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13. oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics13. oscillations1 Mark
MCQ
When a particle of mass $m$ is attached to a vertical spring of spring constant $k$ and released, its motion is described by $y ( t )= y _{0} \sin ^{2} \omega t ,$ where $'y'$ is measured from the lower end of unstretched spring. Then $\omega$ is
  • A
    $\sqrt{\frac{g}{y_{0}}}$
  • ✓
    $\sqrt{\frac{g}{2 y_{0}}}$
  • C
    $\frac{1}{2} \sqrt{\frac{g}{y_{0}}}$
  • D
    $\sqrt{\frac{2 g}{y_{0}}}$
✓

Answer

Correct option: B.
$\sqrt{\frac{g}{2 y_{0}}}$
b
$y = y _{0} \sin ^{2} \omega t$

$y =\frac{ y _{0}}{2}(1-\cos 2 \omega t )$

$y -\frac{ y _{0}}{2}=-\frac{ y _{0}}{2} cos 2 \omega t$

Amplitude : $\frac{y_{0}}{2}$

$\frac{y_{0}}{2}=\frac{m g}{K}$

$2 \omega=\sqrt{\frac{K}{m}}=\sqrt{\frac{2 g}{y_{0}}}$

$\omega=\sqrt{\frac{g}{2 y_{0}}}$

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