Questions
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 a simple harmonic motion. The spring constant is 200 N/m. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g = l0 )
detailed solution
Correct option is B
Let the minimum amplitude of SHM is a.Restoring force on spring, F = kaRestoring force is balanced by weight mg of block.For mass to execute simple harmonic motion of amplitude a.ka = mg or a = mgkHere, m = 2 kg, k = 200 N/m, g = l0 m/s2a = 2×10200 = 10100m = 10100×100 cm = 10 cmHence, minimum amplitude of the motion should be l0 cm, so that the mass gets detached from the pan.Talk to our academic expert!
Similar Questions
Two springs of spring constants are joined in series. The effective spring constant of the-combination is given by:
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests