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Question

A box B of mass M hangs from an ideal spring of force constant k. A small particle A also of mass M is stuck to the ceiling of the box and the system is in equilibrium. The particle gets detached from the ceiling and falls to strike the floor of the box. It takes time t for the particle to hit the floor after it gets detached from the ceiling. The particle on hitting the floor sticks to it and the system thereafter oscillates with a time period of T. Find the height H of the box if it is given that t=T6. Assume that the floor and ceiling of the box always remain horizontal.


A
Mgk[1+π23]
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B
Mgk[1+π29]
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C
Mg2k[1+π23]
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D
Mg2k[1+π29]
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Solution

The correct option is D Mg2k[1+π29]
Originally, the system keeps the spring stretched by a length x0, where, kx0=2Mg .............(1)
As soon as the particle gets detached, it begins to fall with acceleration g and the box begins to perform SHM with amplitude,
A=2MgkMgk=Mgk
The distance travelled (downward) by the particle in time t will be, s1=12gt2
The distance travelled (upward) by the floor in time t will be
s2=AAcos(ωt), where, ω=kM,t=T6 and T=2πω
s2=MgkMgkcos(ωt)
H=s1+s2
=12gt2+MgkMgkcos(ωt)
=12gπ29Mk+MgkMgkcos(π3)=Mg2k[1+π29]

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