262could you explain the question with a proper diag. ?
A mass 'm' is undergoing S.H.M. in the vertical direction about the mean position 'x' with amplitude 'A' and angular freq. \omega . At a distance y from the mean position, the mass detaches from the spring. Assuming that the spring contracts and does not obstruct the motion of 'm' , distance y such that the height 'h' attained by the block is maximum is:
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3 Answers
1sorry. but the pic. is of a spring hanging with a mass from its end
21Let the distance from mean pos be y at pt. of detachment.
Now at this pos. some energy out of the tot. energy is stored in spring and the rest is stored in form of KE of block.
1/2*kA2-1/2*ky2=1/2*mv2 where v=√2gh
Now total height=y+h
=y+v2/2g
=y+k(A2-y2)/2gm
Now to find maxima dH/dy =0.Solve.
we get y=gm/k=g/ω2