Question

A mass of 2 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 an SHM. The spring constant is 200 N/m. What should be the minimum amplitude of motion, so that the mass gets detached from the pan?
[Given, g=10 $m/{{s}^{2}}$]
(A) 10 cm
(B) any value less than 12 cm
(C) 4 cm
(D) 8 cm

Answer 1

The gravitational force that is the weight of the block of mass M, pushes down the spring and due to its tendency, there is a restoring force in the spring which tends to push back to its equilibrium position. If both the forces are equal then the system remains at rest.

Complete step by step answer:
Restoring force on the spring is given by, $F=kx$ which acts in an upward direction. Here, x is the distance by which spring is compressed from its equilibrium position. For the simple harmonic motion to take place, the condition must be;

& Mg=kx \\\ & x=\dfrac{Mg}{k} \\\ \end{aligned}$$ Substituting the values, we get $$x=\dfrac{2\times 10}{200}=0.1m=10cm$$ This shows that the minimum amplitude of the spring motion should be 10 cm to detach the mass from the pan. Hence, the correct option is (A). **Note:** Had it been any of the two forces very very much greater than one, we would have either spring being crumbled under the weight or pan immediately thrown away. The spring constant and other values must be in standard SI units to get the correct answer.