Question

The time period of a simple pendulum is given by ${\text{T}} = 2\pi \sqrt {\dfrac{l}{g}} $, where l is length of the pendulum and g is acceleration due to gravity. Show that this equation is dimensionally correct.

Answer 1

Hint: Here, we will proceed by determining the dimensions of the quantities involved in the given equation (i.e., time, length and acceleration due to gravity) and then, we will take dimensions of the quantities on both the sides of the given equation.

Complete Step-by-Step solution:
Given, Time period of a simple pendulum, ${\text{T}} = 2\pi \sqrt {\dfrac{l}{g}} {\text{ }} \to {\text{(1)}}$ where l is length of the pendulum and g is acceleration due to gravity.
As we know that
Dimension of time period T = [T]
Dimension of length of the pendulum l = Dimension of length = [L]
Since, Acceleration = $\dfrac{{{\text{Velocity}}}}{{{\text{Time}}}} = \dfrac{{\left( {\dfrac{{{\text{Displacement}}}}{{{\text{Time}}}}} \right)}}{{{\text{Time}}}}$
$ \Rightarrow $ Acceleration = $\dfrac{{{\text{Displacement}}}}{{{{\left( {{\text{Time}}} \right)}^2}}}{\text{ }} \to (2{\text{)}}$
As, dimension of displacement (length) is [L] and that of time is [T]
By applying dimensional analysis on equation (2), we get
Dimension of acceleration = $\dfrac{{{\text{Dimension of displacement}}}}{{{{\left( {{\text{Dimension of time}}} \right)}^2}}} = \dfrac{{\left[ {\text{L}} \right]}}{{\left[ {{{\text{T}}^2}} \right]}} = \left[ {\text{L}} \right]\left[ {{{\text{T}}^{ - 2}}} \right] = \left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$
Dimension of acceleration due to gravity (acceleration) is $\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]$
When we will be applying dimensional analysis on equation (1), $2\pi $ is a constant which is getting multiplied so it will be neglected.
By applying dimensional analysis on equation (1), we get
Dimension of T = $\sqrt {\dfrac{{{\text{Dimension of }}l}}{{{\text{Dimension of }}g}}} $
$
   \Rightarrow \left[ {\text{T}} \right] = \sqrt {\dfrac{{\left[ {\text{L}} \right]}}{{\left[ {{\text{L}}{{\text{T}}^{ - 2}}} \right]}}} \\
   \Rightarrow \left[ {\text{T}} \right] = {\left[ {\left[ {\text{L}} \right]\left[ {{{\text{L}}^{ - 1}}{{\text{T}}^2}} \right]} \right]^{\dfrac{1}{2}}} \\
   \Rightarrow \left[ {\text{T}} \right] = {\left[ {\left[ {{{\text{L}}^{1 - 1}}{{\text{T}}^2}} \right]} \right]^{\dfrac{1}{2}}} \\
   \Rightarrow \left[ {\text{T}} \right] = {\left[ {\left[ {{{\text{L}}^0}{{\text{T}}^2}} \right]} \right]^{\dfrac{1}{2}}} \\
   \Rightarrow \left[ {\text{T}} \right] = \left[ {{{\text{T}}^{2 \times \dfrac{1}{2}}}} \right] \\
   \Rightarrow \left[ {\text{T}} \right] = \left[ {\text{T}} \right] \\
 $
In the above equation, the dimensions of both the LHS and the RHS are the same. This means that the given equation is dimensionally correct.

Note- There are seven primary dimensions in units of measurement. The primary dimensions are defined as independent or fundamental dimensions from which other dimensions can be derived. The primary dimensions are mass, length, time , temperature, electric current, quantity of light and quantity of matter.