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13. oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics13. oscillations1 Mark
MCQ
Displacement-time equation of a particle executing $SHM$ is $x\, = \,A\,\sin \,\left( {\omega t\, + \,\frac{\pi }{6}} \right)$ Time taken by the particle to go directly from $x\, = \, - \frac{A}{2}$ to $x\, = \, + \frac{A}{2}$ is
  • ✓
    $\frac{\pi }{{3\omega }}$
  • B
    $\frac{\pi }{{2\omega }}$
  • C
    $\frac{2\pi }{{\omega }}$
  • D
    $\frac{\pi }{{\omega }}$
✓

Answer

Correct option: A.
$\frac{\pi }{{3\omega }}$
a
displacement, $X=A \sin \left(\omega t+\frac{\pi}{6}\right)$

At $X=-\frac{A}{2}$

$-\frac{A}{2}=A \sin \left(\omega t_{1}+\frac{\pi}{6}\right)$

$\omega t_{1}+\frac{\pi}{6}=\sin ^{-1}(-0.5)$

$\omega t_{1}+\frac{\pi}{6}=-\frac{\pi}{6}$

$t_{1}=-\frac{\pi}{3 \omega} \ldots .(1)$

At $X=+\frac{A}{2}$

$\frac{A}{2}=A \sin \left(\omega t_{2}+\frac{\pi}{6}\right)$

$\omega t_{2}+\frac{\pi}{6}=\sin ^{-1}(0.5)$

$\omega t_{2}+\frac{\pi}{6}=\frac{\pi}{6}$

$\omega t_{2}=0$

$t_{2}=0 \ldots \ldots .(2)$

The time taken by the particle,

$T=t_{2}-t_{1}$

$T=0-\left(-\frac{\pi}{3 \omega}\right)$

$T=\frac{\pi}{3 \omega}$

The correct option is $A.$

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