The position, velocity and acceleration of a particle executing simple harmonic motion are found to have magnitudes of and at a certain instant. The amplitude of the motion is where is

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Solution

Q.
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 $\sqrt{x} m$ where $x$ is ________. [2024]

Ans.
(17)    $x = 4 m, v = 2 m / s, a = 16 {m / s}^{2}$

   $| a | = ω^{2} x$

   $\Rightarrow 16 = ω^{2} (4)$

   $ω = 2 rad / s$

   $v = ω \sqrt{A^{2} - x^{2}}$

   $A = \sqrt{\frac{v^{2}}{ω^{2}} + x^{2}} \Rightarrow A = \sqrt{\frac{4}{4} + 16} = \sqrt{17} m$

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