A particle executes simple harmonic motion and is located at $x = a, b$ and $c$ at times $t_{o}, 3 t_{o}$ and $5 t_{o}$ respectively with respect to mean position . At $t = 0$ , the particle is at the mean position and moving toward positive x-axis. The frequency of the oscillation is

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

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

$\frac{1}{4 π t_{o}} \cos^{- 1} (\frac{a + c}{2 b})$

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

answer is C.

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Detailed Solution

$\begin{array}{l} Using y = A \sin ω t \\ a = A \sin ω t_{0} \\ b = A \sin 3 ω t_{0} \\ c = A \sin 5 ω t_{0} \\ a + c = [A \sin ω t_{0} + A \sin 5 ω t_{0}] \\ = 2 A \sin 3 ω t_{0} \cos 2 ω t_{0} \\ = 2 b \cos 2 ω t_{0} \\ \Rightarrow \frac{a + c}{b} = 2 \cos 2 ω t_{0} \\ \Rightarrow \frac{a + c}{2 b} = \cos 2 ω t_{0} \end{array}$

$\begin{array}{l} \Rightarrow ω = \frac{1}{2 t_{0}} \cos^{- 1} (\frac{a + c}{2 b}) (∵ ω = 2 π f) \\ \Rightarrow f = \frac{1}{4 π t_{0}} \times \cos^{- 1} (\frac{a + c}{2 b}) \end{array}$