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 $\mathrm{t}=0 \mathrm{~s}$ are 2 cm and $2 \omega \mathrm{~cm} \mathrm{~s}^{-1}$ respectively, then its amplitude is $x \sqrt{2} \mathrm{~cm}$ where the value of $x$ is ______.

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Elasticity & SHM

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 ______.

Answer: 2

For

x (t) = A sin (ω t + ϕ)

the velocity is

v (t) = A ω cos (ω t + ϕ)

. At

t = 0

A sin ϕ = 2 cm, A ω cos ϕ = 2 ω cm s^{- 1} \Rightarrow A cos ϕ = 2 cm

Squaring and adding removes the phase:

A^{2} = 2^{2} + 2^{2} = 8 \Rightarrow A = 22 cm

Comparing with

A = x 2

cm gives

x = 2

JEE Main · 2021 · 27 Jul 2021 · Shift 2 · Integer answer