The frequency of a particle performing SHM is 12 Hz. Its amplitude is 4 cm. Its initial displacement is 2 cm towards positive extreme position. Its equation for displacement is

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$x = 0 .04 \sin (24 πt)$

$x = 0 .04 \cos (24 πt)$

$x = 0 .04 \sin (24 πt + \frac{π}{6})$

$x = 0 .04 \cos (24 πt + \frac{π}{6})$

answer is C.

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Detailed Solution

$\begin{array}{l} X = Asin (ωt + Φ), ω = 2 πf = 24 π \\ x_{0} = Asin Φ \Rightarrow 2 = 4 \sin Φ \\ \sin Φ = \frac{1}{2} \Rightarrow Φ = \frac{π}{6} \\ x = 4 \times 10^{- 2} \sin [24 πt + \frac{π}{6}] \end{array}$

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