A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with amplitude of 40 cm about mean position located at O. The block just loses contact with the plank when the plank is at momentary rest at the highest point. Then, choose the correct statement(s). $(g = 10 m / \sec^{2})$ .

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the period of oscillation is $(\frac{2 π}{5})$ sec

Block weighs double its weight, when the plank is at one of the positions of momentary rest

Block weighs 0.5 time its weight on the plank when the plank is 20cm above O

Block weighs 1.5 times its weight on the plank when the plank is 20cm below O

answer is A, B, C, D.

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Detailed Solution

Given A= 0.4m, and a=g so $ω^{2} A = g \Rightarrow ω^{2} = \frac{10}{0.4} = 25$
$\Rightarrow ω = s T = \frac{2 π}{ω} = 2 π / 5 \sec$ .
At lowest position acceleration $= ω^{2} A + g + g + g = 2 g$
So weight = m(2g) = 2mg at half distance a=g/2
So weight at upper half distance = m(g-g/2) = mg/2
And weight at lower half distance
$= m (g + g / 2) = \frac{3 m g}{2}$
Actual weight at equilibrium position (maximum v)