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Question

A mass m is undergoing SHM in the vertical direction about the mean position y0 with amplitude A and angular frequency 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 y (measured from the mean position) such that the height h attained by the block is maximum. (Aω)2>g

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A
gω2
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B
2gω2
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C
g2ω2
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D
None of these
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Solution

The correct option is A gω2
Let v be the upward velocity of the block when it detaches from the spring at a distance y above the mean position.
v=wA2y2v2=w2(A2y2) ..........(1)
At the detaching point, the block will continue to move upwards due to inertia and reach the height h above the mean position. At that height the velocity of the block will be zero.
Thus 0v2=2(g)ss=v22g
Now the total height h=s+y=v22g+y
h=w2(A2y2)2g+y
For h to be maximum, dhdy=0
Thus w22g(2y)+1=0
y=gw2

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