JEE Mains · Physics · STD 11 - 13. oscillations
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
- A \(\frac{1}{{2\pi {t_0}}}\cos {\,^{ - 1}}\left( {\frac{{a + b}}{{2c}}} \right)\)
- B \(\frac{1}{{2\pi {t_0}}}\cos {\,^{ - 1}}\left( {\frac{{a + b}}{{3c}}} \right)\)
- C \(\frac{1}{{2\pi {t_0}}}\cos {\,^{ - 1}}\left( {\frac{{2a + 3c}}{{b}}} \right)\)
- D \(\frac{1}{{2\pi {t_0}}}\cos {\,^{ - 1}}\left( {\frac{{a + c}}{{2b}}} \right)\)
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 \(\mathrm{y}=\mathrm{A} \sin \omega \mathrm{t}\) \(\mathrm{a}=\mathrm{A} \sin \omega \mathrm{t}_{0}\) \(b=A \sin 2 \omega t_{0}\) \(c=A \sin 3 \omega t_{0}\) \(\mathrm{a}+\mathrm{c}=\mathrm{A}\left[\sin \omega \mathrm{t}_{0}+\sin 3 \omega t_{0}\right]\)…
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