The position,velocity,and acceleration of a p | vedclass

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(A) Given: Position $x = 4 \ m$,Velocity $v = 2 \ ms^{-1}$,Acceleration $a = 16 \ ms^{-2}$.The magnitude of acceleration in simple harmonic motion is

Vedclass · Accepted Answer

$17$

Vedclass · Answer

(A) Given: Position $x = 4 \ m$,Velocity $v = 2 \ ms^{-1}$,Acceleration $a = 16 \ ms^{-2}$.The magnitude of acceleration in simple harmonic motion is given by $|a| = \omega^2 x$.Substituting the values: $16 = \omega^2(4) \Rightarrow \omega^2 = 4 \Rightarrow \omega = 2 \ rad/s$.The velocity in simple harmonic motion is given by $v = \omega \sqrt{A^2 - x^2}$,where $A$ is the amplitude.Squaring both sides: $v^2 = \omega^2(A^2 - x^2) \Rightarrow \frac{v^2}{\omega^2} = A^2 - x^2$.Rearranging for $A$: $A^2 = \frac{v^2}{\omega^2} + x^2$.Substituting the values: $A^2 = \frac{2^2}{2^2} + 4^2 = \frac{4}{4} + 16 = 1 + 16 = 17$.Therefore,$A = \sqrt{17} \ m$. Comparing this with $\sqrt{x} \ m$,we get $x = 17$.

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

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