Question

Time period of a simple pendulum depends upon the length of pendulum (l) and acceleration due to gravity (g). Using dimensional analysis, obtain an expression for time period of simple pendulum.

Answer 1

Answer:

T=time

m=mass

L= length

k is constant

Answer 2

Answer:

$T\alpha ^\sqrt{\frac{l}{g} }$

Explanation:

Since it is given that time period of a simple pendulum depends upon the length of pendulum (l) and acceleration due to gravity (g).

So,

T=f(l,g)

$T\alpha l^{a} g^{b}$....................(1)

where alpha is the sign of proportionality.

we know the dimension formula for l and g.

for l=L and $g=LT^{-2}$

Now,$T\alpha L^{a} (LT^{-2}) ^{b}$

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

Compare the power of L and T.

a+b=0........................(1),-2b=1.........................(2)

from (2) we get

$b=\frac{-1}{2}$

from (1) we get

$a=\frac{1}{2}$

$b=\frac{-1}{2}$

put a and b in (1) we get

$T\alpha ^\sqrt{\frac{l}{g} }$