Let's check the correctness of the formula [tex]\( t = 2\Omega \sqrt{\frac{L}{g}} \)[/tex] using dimensional analysis to see if it correctly represents the time period of a simple pendulum. In this formula, [tex]\( l \)[/tex] is the length of the simple pendulum and [tex]\( g \)[/tex] is the acceleration due to gravity.
1. Define the dimensions of each variable:
- The length of the pendulum, [tex]\( l \)[/tex], has the dimension of length: [tex]\([l] = L\)[/tex].
- The acceleration due to gravity, [tex]\( g \)[/tex], has the dimension of length per time squared: [tex]\([g] = \frac{L}{T^2}\)[/tex].
2. Analyze the dimensions of the formula:
- We need to analyze the expression inside the square root first, [tex]\(\frac{l}{g}\)[/tex].
- The dimension of [tex]\( l \)[/tex] is [tex]\( L \)[/tex].
- The dimension of [tex]\( g \)[/tex] is [tex]\( \frac{L}{T^2} \)[/tex].
- Thus, the dimension of [tex]\(\frac{l}{g}\)[/tex] is:
[tex]\[
\left[ \frac{l}{g} \right] = \frac{L}{\frac{L}{T^2}} = \frac{L \cdot T^2}{L} = T^2.
\][/tex]
- Next, we take the square root of [tex]\( \frac{l}{g} \)[/tex]:
- The dimension of [tex]\( \sqrt{\frac{l}{g}} \)[/tex] is:
[tex]\[
\left[ \sqrt{\frac{l}{g}} \right] = \sqrt{T^2} = T.
\][/tex]
- Now we multiply by the constant factor [tex]\( 2\Omega \)[/tex] (assuming [tex]\(\Omega = \pi\)[/tex], which is dimensionless):
- Dimensionally, multiplying by a dimensionless constant does not change the dimension. Therefore:
[tex]\[
\left[ 2\Omega \sqrt{\frac{l}{g}} \right] = T.
\][/tex]
3. Conclusion:
- The dimension of the right-hand side of the formula [tex]\( 2\Omega \sqrt{\frac{l}{g}} \)[/tex] is [tex]\( T \)[/tex], which represents time.
- The left-hand side of the formula, [tex]\( t \)[/tex], is also time, which has the dimension [tex]\( T \)[/tex].
Since both sides of the equation have consistent dimensions, the formula [tex]\( t = 2\Omega \sqrt{\frac{l}{g}} \)[/tex] is dimensionally correct as the time period of a simple pendulum.
