Solved Problem on Dimensional Analysis
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In the equation below x has a dimension of length and t has a dimension of time
\[ \begin{gather} x=a\operatorname{e}^{-bt}\cos \left(\theta +b^{2}ct\right) \end{gather} \]
determine the dimensions of the quantities a, b, c, and θ.


Solution

The left side of the equality x has a dimension of length, L, so the right side must have the same dimension.
Starting with the cosine function:

  • θ is the measure of an angle given in radians which is a dimensionless unit.


Note: The measure of an angle is the ratio of the arc length to the radius length, which can be given in millimeters (mm), centimeters (cm), meters (m), inches (in), feet (ft) or any other unit of length
\[ \begin{gather} \theta=\frac{s}{r}=\frac{1\;\cancel{\text{m}}}{1\;\cancel{\text{m}}}=\frac{1\;\cancel{\text{ft}}}{1\;\cancel{\text{ft}}}=1\;\text{rad} \end{gather} \]
the measure of the angle is independent of the measure used for the quantities.

[θ] = 1.
  • The term b2ct must be dimensionless, as t has a dimension of time, T, for the term to be dimensionless, we must have a dimension of c equal to time, T, and b must have a dimension of the reciprocal of time, T−1.
  • .
Note:
\[ \begin{gather} [b]^{2}[c][t]=\left(T^{-1}\right)^{2}TT=T^{-2}T^{2}=1 \end{gather} \]

[c] = T.

[b] = T−1.

Note: With the value found for the quantity b above, the exponential term will also be dimensionless
\[ \begin{gather} \operatorname{e}^{T^{-1}T}=\operatorname{e}^{1}=\operatorname{e} \end{gather} \]
where e is a real number equal to 2.71828...

  • a must have a dimension of length, L, for both sides of the equality to be dimensionally consistent, since the exponential and cosine terms are dimensionless.
[a] = L.

Note:
\[ \begin{gather} \underbrace{[x]}_{L}=[a]\underbrace{\operatorname{e}^{\underbrace{-bt}_{\text{adimensional}}}}_{\text{adimensional}}\underbrace{\cos\left(\underbrace{\theta}_{\text{adimensional}}+\underbrace{b^{2}ct}_{\text{adimensional}}\right)}_{\text{adimensional}}\\[5pt] \underbrace{[x]}_{L}=\underbrace{[a]}_{L} \end{gather} \]
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