Physics-
General
Easy

Question

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 blank N m to the power of negative 1 end exponent. What should be the minimum amplitude of the motion, so that the mass gets detached from the pan? (Take g=10 m s to the power of negative 2 end exponent )

  1. 8.0 cm    
  2. 10.0 cm    
  3. Any value less than 12.0 cm    
  4. 4.0 cm    

The correct answer is: 10.0 cm


    Let the minimum amplitude of SHM is a.
    Restoring force on spring
    F equals k a
    Restoring force is balanced by weight m g of block. For mass to execute simple harmonic motion of amplitude a.
    k a equals m g
    or a equals fraction numerator m g over denominator k end fraction
    Here, m equals 2 blank k g comma blank k equals 200 blank N m to the power of negative 1 end exponent comma blank g equals 10 blank m s to the power of negative 2 end exponent
    therefore a equals fraction numerator 2 cross times 10 over denominator 200 end fraction equals fraction numerator 10 over denominator 100 end fraction m
    equals fraction numerator 10 over denominator 100 end fraction cross times 100 blank c m equals 10 blank c m
    Hence, minimum amplitude of the motion should be 10 cm, so that the mass gets detached from the pan.
     
     

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