqrt{7}}=\tan ^{-1} \frac{\sqrt{3}}{2} \\ & \therefore \quad \mathrm{X}_{\mathrm{R}}=7 \sin (5 \mathrm{t}+\phi) \\ & \quad \mathrm{a}_{\mathrm{R}}=-7 \times 25 \sin (5 \mathrm{t}+\phi) \\ & \therefore \mathrm{a}_{\max }=175 \mathrm{~cm} / \mathrm{sec}=175 \times 10^{-2} \mathrm{~m} / \mathrm{sec}\end{aligned}\)