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The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of 4 m, 2 m s⁻¹ and 16 m s⁻² at a certain instant. The amplitude of the motion is √x m, where x is ______.

Asked in JEE Main 9th April 1st Shift 2024 · v-x and a-x relations

Answer: 17

Step-by-step solution

From |a|=ω²|x|: ω²=(16)/4=4, so ω=2 rad s⁻¹.

A²=x²+(v²)/(ω²)=16+4/4=17, so x=17.

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