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A mass m is suspended by means of two coiled spring which have the same length in unstretched condition as in figure. Their force constant are k subscript 1 end subscript and k subscript 2 end subscript respectively. When set into vertical vibrations, the period will be

  1. 2 pi square root of open parentheses fraction numerator m over denominator k subscript 1 end subscript k subscript 2 end subscript end fraction close parentheses end root    
  2. 2 pi square root of m open parentheses fraction numerator k subscript 1 end subscript over denominator k subscript 2 end subscript end fraction close parentheses end root    
  3. 2 pi square root of open parentheses fraction numerator m over denominator k subscript 1 end subscript minus k subscript 2 end subscript end fraction close parentheses end root    
  4. 2 pi square root of open parentheses fraction numerator m over denominator k subscript 1 end subscript plus k subscript 2 end subscript end fraction close parentheses end root    

The correct answer is: 2 pi square root of open parentheses fraction numerator m over denominator k subscript 1 end subscript plus k subscript 2 end subscript end fraction close parentheses end root


    Given spring system has parallel combination, so
    K subscript e q end subscript equals K subscript 1 end subscript plus K subscript 2 end subscript and time period T equals 2 pi square root of fraction numerator m over denominator left parenthesis K subscript 1 end subscript plus K subscript 2 end subscript right parenthesis end fraction end root

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