To show that the equation for the period of a simple pendulum is dimensionally consistent, we start with the given formula:
[tex]\[
T = 2 \pi \sqrt{\frac{\ell}{g}}
\][/tex]
Here, [tex]\(T\)[/tex] is the period of the pendulum, [tex]\(\ell\)[/tex] is the length of the pendulum, and [tex]\(g\)[/tex] is the acceleration due to gravity.
### Step-by-Step Solution:
1. Identify the dimensions of each variable:
- [tex]\(T\)[/tex]: Time period (measured in units of time, i.e., [tex]\( [T] \)[/tex]).
- [tex]\(\ell\)[/tex]: Length of the pendulum (measured in units of length, i.e., [tex]\( [L] \)[/tex]).
- [tex]\(g\)[/tex]: Acceleration due to gravity (measured in units of length per time squared, i.e., [tex]\( [L/T^2] \)[/tex]).
2. Rewrite the equation using the dimensions:
Substitute the dimensions into the equation:
[tex]\[
[T] = 2 \pi \sqrt{\frac{[\ell]}{[g]}}
\][/tex]
3. Substitute the dimensions of [tex]\(\ell\)[/tex] and [tex]\(g\)[/tex]:
[tex]\[
[T] = 2 \pi \sqrt{\frac{[L]}{[L/T^2]}}
\][/tex]
4. Simplify the fraction inside the square root:
[tex]\[
[T] = 2 \pi \sqrt{\frac{[L]}{[L] / [T^2]}}
\][/tex]
Simplifying the division inside the square root:
[tex]\[
\frac{[L]}{[L] / [T^2]} = \frac{[L] \cdot [T^2]}{[L]} = [T^2]
\][/tex]
5. Take the square root of [tex]\([T^2]\)[/tex]:
[tex]\[
[T] = 2 \pi \sqrt{[T^2]} = 2 \pi [T]
\][/tex]
6. Check the consistency:
Since [tex]\(2 \pi\)[/tex] is a dimensionless constant, it does not affect the overall dimensions. Therefore:
[tex]\[
[T] = [T]
\][/tex]
This shows that the dimensions on both sides of the equation are the same. Therefore, the equation is dimensionally consistent, as both sides have the dimension of time ([tex]\([T]\)[/tex]).
