JEE MAIN - Physics (2020 - 3rd September Evening Slot - No. 23)
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
$${1 \over 2}$$
$$\sqrt 2 $$
$${1 \over {\sqrt 2 }}$$
Explanation
At equilibrium position
V0 = V
$${V_0} = {\omega _1}A = \sqrt {{K \over m}} A$$ .....(i)
$$V = \omega {A^1} = \sqrt {{K \over {{m \over 2}}}} {A^1}$$ .....(ii)
$$ \therefore $$ $${A^1} = {A \over {\sqrt 2 }}$$
V0 = V
$${V_0} = {\omega _1}A = \sqrt {{K \over m}} A$$ .....(i)
$$V = \omega {A^1} = \sqrt {{K \over {{m \over 2}}}} {A^1}$$ .....(ii)
$$ \therefore $$ $${A^1} = {A \over {\sqrt 2 }}$$
Comments (0)
