A particle is subjected to two simple harmonic motions as: x₁ = √(7) sin 5t cm and x₂ = 2√(7) sin (5t + (π)/3) cm where x is displacement and t is time in seconds. The maximum acceleration of the particle is x × 10⁻² m s ⁻². The value of x is:

Oscillations and Waves: JEE Main 2025 Physics question

JEE Main 2025 · 2 Apr, Morning shift · Q27

JEE Main 20252 Apr, Morning shiftPhysicsMediumOscillations and Waves185× since 2002

A particle is subjected to two simple harmonic motions as :

x_1 = \sqrt{7}\sin 5t

cm and

x_2 = 2\sqrt{7}\sin\left(5t + \frac{\pi}{3}\right)

cm where

x

is displacement and

t

is time in seconds. The maximum acceleration of the particle is

x \times 10^{-2}

m s

^{-2}

. The value of

x

is :

JEE Main 2025 Oscillations and Waves Question with Solution (2 Apr Morning shift, Q27) | JEE Arc