A particle executes simple harmonic motion represented by displacement function as x(t) = A sin( ω t + φ) If the position and velocity of the particle at t = 0 s are 2 cm and 2 ω cm s ⁻¹ respectively, then its amplitude is x√(2) cm where the value of x is _________________. Numerical answer type. Work it out, then open the answer.

Answer: 2. At t = 0: Asin φ = 2 v = Aωcos φ = 2ω ⇒ Acos φ = 2 A = √(2² + 2²) = 2√(2) cm So x = 2.

A particle executes simple harmonic motion represented by… | JEE Main 2021 (27 Jul, Shift 2)

Q52JEE Main 2021 · 27 JUL · SHIFT 2Medium

Periodic Motion & SHM Basics

A particle executes simple harmonic motion represented by displacement function as x(t) = A sin( $\omega$ t + $\phi$ )

If the position and velocity of the particle at t = 0 s are 2 cm and 2 $\omega$ cm s ${}^{-1}$ respectively, then its amplitude is $x\sqrt{2}$ cm where the value of x is _________________.

Numerical answer type. Work it out, then open the answer.

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Answer: 2

At $t = 0$ : $A\sin\phi = 2$
$v = A\omega\cos\phi = 2\omega \Rightarrow A\cos\phi = 2$
$A = \sqrt{2^2 + 2^2} = 2\sqrt{2}\text{ cm}$
So $x = 2$ .

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Periodic Motion & SHM Basics question from JEE Main 2021 (27 Jul, Shift 2)

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