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Mechanical Properties Of Fluids

Question
CBSEENPH11020535

A particle is executing SHM along a straight line. Its velocities at distance x1 and x2 from the mean position are v1 and v2, respectively. Its time period is

  • 2 pi square root of fraction numerator straight x subscript 1 superscript 2 plus straight x subscript 2 superscript 2 over denominator straight v subscript 1 superscript 2 plus straight v subscript 2 superscript 2 end fraction end root
  • 2 pi square root of fraction numerator straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 over denominator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 end fraction end root
  • 2 pi square root of fraction numerator straight v subscript 1 superscript 2 plus straight v subscript 2 superscript 2 over denominator straight x subscript 1 superscript 2 plus straight x subscript 2 superscript 2 end fraction end root
  • 2 pi square root of fraction numerator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 over denominator straight x subscript 1 superscript 2 minus straight x subscript 2 superscript 2 end fraction end root

Solution

B.

2 pi square root of fraction numerator straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 over denominator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 end fraction end root

Let A be the amplitude of oscillation then,
straight v subscript 1 superscript 2 space equals space straight omega squared space left parenthesis straight A squared minus straight x subscript 1 superscript 2 right parenthesis space space space space... space left parenthesis straight i right parenthesis
straight v subscript 1 superscript 2 space equals space straight omega squared left parenthesis straight A squared minus straight x squared right parenthesis space space space space.... space left parenthesis ii right parenthesis
subtracting space Eq. space left parenthesis ii right parenthesis thin space from space Eq space left parenthesis straight i right parenthesis comma space we space get
straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 space equals space straight omega squared space left parenthesis straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 right parenthesis
rightwards double arrow space straight omega space equals space square root of fraction numerator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 over denominator straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 end fraction end root
rightwards double arrow space fraction numerator 2 straight pi over denominator straight T end fraction equals space square root of fraction numerator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 over denominator straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 end fraction end root

straight T equals space 2 straight pi square root of fraction numerator straight x subscript 2 superscript 2 minus straight x subscript 1 superscript 2 over denominator straight v subscript 1 superscript 2 minus straight v subscript 2 superscript 2 end fraction end root