JEE Main 2018 · Physics · Simple Harmonic Motion · medium

A block of mass 0.5 kg oscillates on a smooth horizontal surface attached to a spring of constant 200 N m^-1 with amplitude 5 cm. As it passes…

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A block of mass

0.5\ \mathrm{kg}

oscillates on a smooth horizontal surface attached to a spring of constant

200\ \mathrm{N\,m^{-1}}

with amplitude

5\ \mathrm{cm}

. As it passes through the mean position, a lump of putty of mass

0.5\ \mathrm{kg}

drops vertically onto it and sticks. The new amplitude of oscillation is

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Answer: (A) $\dfrac{5}{\sqrt2}\ \mathrm{cm}$

At the mean position the speed is maximum,

v=\omega A

with

\omega=\sqrt{k/m}=20\ \mathrm{rad\,s^{-1}}

, so

v=1\ \mathrm{m\,s^{-1}}

. The putty lands vertically, so horizontal momentum is conserved:

0.5(1)=1.0\,v'

, hence

v'=0.5\ \mathrm{m\,s^{-1}}

. The new angular frequency is

\omega'=\sqrt{200/1.0}=10\sqrt2

. The new amplitude is

A'=\dfrac{v'}{\omega'}=\dfrac{0.5}{14.14}=3.54\ \mathrm{cm}=\dfrac{5}{\sqrt2}\ \mathrm{cm}

. Had the putty landed at an extreme position, the amplitude would remain

5\ \mathrm{cm}

Original question written in the JEE Main 2018 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake