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
JEE MAIN 2018, Diffcult
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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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