Question

The time period of a single pendulum t depends on length of the pendulum l and acceleration due to gravity g. derive the formula for t

Answer 1

Let Time period =T
      Mass of the bob = m
      Acceleration due to gravity = g
     Length of string = L

Let TαmagbLcT \alpha m ^{a}g ^{b}L ^{c}TαmagbLc 
      [T]α[m]a[g]b[L]c[T] \alpha [m] ^{a}[g] ^{b}[L] ^{c}[T]α[m]a[g]b[L]c 
      M0L0T1=MaLbT−2bLcM^{0}L^{0}T^{1}=M^{a}L^{b}T^{-2b}L^{c}M0L0T1=MaLbT−2bLc 
      M0L0T1=MaLb+cT−2bM^{0}L^{0}T^{1}=M^{a}L^{b+c}T^{-2b}M0L0T1=MaLb+cT−2b 
      ⇒a=0 ⇒ Time period of oscillation is independent of mass of the bob
      
      -2b=1
      ⇒b=-12\frac{1}{2}21​ 
      
      b+c = 0
      -12\frac{1}{2}21​ + c =0
      c=12\frac{1}{2}21​ 
      
Giving values to a,b and c in first equation
      Tαm0g−12L12T \alpha m ^{0}g ^{- \frac{1}{2} }L ^{ \frac{1}{2} }Tαm0g−21​L21​ 
      TαLgT \alpha \sqrt{ \frac{L}{g} }TαgL​​ 

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

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