Question

The time period. of simple pendulam it is depands on lenght of the thread (l) and gravitational acceration (g). Find the relation between them.​

Answer 1

Solution:

Given relation:

$\sf{T \propto (l)^{x}\;\;(g)^{y}\;\;\;\;..........(1)}$

$\sf{T= k(l)^{x}\;\;(g)^{y}\;\;\;\;..........(2)}$

Where,

T = time

l = length of the thread

g = gravitational acceleration

k = constant

Putting the following dimensions in equation (2), we get

$\sf{[M^{0}L^{0}T_{1}]=k[M^{0}L^{1}T_{0}]^{x}\;[M^{0}L^{1}T^{-2}]^{y}\;\;\;\;.........(3)}$

Now we compair the following values,

$'M'\;\;\;\;0 = 0+0\;\;\;\;......(4)\\ \\ 'L'\;\;\;\;0 = x+y\;\;\;\;.......(5)\\ \\ 'T'\;\;\;\;1 = 0-2y\;\;\;\;.......(6)$

From eq (6)

$\sf{\implies 1 = -2y}$

$\sf{\implies y =-\dfrac{1}{2}\;\;\;\;......(7)}$

put the value of y in eq (5), we get

$\sf{\implies x+\bigg(-\dfrac{1}{2}\bigg)=0}$

$\sf{\implies x = \dfrac{1}{2}}$

Put the values of x and y in eq (2), we get

$\sf{T= k(l)^{x}\;\;(g)^{y}}$

$T = k(l)^{\frac{1}{2}}\;\;(g)^{-\frac{1}{2}}$

${\boxed{\boxed{\sf{\implies T=k\sqrt{\dfrac{l}{g}}}}}}$