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13. oscillations — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysics13. oscillations1 Mark
MCQ
A ball suspended by a thread swings in a vertical plane so that its magnitude of acceleration in the extreme position and lowest position are equal. The angle $(\theta)$ of thread deflection in the extreme position will be :
  • A
     $\tan ^{-1}(\sqrt{2})$
  • ✓
     $2 \tan ^{-1}\left(\frac{1}{2}\right)$
  • C
     $\tan ^{-1}\left(\frac{1}{2}\right)$
  • D
     $2 \tan ^{-1}\left(\frac{1}{\sqrt{5}}\right)$
✓

Answer

Correct option: B.
 $2 \tan ^{-1}\left(\frac{1}{2}\right)$
b
Loss in kinetic energy $=$ Gain in potential energy

$ \Rightarrow \frac{1}{2} \mathrm{mv}^2=\mathrm{mg} \ell(1-\cos \theta)$

$ \Rightarrow \frac{\mathrm{v}^2}{\ell}=2 \mathrm{~g}(1-\cos \theta)$

Acceleration at lowest point $=\frac{\mathrm{v}^2}{\ell}$

Acceleration at extreme point $=g \sin \theta$

Hence, $\frac{\mathrm{v}^2}{\ell}=\mathrm{g} \sin \theta$

$ \therefore \sin \theta=2(1-\cos \theta) $

$ \Rightarrow \tan \frac{\theta}{2}=\frac{1}{2} \Rightarrow \theta=2 \tan ^{-1}\left(\frac{1}{2}\right)$

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