Question

Using dimensional analysis derive an expression for the time period of a simple pendulum

Answer 1

Answer:

i want to clear one thing that b/c means beacuse and this explanation which is written by colur is regarded to above equation which is underlined with blue colour..

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and one more thing...

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Answer 2

Time period of pendulum using dimensional analysis :

Let the time period of the pendulum be dependent on two quantities:

• L - length of pendulum
• g - gravity

Now, let the dependence be as follows :

$t \propto \: {L}^{x} \times {g}^{y}$

Now, expressing length and gravity in terms of fundamental physical quantities :

$\implies t \propto \: {(L)}^{x} \times {(L{T}^{ - 2} )}^{y}$

$\implies t \propto \: {L}^{x + y} \times {T}^{ - 2y}$

• Comparing both sides, we get :

$1) \: x + y = 0$

$2) \: - 2y = 1$

• After solving the equations , we get :

$\boxed{x = \dfrac{1}{2} \: \: \: \: and \: \: \: \: y = - \dfrac{1}{2} }$

So, putting the values of x and y :

$t \propto \: {L}^{ \frac{1}{2} } \times {g}^{ \frac{1}{2} }$

$\implies t \propto \: \dfrac{{L}^{ \frac{1}{2} }}{ {g}^{ \frac{1}{2} } }$

$\implies t \propto \: \sqrt{ \dfrac{L}{ g }}$

$\boxed{\implies t = k \sqrt{ \dfrac{L}{ g }}}$

• 'k' is a constant.

[Hence derived]