2.B Dimensional Analysis

Performing a Dimensional Analysis

Given independent variables \(x,y,z\) and dependent variable \(w\):

  1. Write down the hypothesized “generalized product” \(w = k x^a y^b z^c\)
  2. Write & simplify the dimensional version \([w] = [x]^a [y]^b [z]^c\)
  3. List the dimensions of all quantities
  4. Invoke dimensional compatibility to solve.
  5. Rewrite your equation, marvel at your awesomeness!

Activities

Vocal Chord Frequency

The frequency \(f\) of the sound of an organism’s vocal chords depend on their length \(\ell\), tension \(s\) (a force) and mass density \(\mu\) (mass per unit length). Find a formula for the frequency \(f = f(\ell, s, \mu).\)

Mixer Power

A mixer with power \(P\) turns a mixer with wings of length \(D\) through a liquid of viscosity \(H\). Find a formula for the angular velocity \(A = A(P,D,H)\). Note that since angles are dimensionless, we have \([A] = 1/T\). The dimensions of viscosity are \([H] = M/(LT)\).

Velocity of an Ocean Wave

The velocity \(V\) of a large ocean wave depends upon the period \(P\), the acceleration of gravity \(g\), and the density \(\mu\) of the water. Find a formula for the velocity \(V=V(P,g,\mu)\).

Solutions

  1. We have \[ \begin{array}{rcl} f &=& k \ell^a s^b \mu^c \\ [f] &=& [\ell]^a [s]^b [\mu]^c \\ T^{-1} &=& \left( L\right)^a \left(MLT^{-2}\right)^b \left( M L^{-1} \right)^c \\ M^0L^0T^{-1} &=& M^{b+c} L^{a+b-c} T^{-2b} \end{array} \] and so we have \[ 0 = b+c \qquad 0 = a+b-c \qquad -1=-2b. \] Therefore \(b=1/2\) and \(c=-1/2\) and \(a=-1\). Our formula is \[ f = k \ell^{-1} s^{1/2} \mu^{-1/2} = k \frac{1}{\ell} \sqrt{\frac{s}{\mu}}. \]

  2. We have \[ \begin{array}{rcl} A &=& k P^a D^b H^c \\ [A] &=& [P]^a [D]^b [H]^c \\ T^{-1} &=& \left( ML^2T^{-3}\right)^a L^b \left( M L^{-1} T^{-1} \right)^c \\ M^0L^0T^{-1} &=& M^{a+c} L^{2a+b-c} T^{-3a-c} \end{array} \] and so we have \[ 0 = a+c \qquad 0 = 2a+b-c \qquad -1 = -3a-c \] which means that \(a=1/2\) and \(c=-1/2\) and \(b=-3/2\). So our final equation is \[ A = k a^{1/2} b^{-3/2} c^{-1/2} = k \sqrt{\frac{P}{D^3H}} \]

  3. We have \[ \begin{array}{rcl} V &=& k P^a g^b \mu^c \\ [V] &=& [P]^a [g]^b [\mu]^c \\ LT^{-1} &=& (T)^a (LT^{-2})^b (ML^{-3})^c \\ L^1 T^{-1} M^0 &=& L^{b-3c} T^{a-2b} M^{c} \end{array} \] so we have the equations \[ 1 = b - 3c \qquad -1 = a-2b \qquad 0 = c \] which means that \(c=0\) and \(b=1\) and \(a=1\). So our final equation is \[ V = k Pg \]