JEE PYQ: Oscillations Question 23
Question 23 - 2020 (03 Sep 2020 Shift 2)
A block of mass $m$ attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become $fA$. The value of $f$ is:
(1) $\frac{1}{\sqrt{2}}$
(2) 1
(3) $\frac{1}{2}$
(4) $\sqrt{2}$
Show Answer
Answer: (1)
Solution
At equilibrium, half mass breaks off. Remaining energy = $\frac{1}{4}kA^2$. $A’ = \frac{A}{\sqrt{2}}$. $f = \frac{1}{\sqrt{2}}$.
BETA
