Question

A horizontal plank has a rectangular block placed on it. The plank starts oscillating vertically and simple harmonically with an amplitude of $40\text{cm}$. The block just loses contact with the plank when the later is at momentary rest. Then:

• A. The period of oscillation is $\left(\frac{2\pi }{5}\right)$
• B. The block weighs double its weight, when the plank is at one of the positions of momentary rest.
• C. The block weights 0.5 times its weight on the plank halfway up
• D. The block weights 1.5 times its weight on the plank halfway down

Answer 1

Answer: (A), (B), (C), (D)

The block weights 1.5 times its weight on the plank halfway down

Given $A=0.4m$, and $a=g$
So ${\omega }^{2}A=g\Rightarrow {\omega }^2=\frac{10}{0.4}=25$
$\Rightarrow \omega =5,T=\frac{2\pi }{\omega }=\frac{2\pi }{5}\sec$
At lowest position acceleration. $={\omega }^{2}A+g=g+g=2g$
So weight $=m\left(2g\right)=2mg$
at half distance $a=\frac{g}{2}$
So weight at upper half distance $=m(g-\frac{g}{2})=\frac{mg}{2}$
and weight at lower half distance
$=m(g+\frac{g}{2})=\frac{3mg}{2}$
actual weight at equilibrium position (maximum v)

Why this question? Tip: The body looses contact with the surface on which it is palced at the instant the acceleration of 2 differs! We can equate the two in limiting case. Caution: Weight in a frame is equal to normal reaction!