JEE PYQ: Oscillations Question 12
Question 12 - 2021 (24 Feb 2021 Shift 2)
A particle is projected with velocity $v_0$ along $x$-axis. A damping force is acting on the particle which is proportional to the square of the distance from the origin i.e. $ma = -\alpha x^2$. The distance at which the particle stops:
(1) $\left(\frac{3v_0^2}{2\alpha}\right)^{1/3}$
(2) $\left(\frac{3v_0}{2\alpha}\right)^{1/3}$
(3) $\left(\frac{3v_0^2}{2\alpha}\right)^{1/2}$
(4) $\left(\frac{2v_0^2}{3\alpha}\right)^{1/3}$
Show Answer
Answer: (3)
Solution
$v,dv = -\alpha x^2,dx$. $\int_{v_0}^{0} v,dv = -\int_0^x \alpha x^2,dx$. $\frac{v_0^2}{2} = \frac{\alpha x^3}{3}$. $x = \left(\frac{3v_0^2}{2\alpha}\right)^{1/3}$.
BETA
