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

Answer: 17. |a| = ω² x ⇒ 16 = 4ω² ⇒ ω = 2 rad/s. A² = x² + (v²)/(ω²) = 16 + 4/4 = 17. A = √(17) m, so x = 17.

The position, velocity and acceleration of a particle executing… | JEE Main 2024 (9 Apr, Shift 1)

Q35JEE Main 2024 · 9 APR · SHIFT 1Medium

Velocity & Acceleration in SHM

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of $4\text{ m}, 2\text{ m s}^{-1}$ and $16\text{ m s}^{-2}$ at a certain instant. The amplitude of the motion is $\sqrt{x}\text{ m}$ where $x$ is _________.

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

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Answer: 17

$|a| = \omega^2 x \Rightarrow 16 = 4\omega^2 \Rightarrow \omega = 2$ rad/s.
$A^2 = x^2 + \frac{v^2}{\omega^2} = 16 + \frac{4}{4} = 17$ .
$A = \sqrt{17}$ m, so $x = 17$ .

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