Question

The position and velocity of a particle executing simple harmonic motion at $t\, = \,0$ are given by $3\,cm$ and $8 \,cm/s^{-1}$ respectively. If the angular frequency of the particle is $2 \,rad\,s^{-1}$ then the amplitude of oscillation, in centimeters, is

• A. 3
• B. 4
• C. 5
• D. 6
• E. 8

Answer 1

Answer: (C)

Given, the position and velocity of the particle executing SHM.
$y=3\, cm $
$V=8 \,cm / s$
Angular frequency, $\omega=2$ rad's
The velocity,
$ V =\omega \sqrt{a^{2}-y^{2}} $
$8 =2 \sqrt{a^{2}-(3)^{2}} $
$4 =\sqrt{a^{2}-(3)^{2}} $
$ 16 =a^{2}-9$
$ a^{2} =25 $
$ a =5\, cm $