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Initially spring is compressed by distance x subscript 0 end subscript from equilibrium position. At this compression block is given velocity square root of fraction numerator 3 k over denominator m end fraction end root x subscript 0 end subscript comma so that compression in the spring increases and block start SHM (Spring constant K). Equation of motion of the block is

  1. y equals 3 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator k over denominator m end fraction end root t plus fraction numerator pi over denominator 3 end fraction close square brackets    
  2. y equals 2 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator k over denominator m end fraction end root t plus fraction numerator pi over denominator 6 end fraction close square brackets    
  3. y equals 3 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator k over denominator m end fraction end root t plus fraction numerator pi over denominator 6 end fraction close square brackets    
  4. y equals 2 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator k over denominator m end fraction end root t plus fraction numerator pi over denominator 3 end fraction close square brackets    

The correct answer is: y equals 2 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator k over denominator m end fraction end root t plus fraction numerator pi over denominator 6 end fraction close square brackets


    From conservation of energy
    fraction numerator 1 over denominator 2 end fraction K x subscript 0 end subscript to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction m open square brackets square root of fraction numerator 3 K over denominator m end fraction end root x subscript 0 end subscript close square brackets to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction K x to the power of 2 end exponent
    fraction numerator 1 over denominator 2 end fraction K x subscript 0 end subscript to the power of 2 end exponent plus fraction numerator 3 over denominator 2 end fraction K x subscript 0 end subscript to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction K x to the power of 2 end exponent
    therefore x equals 2 x subscript 0 end subscript left parenthesis A m p l i t u d e right parenthesis
    Equation y equals 2 x subscript 0 end subscript sin invisible function application open square brackets square root of fraction numerator K over denominator m end fraction end root t plus fraction numerator pi over denominator 6 end fraction close square brackets satisfies the given condition

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