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Q. A block of mass $m$ attached to 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:

JEE MainJEE Main 2020Oscillations

Solution:

At equilibrium position
$V _{0}=\omega_{0} A =\sqrt{\frac{ K }{ m }} A $ .....(i)
$V =\omega A ^{1}=\sqrt{\frac{ K }{\frac{ m }{2}} A ^{1}} $......(ii)
$\therefore A^{1}=\frac{A}{\sqrt{2}}$