Question

The time period of oscillation depends upon mass of pendulum length of pendulum and acceleration due to gravity derived the relation between them?

Answer 1

Answer:-

$\mathsf{T = 2 \pi \sqrt{\dfrac{l}{g}}}$

Given :-

$\mathsf{ T \propto m lg}$

m = mass

l = length

g = acceleration due to gravity.

→$\mathsf{ T \propto m^a l^b g^c}$

→$\mathsf{ T = K m^a l^b g^c }$

• Dimension formula of mass is M.
• Dimension formula of length is L.
• Dimension formula of gravity is $LT^{-2}$

→$\mathsf{T = K [M]^a [L]^b [LT^{-2}]^c}$

→$\mathsf{T = K M^a L^b L^c T^{-2c}}$

→$\mathsf{M^0L^0T = K M^a L^{b+c}T^{-2c}}$

• On comparing with L. H. S

→$\mathsf{M^0 = M^a }$

→$\mathsf{a = 0}$

→$\mathsf{L^0 = L^{b+c}}$

→$\mathsf{0 = b + c}$

→$\mathsf{ b = -c}$

→$\mathsf{T = T^{-2c}}$

→$\mathsf{1 = -2c}$

→$\mathsf{ c = \dfrac{-1}{2}}$

also,

→$\mathsf{ b = -c }$

• put the value of c.

→$\mathsf{b = \dfrac{-(-1)}{2}}$

→$\mathsf{b = \dfrac{1}{2}}$

Now,

• put the value of a, b and c.

→$\mathsf{ T = K m^0 l^{\frac{1}{2}}g^{\frac{-1}{2}}}$

→$\mathsf{T = K \sqrt{\dfrac{l}{g}}}$

• where k is the constant of proportionality whose value is 2π.

→$\mathsf{T = 2 \pi \sqrt{\dfrac{l}{g}}}$

hence,

The relation between them is $\mathsf{ T = 2 \pi \sqrt{\dfrac{l}{g}}}$.