Answers

2016-04-22T20:39:28+05:30
Let Time period =T
      Mass of the bob = m
      Acceleration due to gravity = g
     Length of string = L

Let T \alpha m ^{a}g ^{b}L ^{c}
      [T] \alpha [m] ^{a}[g] ^{b}[L] ^{c}
      M^{0}L^{0}T^{1}=M^{a}L^{b}T^{-2b}L^{c}
      M^{0}L^{0}T^{1}=M^{a}L^{b+c}T^{-2b}
      ⇒a=0 ⇒ Time period of oscillation is independent of mass of the bob
     
      -2b=1
      ⇒b=-\frac{1}{2}
      
      b+c = 0
      -\frac{1}{2} + c =0
      c=\frac{1}{2}
      
Giving values to a,b and c in first equation
      T \alpha m ^{0}g ^{- \frac{1}{2} }L ^{ \frac{1}{2} }
      T \alpha  \sqrt{ \frac{L}{g} }

The real expression for Time period is
      T =2 \pi  \sqrt{ \frac{L}{g} }

Therefore time period of oscillation depends only on gravity and length of the string.
Not on mass of the bob.
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