A particle executes simple harmonic motion and is located at $x = a, b$ and $c$ at times $t_{0}, 2 t_{0} and 3 t_{0}$ respectively. The frequency of the oscillation is

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$\frac{1}{2 π t_{o}} \cos^{- 1} (\frac{a + c}{2 b})$

$\frac{1}{2 π t_{o}} \cos^{- 1} (\frac{a + b}{3 c})$

$\frac{1}{2 π t_{o}} \cos^{- 1} (\frac{a + b}{2 c})$

$\frac{1}{2 π t_{o}} \cos^{- 1} (\frac{2 a + 3 c}{b})$

answer is D.

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Detailed Solution

Using $y = A \sin ω t$

$a = A \sin ω t_{0}$

$\begin{aligned} b = A \sin 2 ω t_{0} \\ c = A \sin 3 ω t_{0} \\ a + c = A [\sin ω t_{0} + \sin 3 ω t_{0}] \\ a + c = 2 A \sin 2 ω t_{0} \cos ω t_{0} \\ a + c = 2 b \cos ω t_{0} \\ ω t_{0} = \cos^{- 1} (\frac{a + c}{2 b}) \\ f = \frac{1}{2 π t_{0}} \cos^{- 1} (\frac{a + c}{2 b}) \end{aligned}$