JEE Main Physics Mechanics #7 Simple Harmonic Motion

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In the given figure, a mass M is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is k, The mass oscillates on a frictionless surface with the time period T and amplitude A. When the mass is in equilibrium position as shown in the figure, another mass m is gently fixed upon it. The new amplitude of oscillation will be:
A) A SQRT[M/(M+m)]
B) A SQRT[M/(M-m)]
C) A SQRT[(M-m)/M]
D) A SQRT[(M+m)/M]

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[anshumaangangwar]

We can also use Vmax=A*omega

1. By conserving momentum:
(New Vmax)*(M+m) = (Old Vmax)*(M). New Vmax can be written in terms of Old Vmax

2. Old Vmax=A*omega

3. New omega =(k/M+m)^1/2

Using all the equation, New amplitude can be related to Old amplitude

[Peter_]

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[abhijeet]

This channel is such a gem. I wish I had known about this a long time ago. You make topics so easy to understand and that too within a few minutes. Thank you so much.

[LonesomeTraveller]

I missed this video, but thankfully opened your channel to see this one, albeit a bit late :-)

[ayushtripathy]

Sir Your videos are very helpful for a Jee aspirant.

[shashankdesai]

sir we can eliminate option c also . if we put m=M the amplitude becomes 0 which is not possible when energy is conserved. So we can mark A and save time

[itachiUchiHa-jqno]

Sir i would like to ask do you make a video about radar system cuz i want to ask you sir suggest and explain how to produce a radar that can determine the position of a plane effectively.

[nagavalli]

Sir, Can we solve this using problem assuming omega (angular frequency) remains same after the block is placed.