Question

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 amplitude of oscillations. (b = 2 units)

Answer 1

Amplitude of oscillations is the separation of the particle from mean position to the extreme positions. Also the speed of the particle becomes zero at extreme position. Let $x$ represents extreme positions, then
$8bx-x^2-12b^2=0$ or $x^2-8bx+12b^2=0$ or (x - 6b) (x - 2b) = 0
It shows that particle moves long x-axis from x = 2b to 6b.
If A is the amplitude of oscillations, then 2A = 6b - 2b = 4b $\therefore$ A = 2b