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Question

A small body attached to one end of a vertically hanging spring is performing SHM about it's mean position with angular frequency ω and amplitude a. If at a height y from the mean position the body gets detached from the spring, calculate the value of y so that the height H attained by the mass is maximum. The body does not interact with the spring during it's subsequent motion after detachment. (aω2>g).
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A
y=2gω2
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B
y=3gω2
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C
y=5gω2
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D
y=gω2
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Solution

The correct option is D y=gω2
Angular frequency =ω
Amplitude =a
Under SHM velocity, v=ωa2y2
After detaching from spring, net downward acceleration of block =g
Height attained by block, h=y+v22g =y+ωa2y22g
For h to be maximum, dhdy=0,y=y
dhdy=1+ω22g(2y2)=0
12y2ω22g=0
or y=gω2
Since, aω2>g (given)
a>gω2
a>y;
y from mean position <a
Hence, y=gω2

1103278_1010248_ans_4409265fa8cd4910b14150967d170904.png

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