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STD 11 - 13. oscillations — NEET STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceNEETSTD 11 - 13. oscillations4 Marks
MCQ
A particle performs simple harmonic motion with amplitude $A$. Its speed is trebled at the instant that it is at a distance $\frac{{2A}}{3}$ from equilibrium position. The new amplitude of the motion is
  • A
    $A$$\sqrt 3 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
  • ✓
    $\;\frac{{7A}}{3}$
  • C
    $\;\frac{A}{3}\sqrt {41} $
  • D
    $3A$
✓

Answer

Correct option: B.
$\;\frac{{7A}}{3}$
b
Let new amplitude is $A'$

initial velocity

$v^{2}=\omega^{2}\left(A^{2}-\left(\frac{2 A}{3}\right)^{2}\right)$                $...(1)$

Where $A$ is initial amplitude $\&\, \omega$ is angular frequency.

Final velocity

$(3 v)^{2}=\omega^{2}\left(A^{\prime 2}-\left(\frac{2 A}{3}\right)^{2}\right)$                $...(2)$

From equation $\&$ equation $( 2)$

$\frac{1}{9}=\frac{A^{2}-\frac{4 A^{2}}{9}}{A^{2}-\frac{4 A^{2}}{9}}$

$A^{\prime}=\frac{7 A}{3}$

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