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.

Options

A. the period of oscillation is `((2pi)/(5))`
B. the block weight double of its weight, when the plank is at opne of the positions of momentary rest
C. the block weight `0.5` times its weight on the plank halfway up
D. the block weighs `1.5` times its weight on the plank halfway down

Answer

A, B, C, D

Explanation

The position of momentry rest in S.H.M. is exreme position where velocity of particle is zero.

As the block loses contact with the plank at this position i.e normal force become where accleration of the block will be g downwards.
`:. omega^(2)A = g rArr omega^(2) = (10)/(0.4) = 25 rArr omega = 5 rad//s`
There period `T = (2pi)/(omega) = (2pi)/(5)s`
Acceleration in `S.H.M.` is given by ` a = omega^(2)x`
From the figrue we an see that, At lower extreme acceleration is g upwards
`:. N - mg = ma rArr N = m(a + g) = 2mg`
At halfway up, acceleration is `g//2` downwards
`:. mg - N = ma rArr N = m (g- g/2) = 1/2 mg`
At halfway down acceleration is `g//2` upwards
`:. N - mg = ma rArr N = m (g+g/2) = 3/2 mg`