The speed $v$ of a particle moving along $x$ - axis is given by, $v^2=8bx-x^2-12b^2$ , where $b$ is a constant. Find the amplitude of oscillations.

Amplitude of oscillations is the displacement of the particle from mean position to the extreme position. Also the speed of the particle becomes zero at extreme positions. Let $x$ represents these positions, then

                       $8bx-x^2-12b^2=0$

or                   $x^2-8bx+12b^2=0$

or                  $(x-6b)(x-2b)=0$

$\therefore$                 $x=2b$ and  $6b$ 

It shows that particle moves along $x$ - axis from $x= 2b$ to $6b$ . If $A$ is the amplitude of oscillations, then $2A=6b-2b=4b$

or                  $A = 2b$