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Question: Answered & Verified by Expert

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 N m-1. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g=10 ms-2 )

PhysicsOscillationsNEET
Options:
  • A 8 cm
     
  • B 10 cm
     
  • C Any value less than 12 cm
  • D 4 cm
Solution:
1030 Upvotes Verified Answer
The correct answer is: 10 cm
 
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=mgk

Here, m=2 kg, k=200 m-1, g=10 s-2

a=2×10200=10100 m

= 10100×100 cm=10 cm

Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.

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