A coin is placed on a horizontal platform which undergoes vertical simple harmonic motion of angular frequency w. The amplitude of oscillation is gradually increased. The coin will leave the contact with the platform for the first time
(a) at the highest position of the platform.
(b) at the mean position of the platform.
(c) for an amplitude of g/w2.
(d) for an amplitude of g2/w2.
is the answer c-g/w2???? if yes then i can post the soln!!!
Yes, UR right, Please post the solution.
hi sooraj,
soln: for the coin to leave the platform normal force on it must become zero!
f=mg-N;
kx=mg-N;
as N=0;when coin leaves platform,
kx=mg
we know that k=mw2,
mw2x=mg;
x=g/w2 which is the answer!!!!
have a nice day!!!!