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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2018
  • Hard
  • MCQ

A particle executes simple harmonic motion and is located at \(x = a, b\)  and  \(c\) at times \(t_0, 2t_0\) and \(3t_0\) respectively. The frequency of the oscillation is

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Answer & Solution

Correct Answer

(D) \(\frac{1}{{2\pi {t_0}}}\cos {\,^{ - 1}}\left( {\frac{{a + c}}{{2b}}} \right)\)

Step-by-step Solution

Detailed explanation

Using \(\mathrm{y}=\mathrm{A} \sin \omega \mathrm{t}\)

\(\mathrm{a}=\mathrm{A} \sin \omega \mathrm{t}_{0}\)

\(b=A \sin 2 \omega t_{0}\)

\(c=A \sin 3 \omega t_{0}\)

\(\mathrm{a}+\mathrm{c}=\mathrm{A}\left[\sin \omega \mathrm{t}_{0}+\sin 3 \omega t_{0}\right]\)\(=2 \mathrm{A} \sin 2 \omega \mathrm{t}_{0}\cos \omega t_{0}\)

\(\frac{a+c}{b}=2 \cos \omega t_{0}\)

\(\Rightarrow \omega=\frac{1}{t_{0}} \cdot \cos ^{-1}\left(\frac{a+c}{2 b}\right)\)

\( \Rightarrow f=\frac{1}{2 \pi t_{0}} \cos ^{-1}\left(\frac{a+c}{2 b}\right)\)

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