Question

A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of 40 cm. The block just loses contact with the plank when the latter is at momentary rest Then.
A. the period of oscillation is `((2pi)/(5))`
B. `the block weighs double its actual weght, then the plank is at one of the positions of momentary rest.
C. the block weighs 1.5 times its weight on the plank halfway down
D. the block weghs its true weight on the plank when the later moves fastest.

Answer 1

Correct Answer - A::B::C::D

The position of momentary rest in SHM is the extreme position where velocity of particle is zero.
As the block loses contact with thw plank at this position i.e, normal force becomes zero, it has to be the upper extreme where acceleration of the block will be g downwards.
`omega^2A=gimpliesomega^2=(10)/(0.4)=25`
`omega=5(rad)/(s)impliesT=(2pi)/(omega)=(2pi)/(5)s`
Acceleration in SHM is given as `a=omega^2x`.
From the figure, we can see that at lower extreme, acceleration is g upwards
`N-mg=ma`
or `N=m(a+g)=2mg`
at halfway up acceleration is `(g)/(2)` downwards
`mg-N=ma` or `N=m(g-(g)/(2))=(1)/(2)mg`
At halfway down, acceleration is `(g)/(2)` upwards.
`N-mg=ma` or `N=m(g+(g)/(2))=(3)/(2)mg`
At mean position velocity is maximum and acceleration is zero.