Equation of motion of particle A can be written as (in cm):

x = 2 c o s 2

π t

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x = 3 + 2 c o s 2

π t

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x = 2 s i n 2

π t

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x = 3 - 2 c o s 2

π t

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Solution

The correct option is C $x = 3 - 2 c o s 2$ $π t$
Particle A is executing $S H M$ about mean position $x = 3 c m$ whose amplitude, $A = 2 c m$ and frequency of oscillation $ν = 1 H z$
$⟹ w = 2 π ν = 2 π r a d / s$
Thus the displacement equation of particle A can be written as $x = 3 c m + A s i n (w t + ϕ)$
Putting the given values, $x = (3 c m) + (2 c m) s i n (2 π t + ϕ)$
As it is given that at $t = 0$ , the particle is at $x = 1 c m$ moving away from the origin .
Thus at $t = 0$ , $1 c m = (3 c m) + (2 c m) s i n (ϕ) ⟹ s i n ϕ = - 1$
$∴ ϕ = - \frac{π}{2}$
$⟹$ $x = (3 c m) - (2 c m) c o s 2 π t$

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