Question

A block is placed on a horizontal plank. The plank is performing $SHM$ along a vertical line with an amplitude of $40cm$. The block just loses contact with the plank when the plank is momentarily at rest. Then:

• A. The period of its oscillations is $2\pi /5sec$.
• B. The block weights on the plank double its weight when the plank is at one of the positions ofmomentary rest.
• C. The block weights $1.5$ times its weight on the plank halfway down from the mean position.
• D. The block weights its true weight on the plank when the velocity of the plank is maximum.

Answer 1

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

The correct options are
A The period of its oscillations is $2\pi /5sec$.
B

The block weights on the plank double its weight when the plank is at one of the positions of

momentary rest.
C The block weights $1.5$ times its weight on the plank halfway down from the mean position.
D The block weights its true weight on the plank when the velocity of the plank is maximum.

Given:

$A=40cm=0.4m$

Since the block loses contact $(N=0)$ when the plank is momentarily at rest at the top. Then the force balance equation on the block at the top is:

$N-mg=m{a}_{max}$

$\Rightarrow a_{max}=-g=-10m/s^2$

But $a=-{\omega }^{2}x$, then at the top:

$a_{max}=-{\omega }^{2}A$

$\Rightarrow \omega =\sqrt{-\frac{a_{max}}{A}}=\sqrt{\frac{10m/s^2}{0.4}}$

$\Rightarrow \omega =5rad/s$

(A) Thus, $T=\frac{2\pi }{\omega }=\frac{2\pi }{5}s$

(B) Now, at the bottom-most point, $(x=-A)$:

$a=-{\omega }^{2}x={\omega }^{2}A=g$

Then the force balance equation on the block:

$N-mg=ma$

$\Rightarrow N=m(g+a)=2mg$

Thus, the weight of block is doubled at the bottom point.

(C) At the halfway down position, $(x=\frac{A}{2})$:

$a=-{\omega }^{2}x=-{\omega }^2\frac{A}{2}=-\frac{g}{2}$

Then the force balance equation on the block:

$N-mg=ma$

$\Rightarrow N=m(g+a)=\frac{1}{2}mg$

Thus, the weight of block is halved at the halfway down point.

(D) The velocity of plank is maximum at zero position $(x=0)$:

$a=-{\omega }^{2}x=0$

Then the force balance equation on the block:

$N-mg=ma$

$\Rightarrow N=m(g+a)=mg$

Thus, the weight of block remains the same.