A particle in $SHM $ is described by the displacement equation $x(t) = A\cos (\omega t + \theta ).$ If the initial $(t = 0)$ position of the particle is $1 \,cm$ and its initial velocity is $\pi $cm/s, what is its amplitude? The angular frequency of the particle is $\pi {s^{ - 1}}$
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(b) Given, $v = \pi \,cm/\sec ,$ $x = 1\,cm$ and $\omega = \pi {s^{ - 1}}$
using $v = \omega \sqrt {{a^2} - {x^2}} $
using $v = \omega \sqrt {{a^2} - {x^2}} $
==> $\pi = \pi \sqrt {{a^2} - 1} $
$ \Rightarrow 1 = {A^2} - 1$
$ \Rightarrow A = \sqrt 2 \,cm.$
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