Question

Consider a simple pendulum, having a bob attached to a string that
oscillates under the action of the force of gravity. Suppose that the
period of oscillations of the simple pendulum depends on its length
(1) mass of the bob (m) and accelerations due to gravity (g). Derive
the expressions for its time period using the method of dimensions.
131​

Answer 1

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

$\huge{\textbf{\underline{As per Question:-}}}$

Period of oscillation depends upon :-

• mass of the bob.
• length of the string.
• acceleration due to gravity.

$\huge{\textbf{\underline{Solution:-}}}$

$T \propto m^a l^b g^c$

$T = k m^a l^b g^c$

• Dimension of mass is M length is L and acceleration due to gravity is LT-².

$\text{T = M^a L^b (LT^{-2})^c }$

$\text{T = M^a L^b L^c T^{-2c} }$

$M^0L^0T = M^a + L^{b+c}T^{-2c}$

• On comparing L.H.S and R.H.S.

$M^0 = M^a$

$a = 0$

$T = T^{-2c}$

$c = \dfrac{-1}{2}$

$L^0 = L^{b+c}$

$0 = b + \dfrac{(-1)}{2}$

$b = \dfrac{1}{2}$

put the value of a , b and c.

$\text{ T = k m^0 l^{\dfrac{1}{2}} g^{\dfrac{-1}{2}} }$

$\textbf{ T = k \sqrt{\dfrac{l}{g}} }$

Where k is constant and it's value is 2π.

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

Answer 2

Answer:

T = K√(l/g)

Explanation: