Physics-
General
Easy

Question

Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is

  1. fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator k over denominator 4 M end fraction end root    
  2. fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator 4 k over denominator M end fraction end root    
  3. fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator k over denominator 7 M end fraction end root    
  4. fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator 7 k over denominator M end fraction end root    

The correct answer is: fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator 4 k over denominator M end fraction end root


    The two springs on left side having spring constant of 2k each are in series, equivalent constant is fraction numerator 1 over denominator open parentheses fraction numerator 1 over denominator 2 k end fraction plus fraction numerator 1 over denominator 2 k end fraction close parentheses end fraction equals k. The two springs on right hand side of mass M are in parallel. Their effective spring constant is left parenthesis k plus 2 k right parenthesis equals 3 k.
    Equivalent spring constants of value k and 3k are in parallel and their net value of spring constant of all the four springs is k plus 3 k equals 4 k
    therefore Frequency of mass is n equals fraction numerator 1 over denominator 2 pi end fraction square root of fraction numerator 4 k over denominator M end fraction end root

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