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Question

A mass m is undergoing SHM in vertical direction about the mean position with amplitude A and angular velocity ω . At a distance y from the mean position, the mass detaches from the spring. Assume that the spring contracts and does not obstruct the motion of m. Find the distance y0 (measured from the mean position) such that the height attained by the block is maximum (Aω2>g)


A

gω2

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B

2gω2

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C

g2ω2

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D

22gω2

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Solution

The correct option is A

gω2


Total height attained by block after detaching from spring above the mean position,

h=y+v22g

where v is the velocity of block at a distance y above the mean position.

v=ω(A2y2)

h=y+ω2(A2y2)2g

for h to be maximum, dhdy=0

or y0=gω2


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