Question

A block of $0.5kg$ is placed on a horizontal platform. The system is making vertical oscillations about a fixed point with a frequency of $0.5Hz$, the maximum amplitude of oscillation if the block is not lost contact with the horizontal platform is nearly:

• A. $0.5m$
• B. $0.75m$
• C. $1m$
• D. $1.2m$

Answer 1

The correct option is A $0.5m$

As we know $T=\frac{1}{f}=\frac{2\pi }{\sqrt{k}}$

$k=4\times {\pi }^2\times f^2=4\times {\pi }^2\times {0.25}^2=9.86$

${\omega }^{2}A=g$

$kx=mg$
condition of just losing the contact
$\therefore {\omega }^{2}A=g$
$t=0.5cm$
$\omega =2\pi f$
${\pi }^{2}A=g$
$A=1$
$mg=kx+N$
For losing contact $N=0$;
$\therefore mg=Kx$

$x=\frac{0.5\times 9.8}{9.86}=0.5$