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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 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:
Option: 1 A\sqrt{\frac{M+m}{M}}
 
Option: 2 A\sqrt{\frac{M}{M-m}}
Option: 3 A\sqrt{\frac{M}{M+m}}  
Option: 4 A\sqrt{\frac{M-m}{M}}

Answers (1)

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As there is no external force acting on the system, the momentum can be conserved.

At mean/equilibrium position, initial velocity, v=Aw and final velocity, v'=A'w'

So, 

\\ p_i=p_f\\ \Rightarrow Mv=(m+M)v'\\ \Rightarrow MA\omega=(m+M)A'\omega'\\ \Rightarrow MA\sqrt{\frac{k}{M}}=(m+M)A'\sqrt{\frac{k}{(m+M)}}\\ \Rightarrow A'=A\sqrt{\frac{M}{(m+M)}}

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avinash.dongre

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