To show that the equation [tex]\(C = \sqrt{\frac{E}{\rho}}\)[/tex] is dimensionally correct, we need to verify that both sides of the equation have the same dimensions.
1. Identify the Dimensions of Each Quantity:
- Speed, [tex]\(C\)[/tex]: The dimension of speed is length per time, denoted as [tex]\([C] = \text{[L T\(^{-1}\)[/tex]]}\).
- Young's Modulus, [tex]\(E\)[/tex]: Young's modulus is a measure of stiffness and has the same units as pressure, which is force per unit area. Therefore, the dimensional formula for Young's modulus is:
[tex]\[
[E] = \left[\frac{\text{Force}}{\text{Area}}\right] = \frac{\text{[M L T\(^{-2}\)]}}{\text{[L\(^{2}\)]}} = \text{[M L\(^{-1}\) T\(^{-2}\)]}
\][/tex]
- Density, [tex]\(\rho\)[/tex]: Density is mass per unit volume, and its dimensional formula is:
[tex]\[
[\rho] = \left[\frac{\text{Mass}}{\text{Volume}}\right] = \frac{\text{[M]}}{\text{[L\(^{3}\)]}} = \text{[M L\(^{-3}\)]}
\][/tex]
2. Analyze the Right-Hand Side ([tex]\(\sqrt{\frac{E}{\rho}}\)[/tex]):
- First, calculate the dimensional formula for [tex]\(\frac{E}{\rho}\)[/tex]:
[tex]\[
\left[\frac{E}{\rho}\right] = \frac{\text{[M L\(^{-1}\) T\(^{-2}\)]}}{\text{[M L\(^{-3}\)]}} = \text{[M L\(^{-1}\) T\(^{-2}\)]} \cdot \text{[M\(^{-1}\) L\(^{3}\)]}
\][/tex]
[tex]\[
= \left[\frac{L^{3}}{L} \cdot T^{-2}\right] = \left[\frac{L^{2}}{T^{2}}\right] = \text{[L\(^{2}\) T\(^{-2}\)]}
\][/tex]
- Now, take the square root of the result:
[tex]\[
\sqrt{\left[\frac{E}{\rho}\right]} = \sqrt{\text{[L\(^{2}\) T\(^{-2}\)]}} = \text{[L T\(^{-1}\)]}
\][/tex]
3. Compare Dimensions:
- The dimensions on the left-hand side of the equation (speed, [tex]\(C\)[/tex]) are [tex]\(\text{[L T\(^{-1}\)[/tex]]}\).
- The calculated dimensions on the right-hand side ([tex]\(\sqrt{\frac{E}{\rho}}\)[/tex]) are also [tex]\(\text{[L T\(^{-1}\)[/tex]]}\).
Since both sides of the equation have the same dimensions, the equation [tex]\(C = \sqrt{\frac{E}{\rho}}\)[/tex] is dimensionally correct.
