Question

The position and velocity of a particle executing simple harmonic motion at $t=0$ are given by $3cm$ and $8cm/s^{-1}$ respectively. If the angular frequency of the particle is $2rad{s}^{-1}$ 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.
$y=3cm$
$V=8cm/s$
Angular frequency, $\omega =2$ rad's
The velocity,
$V=\omega \sqrt{a^2-y^2}$
$8=2\sqrt{a^2-(3{)}^2}$
$4=\sqrt{a^2-(3{)}^2}$
$16=a^2-9$
$a^2=25$
$a=5cm$