A particle in SHM is described by the displacement function $x\ (t) = A\cos{(\omega t + heta).}\ $If the initial (t = 0) position of the particle is 1 cm and its initial velocity is $\pi cms^{- 1}$ , what is its amplitude? The angular frequency of the particle is $\pi\ s^{- 1}.$

Question

A particle in SHM is described by the displacement function $x\ (t) = A\cos{(\omega t + 	heta).}\ $If the initial (t = 0) position of the particle is 1 cm and its initial velocity is $\pi cms^{- 1}$ , what is its amplitude? The angular frequency of the particle is $\pi\ s^{- 1}.$

Dr. Abubacker · Accepted Answer

(b) $x = A\cos(\omega t + 	heta);$
Velocity, $v = \frac{dx}{dt} = - A\omega\sin{(\omega t + 	heta)}$
$= - A\omega\sqrt{1 - \cos^{2}{(\omega t + 	heta)}}$
$= - A\omega\sqrt{1 - \frac{x^{2}}{A^{2}}} = - \omega\sqrt{A^{2} - x^{2}}$
Here, $v = \pi\ cms^{- 1},\ x = 1\ cm,\ \omega = \pi s^{- 1}$
So $\pi = - \pi\sqrt{A^{2} - 1^{2}}$
or $( - 1)^{2} = A^{2} - 1\ \ or\ \ \ A^{2} = 2$
or $A = \sqrt{2}\ cm$

Q. A particle in SHM is described by the displacement function $x\ (t) = A\cos{(\omega t + \theta).}\ $If the initial (t = 0) position of the particle is 1 cm and its initial velocity is $\pi cms^{- 1}$ , what is its amplitude? The angular frequency of the particle is $\pi\ s^{- 1}.$

Explanation

Reviewed by Dr. Abubacker

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