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

\frac{1}{2 π t_{0}} c o s^{- 1} (\frac{2 a + 3 c}{b})

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\frac{1}{2 π t_{0}} c o s^{- 1} (\frac{a + c}{2 b})

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\frac{1}{2 π t_{0}} c o s^{- 1} (\frac{a + c}{2 c})

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\frac{1}{2 π t_{0}} c o s^{- 1} (\frac{a + 2 b}{2 c})

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Solution

The correct option is B $\frac{1}{2 π t_{0}} c o s^{- 1} (\frac{a + c}{2 b})$
Let $,$
$a = A s i n (w t_{0})$
$b = A c o s (2 w t_{0})$
$c = A c o s (3 w t_{0})$
Now take $,$
$a + c = A s i n (w t_{0}) + A s i n (3 w t_{0})$
$a + c = 2 A s i n (2 w t_{0}) . c o s (w t_{0})$
$a + c = 2 b . c o s (w t_{0})$
$c o s (w t_{0}) = (a + c) / 2 b$
Taking inverse on both sides $,$
$w t - 0 = c o s i n v [(a + c) / 2 b]$
$2 π f t_{0} = c o s i n v [(a + c) / 2 b]$
$f = c o s i n v [(a + c) / 2 b] / 2 π t_{0}$
Therefore $,$ frequency of oscillation is $1 / 2 π t 0 c o s i n v [(a + c) / 2 b] .$
Hence,
option $(B)$ is correct answer.

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