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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2024
  • Hard
  • Numerical

The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of \(4 \mathrm{~m}, 2 \mathrm{~ms}^{-1}\) and \(16 \mathrm{~ms}^{-2}\) at a certain instant. The amplitude of the motion is \(\sqrt{\mathrm{x}} \mathrm{m}\) where \(\mathrm{x}\) is _______.

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Answer & Solution

Correct Answer

(A) \(17\)

Step-by-step Solution

Detailed explanation

\(x=4 \mathrm{~m}, \mathrm{~V}=2 \mathrm{~m} / \mathrm{s}, \mathrm{a}=16 \mathrm{~m} / \mathrm{s}^2\)

\(|\mathrm{a}|=\omega^2 \mathrm{x}\)

\(\Rightarrow 16=\omega^2(4)\)

\(\omega=2 \mathrm{rad} / \mathrm{s}\)

\(\mathrm{v}=\omega \sqrt{\mathrm{A}^2-\mathrm{x}^2}\)

\(A=\sqrt{\frac{\mathrm{v}^2}{\omega^2}+\mathrm{x}^2} \Rightarrow \mathrm{A}=\sqrt{\frac{4}{4}+16}\)

\(A=\sqrt{17} \mathrm{~m}\)

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