A block of mass $m$ is at rest on an another block of same mass as shown in figure. Lower block is attached to the spring, then the maximum amplitude of motion so that both the block will remain in contact is
A$\frac{{mg}}{{2K}}$
B$\frac{{mg}}{{K}}$
C$\frac{{2mg}}{{K}}$
D$\frac{{3mg}}{{K}}$
Medium
Download our app for free and get started
C$\frac{{2mg}}{{K}}$
c The block will remain in contact till the block do not go above the natural length of spring because after this decleration of lower block becomes more than upper block due to spring force so they will get separated
So, $A=\frac {2mg}{K}$
Download our app
and get started for free
Experience the future of education. Simply download our apps or reach out to us for more information. Let's shape the future of learning together!No signup needed.*
A particle has simple harmonic motion. The equation of its motion is $x = 5\sin \left( {4t - \frac{\pi }{6}} \right)$, where $x$ is its displacement. If the displacement of the particle is $3$ units, then it velocity is
One end of a spring of force constant k is fixed to a vertical wall and the other to a block of mass m resting on a smooth horizontal surface. There is another wall at a distance ${x_0}$ from the black. The spring is then compressed by $2{x_0}$ and released. The time taken to strike the wall is
Assume that the earth is a solid sphere of uniform density and a tunnel is dug along its diameter throughout the earth. It is found that when a particle is released in this tunnel, it executes a simple harmonic motion. The mass of the particle is $100 g$. The time period of the motion of the particle will be (approximately) (take $g =10\,ms ^{-2}$, radius of earth $=6400\,km$ )
A block is placed on a horizontal plank. The plank is performing $SHM$ along a vertical line with amplitude of $40\,cm.$ The block just loses contact with the plank when the plank is momentarily at rest. Then :