Question

Check the correctness of T = 2π\sqrt{\frac{l}{g} }

Answer 1

The Time period of a Simple pendulum is given by,

$T = 2 \pi \sqrt{ \frac{l}{g} }$

Here, l is the length of pendulum, and g is the Acceleration due to gravity.

Checking the correctness of the above equation through dimensional analysis

We need to prove that, The both sides of the equation have same dimensions.

2π is a dimensionless constant

On the Left hand side

Time period = [T]

On the right hand side

2π = No dimensions

l = [L]

g = [LT^-2]

Now,

T = 2π√(l/g)

$[T] = \sqrt{ \frac{ [L] }{[LT ^{ - 2} ]}} \\ \\ [T] = \sqrt{ [L] ^{1 - 1} \times [T ]^{ - ( - 2)}} \\ \\ [T] = \sqrt{[T] ^{2} } \\ \\ [T] = [T] \\ \\ </strong></p><p><strong>[</strong><strong>T] = \sqrt{ \frac{ [L] }{[LT ^{ - 2} ]}} \\ \\ [T] = \sqrt{ [L] ^{1 - 1} \times [T ]^{ - ( - 2)}} \\ \\ [T] = \sqrt{[T] ^{2} } \\ \\ [T] = [T]$

Therefore, The dimensions on both sides are equal.

Hence, $T = 2 \pi \sqrt{ \frac{l}{g} }$ is dimensionally correct.