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A particle in S.H.M. is described by the displacement function $x (t) = a \cos (ω t + ϕ)$ . If the initial (t=0) position of the particle is 1 cm and its initial velocity is π cm/s . The angular frequency of the particle is π rad/s , then it’s amplitude is:

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$\sqrt{2}$ cm

2.5 cm

1 cm

2 cm

answer is B.

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Detailed Solution

$x = a \cos (ω t + θ) . . . (i) v = \frac{d x}{d t} = - a ω \sin (ω t + θ) . . . (i i)$

Given at t = 0 , x = 1 cm and v = π and ω = π
Putting these values in equation (i) and (ii) we will get

$\sin θ = \frac{- 1}{a} a n d \cos θ = \frac{1}{a} \Rightarrow \sin^{2} θ + \cos^{2} θ = {(\frac{- 1}{a})}^{2} + {(\frac{1}{a})}^{2} \Rightarrow a = \sqrt{2} c m$

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