Exam

JEE Main

Year

2025

Date

2 April

Shift

Morning Shift

Subject

Physics

Chapter

Oscillations

JEE Main 2025 · 2 April · Morning Shift · Physics · Oscillations

A particle is subjected two simple harmonic motions as : x_1= 7 5...

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 :

Step-by-step solution

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

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