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Question

(D) In simple harmonic motion,the velocity $v$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency

Vedclass · Accepted Answer

$\sqrt{4A^2-3x^2}$

Vedclass · Answer

(D) In simple harmonic motion,the velocity $v$ at a displacement $x$ is given by $v = \omega \sqrt{A^2 - x^2}$,where $\omega$ is the angular frequency and $A$ is the amplitude.Initially,the velocity is $v_1 = \omega \sqrt{A^2 - x^2}$.After the blow,the velocity becomes $v_2 = 2v_1 = 2\omega \sqrt{A^2 - x^2}$.Let the new amplitude be $A'$. Since the frequency $\omega$ remains constant,the new velocity at the same position $x$ is $v_2 = \omega \sqrt{A'^2 - x^2}$.Equating the two expressions for $v_2$: $\omega \sqrt{A'^2 - x^2} = 2\omega \sqrt{A^2 - x^2}$.Squaring both sides: $A'^2 - x^2 = 4(A^2 - x^2)$.$A'^2 - x^2 = 4A^2 - 4x^2$.$A'^2 = 4A^2 - 3x^2$.Therefore,the new amplitude is $A' = \sqrt{4A^2 - 3x^2}$.

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