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A particle is executing SHM represented by the equation x(t) = Acos⁡(ωt + ϕ). Given that the initial distance of the particle from its equilibrium position is 1 cm and the initial velocity is π cms $s^{- 1}$ and its angular frequency is $π s^{- 1} .$ Using this, data, calculate the amplitude of the motion.

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$π cm$

2 cm

$\sqrt{2} cm$

1 cm

answer is C.

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Detailed Solution

$x = Acos (ω t + ϕ)$ where A is amplitude

At t = 0, x = 1 cm

$∴ 1 = Acos ϕ \dots (i)$

Velocity,

$v = \frac{dx}{dt} = \frac{d}{dt} (Acos (ω t + ϕ)) =$

$- A ω \sin (ω t + ϕ)$

At t = 0, $v = π {cms}^{- 1}$

$\begin{array}{l} ∴ π = - A ω \sin ϕ or \frac{π}{ω} = - Asin ϕ \\ ∵ ω = π s^{- 1} \\ ∴ 1 = - Asin ϕ . . . . . . (i i) \end{array}$

Squaring and adding (i) and (ii), we get

$\begin{array}{l} A^{2} \cos^{2} ϕ + A^{2} \sin^{2} ϕ = 2 \\ A^{2} = 2 (∵ \sin^{2} ϕ + \cos^{2} ϕ = 1) \\ ∴ A = \sqrt{2} cm \end{array}$

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