Q 11-13-062JEE MainJEE Main 2024 (9 Apr, Shift 1)Easy
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 value type. Enter your answer.
Answer: 17
$a = \omega^2x \Rightarrow \omega^2 = \dfrac{16}{4} = 4$.
$$v^2 = \omega^2(A^2 - x^2) \;\Rightarrow\; 4 = 4(A^2 - 16) \;\Rightarrow\; A^2 = 17$$
So $A = \sqrt{17}\ \text{m}$ and $x = 17$.
Solution by Sreeraj P, M.Sc Physics