A) at the mean position of the platform
B) for an amplitude of \[g/{{\omega }^{2}}\]
C) for an amplitude of \[g/{{\omega }^{2}}\]
D) at the highest position of the platform
Correct Answer: B
Solution :
As the amplitude is increased, the maximum acceleration of the platform (along with coin as long as they doesnt get separated) increases. If we draw the FBD for coin at one of the extreme positions as shown then from Newtons law, \[\Rightarrow \] For loosing contact with the platform, N = 0 So, \[{{v}_{T}}(Ag)=\frac{10.5-1.5}{19.5-1.5}\times 0.2=\frac{9}{18}\times 0.2\]You need to login to perform this action.
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