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13. oscillations — Physics STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 SciencePhysics13. oscillations1 Mark
MCQ
A simple harmonic oscillator has an amplitude $A$ and time period $6 \pi$ second. Assuming the oscillation starts from its mean position, the time required by it to travel from $x=A$ to $x=\frac{\sqrt{3}}{2} A$ will be $\frac{\pi}{x}$ s, where $x=$__________.
  • ✓
    $2$
  • B
    $12$
  • C
    $4$
  • D
    $9$
✓

Answer

Correct option: A.
$2$
a
From phasor diagram particle has to move from $\mathrm{P}$ to $Q$ in a circle of radius equal to amplitude of $SHM.$

$ \cos \phi=\frac{\frac{\sqrt{3} \mathrm{~A}}{2}}{\mathrm{~A}}=\frac{\sqrt{3}}{2} $

$ \phi=\frac{\pi}{6}$

Now, $\frac{\pi}{6}=\omega t$

$\frac{\pi}{6}=\frac{2 \pi}{T} t$

$\frac{\pi}{6}=\frac{2 \pi}{6 \pi} t$

$\mathrm{t}=\frac{\pi}{2}$

So, $\mathrm{x}=2$

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