Dimensional Analysis: Speed of Sound Equation
Question: The speed of sound in a medium is given empirically by
\[
v = k\,P^a\,\rho^b,
\]
where
\(v\) is the speed of sound (with SI dimensions \(LT^{-1}\)),
\(P\) is the pressure (SI dimensions \(ML^{-1}T^{-2}\)),
\(\rho\) is the density (SI dimensions \(ML^{-3}\)),
\(k\) is a dimensionless constant,
and \(a\) and \(b\) are exponents.
Using dimensional analysis, determine the values of \(a\) and \(b\).
Select the correct option from the choices below:
- (1) \(a = \frac{1}{2},\; b = -\frac{1}{2}\)
- (2) \(a = 1,\; b = -1\)
- (3) \(a = \frac{1}{2},\; b = \frac{1}{2}\)
- (4) \(a = -\frac{1}{2},\; b = \frac{1}{2}\)
Select the Correct Option
Step‑by‑Step Explanation
-
Step 1: Write down the dimensions of each quantity:
- Speed of sound: \( [v] = LT^{-1} \).
- Pressure: \( [P] = ML^{-1}T^{-2} \).
- Density: \( [\rho] = ML^{-3} \). - Step 2: Write the given relation dimensionally: \[ v = k\,P^a\,\rho^b. \] Since \(k\) is dimensionless, equate the dimensions of both sides: \[ LT^{-1} = (ML^{-1}T^{-2})^a\,(ML^{-3})^b. \]
- Step 3: Expand the dimensions on the right-hand side: \[ (ML^{-1}T^{-2})^a = M^a L^{-a} T^{-2a}, \quad (ML^{-3})^b = M^b L^{-3b}. \] Multiplying these together gives: \[ M^{a+b} L^{-a-3b} T^{-2a}. \]
-
Step 4: Match the exponents for each fundamental dimension:
For mass: \(a + b = 0\) (1)
For length: \(-a - 3b = 1\) (2)
For time: \(-2a = -1\) (3) - Step 5: Solve for \(a\) using equation (3): \[ -2a = -1 \quad \Longrightarrow \quad a = \frac{1}{2}. \]
- Step 6: Substitute \(a = \frac{1}{2}\) into equation (1) to find \(b\): \[ \frac{1}{2} + b = 0 \quad \Longrightarrow \quad b = -\frac{1}{2}. \]
- Step 7: Verify the length equation (2): \[ -a - 3b = -\frac{1}{2} - 3\left(-\frac{1}{2}\right) = -\frac{1}{2} + \frac{3}{2} = 1, \] which is correct.
- Final Answer: The exponents are \(a = \frac{1}{2}\) and \(b = -\frac{1}{2}\). (Option (1))
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