Question

A particle moves according to the law $a=-ky$ (where $a$ is acceleration of the particle). Find the velocity as a function of displacement $y.\text{(if}{v}_0$ is the initial velocity$)$

• A. $v^2=v_o^2-k{y}^2$
• B. $v^2=v_o^2-2ky$
• C. $v^2=v_o^2-2k{y}^2$
• D. None

Answer 1

The correct option is A $v^2=v_o^2-k{y}^2$

We know that
$a=\frac{dv}{dt}$
$a=\frac{dv}{dy}.\frac{dy}{dt}$
According to question
$a=-ky$
$\Rightarrow \frac{dv}{dy}.\frac{dy}{dt}=-ky$
$\Rightarrow v\frac{dv}{dy}=-ky$ $(\because \frac{dy}{dt}=v,\text{velocity of the particle})$
$\Rightarrow {\int }_{v_o}^{v}vdv={\int }_o^y-kydy$
$\therefore v_o^2-v^2=k{y}^2$