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 latter is at momentary rest. Then, the period of oscillations is (in seconds):
[Take $g=10{\text{m/s}}^2$]

• A. $\frac{2\pi }{5}$
• B. $\frac{2\pi }{10}$
• C. $\frac{4\pi }{5}$
• D. $\frac{3\pi }{10}$

Answer 1

The correct option is A $\frac{2\pi }{5}$

Let the mass of the block be $m$.
The block and plank are at momentary rest when the plank reaches the extreme position.
To lose contact, normal reaction = 0.
Taking the plank as a frame of reference, the block will face a pseudo force $m{a}_{max}$ in the upward direction at the top extreme position.


At extreme top position, applying Newton's 2nd law,
$\sum F_y=0$
$\Rightarrow N+m{a}_{max}=mg$
$N=0\Rightarrow a_{max}=g$

We know, at extreme position, $a_{max}=A{\omega }^2$
$\therefore A{\omega }^2=g\Rightarrow \omega =\sqrt{\frac{g}{A}}=\sqrt{\frac{10}{0.4}}=\sqrt{25}$
or $\omega =5\text{rad/s}$

Time period of oscillation $=\frac{2\pi }{\omega }=\frac{2\pi }{5}\text{s}$
$\therefore$ option $\left(a\right)$ is correct.