Q.
A mass of 2.0 kg is put on a flat pan attached to a vertical spring fixed on the ground as shown in the figure. The mass of the spring and the pan is negligible. When pressed slightly and released the mass executes slightly and released the mass executes a simple harmonic motion. The spring constant is $200\ Nm^{- 1}$. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g=10 $ms^{- 2}$ )
A) $8.0$ cm
B) $10.0$ cm
C) Any value less than 12.0 cm
D) $4.0$ cm
Explanation
(b) Let the minimum amplitude of SHM is $a$.
Restoring force on spring
$F = ka$
Restoring force is balanced by weight $mg$ of block. For mass to execute simple harmonic motion of amplitude $a$.
$ka = mg$
or $a = \frac{mg}{k}$
Here, $m = 2\ kg,\ k = 200\ Nm^{- 1},\ g = 10\ ms^{- 2}$
$\therefore$ $a = \frac{2 \times 10}{200} = \frac{10}{100}$ m
$= \ \frac{10}{100} \times 100\ cm = 10\ cm$
Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.