Solved Problem on Dimensional Analysis
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During the presentation of a project on an acoustic system, a student forgot the expression of the intensity of a sound wave. However, using intuition, he concluded that the average intensity (I) is a function of the amplitude of the air movement (A), the frequency (f), the air density (%rho;), and the speed of sound (c), obtaining at the expression \( I=A^{x}.f^{y}.\rho ^{z}.c \). Considering the fundamental quantities, mass, length, and time, find the values of the exponents x, y, and z.


Solution

From the problem statement, we have that the Dimensional Equation will be
\[ \begin{gather} [I]=[A]^{x}.[f]^{y}.[\rho ]^{z}.[c] \tag{I} \end{gather} \]
The left side of the equation can be written in terms of basic quantities
The intensity (I) is equal to the power (\( \mathscr{P} \)) divided by the area (A).
\[ \begin{gather} I=\frac{\mathscr{P}}{A} \quad , \quad [\text{intensity}]=\frac{[\text{power}]}{[\text{area}]} \tag{II} \end{gather} \]
The power (X) is equal to the work (W) divided by the time interval (Δt). The area (A) is equal to the length squared (S2)
\[ \begin{align} & \mathscr{P}=\frac{W}{\Delta t} \quad , \quad [\text{power}]=\frac{[\text{work}]}{[\text{time interval}]} \tag{III} \\[10pt] & A=S^{2} \quad , \quad [\text{area}]=[\text{length}]^{2} \tag{IV} \end{align} \]
substituting relations (III) and (IV) into the relationship (II)
\[ \begin{gather} I=\frac{W}{\Delta t.S^{2}} \quad , \quad [\text{intensity}]=\frac{[\text{work}]}{[\text{time interval}].[\text{length}]^{2}} \tag{V} \end{gather} \]
The work (W) is equal to the force (F) multiplied by displacement (d)
\[ \begin{gather} W=F.d \quad , \quad [\text{work}]=[\text{force}].[\text{length}] \tag{VI} \end{gather} \]
substituting relation (VI) into the relationship (V)
\[ \begin{gather} I=\frac{F.d}{\Delta t.S^{2}} \quad , \quad [\text{intensity}]=\frac{[\text{force}].[\text{length}]}{[\text{time interval}].[\text{length}]^{2}} \tag{VII} \end{gather} \]
The force (F) is equal to the mass (m) multiplied by acceleration (a)
\[ \begin{gather} F=m.a \quad , \quad [\text{force}]=[\text{mass}].[\text{acceleration}] \tag{VIII} \end{gather} \]
substituting relation (VIII) into the relationship (VII)
\[ \begin{gather} I=\frac{m.a.d}{\Delta t.S^{2}} \quad , \quad [\text{intensity}]=\frac{[\text{mass}].[\text{acceleration}].[\text{length}]}{[\text{time interval}].[\text{length}]^{2}} \tag{IX} \end{gather} \]
The acceleration (a) is equal to the change in velocity (Δv) divided by the time interval (Δt), which equals the change in displacement (ΔS) divided by the time interval squared (Δt2)
\[ \begin{gather} a=\frac{\Delta v}{\Delta t}=\frac{\Delta S}{\Delta t^{2}} \quad , \quad [\text{acceleration}]=\frac{[\text{speed}]}{[\text{time interval}]}=\frac{[\text{length}]}{[\text{time interval}]^{2}} \tag{X} \end{gather} \]
substituting relation (X) into the relationship (IX)
\[ \begin{gather} I=\frac{m.d^{2}}{\Delta t^{3}.A} \quad , \quad [\text{intensity}]=\frac{[\text{mass}].[\text{length}]^{2}}{[\text{time interval}]^{3}.[\text{area}]} \end{gather} \]
in terms of M, L, and T for the dimensions of mass, length, and time
\[ \begin{gather} I=\frac{M.\cancel{L^{2}}}{T^{3}.\cancel{L^{2}}}=\frac{M}{T^{3}}=M.T^{-3} \tag{XI} \end{gather} \]
The right side of the equation (I) will have the following dimensions
  • Amplitude of air movement, given in length dimension:    \( [A]=L \);
  • Frequency, given in inverse time dimension:    \( [f]=\dfrac{1}{T}=T^{-1} \);
  • Air density, given in mass per unit volume dimension:    \( [\rho ]=\dfrac{M}{L^{3}}=M.L^{-3} \);
  • Speed of sound, given in length per unit of time dimension:    \( [c]=\dfrac{L}{T}=L.T^{-1} \).
Thus the right side of (I) will have the dimensions
\[ \begin{gather} [A]^{x}.[f]^{y}.[\rho]^{z}.[c]=L^{x}.\left(T^{-1}\right)^{y}.\left(M . L^{-3}\right)^{z}. L . T^{-1} \tag{XII} \end{gather} \]
substituting expressions (II) and (III) into equation (I)
\[ M .T^{-3}=L^{x}.\left(T^{-1}\right)^{y}.\left(M .L^{-3}\right)^{z}.L .T^{-1} \]
on the left side, there is no length dimension present, so we write it as L0
\[ M . T^{-3}.L^{0}=L^{x}.T^{-y}.M^{z}.L^{-3z}.L^{1}.T^{-1} \]
on the right side of the equation, we write the terms L, M, and T as
\[ M . T^{-3}.L^{0}=L^{x-3z+1}.T^{-y-1}.M^{z} \]
equating the exponents of equal magnitudes on the left and right sides
\[ \left\{ \begin{array}{l} z=1\\ -y-1=-3\\ x-3z+1=0 \end{array} \right. \tag{IV} \]
From the first equation of the system (IV), we have the value of z
\[ \bbox[#FFCCCC,10px] {z=1} \]
From the second equation of system (IV)
\[ \begin{gather} -y-1=-3\\ y=3+1 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {y=2} \]
Substituting the value of z found above into the third equation of the system (IV)
\[ \begin{gather} x-3.1+1=0\\ x-3+1=0\\ x-2=0 \end{gather} \]
\[ \bbox[#FFCCCC,10px] {x=2} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .