Certainly! Let's solve the dimensional analysis problem using the **Buckingham Pi Theorem** (method of repeating variables) step by step, as requested. --- # **Step-by-Step Solution** ### **Step 1: List All Variables** Given: - Frequency of oscillation: \( \omega \) - Sign width: \( b \) - Sign height: \( h \) - Wind velocity: \( V \) - Air density: \( \rho \) - Elastic constant: \( k \) (dimensions \( FL \)) Variables: - \( \omega, b, h, V, \rho, k \) (6 variables) --- ### **Step 2: Write Dimensions of Each Variable** Write dimensions in terms of fundamental quantities (M, L, T, F): | Variable | Description | Dimension | |------------|----------------------|-------------------| | \( \omega \) | Frequency | \( T^{-1} \) | | \( b \) | Width (length) | \( L \) | | \( h \) | Height (length) | \( L \) | | \( V \) | Velocity | \( LT^{-1} \) | | \( \rho \) | Density | \( ML^{-3} \) | | \( k \) | Elastic constant | \( FL \) | > **Note:** Force \( F = MLT^{-2} \Rightarrow k = FL = MLT^{-2} \cdot L = ML^2T^{-2} \) So, \( k \) has dimension \( ML^2T^{-2} \). --- ### **Step 3: Number of Fundamental Dimensions** From above, we have: - **M** (mass) - **L** (length) - **T** (time) Number of fundamental dimensions = **3** --- ### **Step 4: Number of Pi Terms** Number of variables (\( n \)) = 6 Number of fundamental dimensions (\( r \)) = 3 Number of Pi terms = \( n - r = 6 - 3 = 3 \) --- ### **Step 5: Choose Repeating Variables** Choose 3 variables (equal to number of fundamental dimensions), which together cover all fundamental units. **Common good choices:** - Must not include the dependent variable (\( \omega \)) - Must include at least one with each dimension Let's pick: **\( b, V, \rho \)** - \( b \): \( L \) - \( V \): \( LT^{-1} \) - \( \rho \): \( ML^{-3} \) They cover \( M, L, T \). --- ### **Step 6: Form the Pi Terms** Each Pi term will look like: \[ \pi = (\text{Non-repeating variable}) \times b^a V^b \rho^c \] #### **First Pi Term (\( \pi_1 \)) — Involving \( \omega \):** Let: \[ \pi_1 = \omega b^a V^b \rho^c \] Replace with dimensions: - \( \omega \): \( T^{-1} \) - \( b \): \( L \) - \( V \): \( LT^{-1} \) - \( \rho \): \( ML^{-3} \) So, in terms of dimensions: \[ [T^{-1}] [L]^a [L T^{-1}]^b [M L^{-3}]^c = M^c L^{a+b-3c} T^{-b-1} \] Set exponents for \( M, L, T \) to zero: - \( M \): \( c = \) - \( L \): \( a + b - 3c = \) - \( T \): \( -b - 1 = \) Solve: - \( c = \) - \( -b - 1 = \Rightarrow b = -1 \) - \( a + b = \Rightarrow a = 1 \) Plug in values: \[ \pi_1 = \omega b^1 V^{-1} = \frac{\omega b}{V} \] --- #### **Second Pi Term (\( \pi_2 \)) — Involving \( h \):** Let: \[ \pi_2 = h b^a V^b \rho^c \] Dimensions: - \( h \): \( L \) So, \[ [L][L]^a[LT^{-1}]^b[M L^{-3}]^c = M^c L^{a + b + 1 - 3c} T^{-b} \] Set exponents to zero: - \( M \): \( c = \) - \( L \): \( a + b + 1 - 3c = \) - \( T \): \( -b = \Rightarrow b = \) So, - \( c = \) - \( b = \) - \( a + 1 = \Rightarrow a = -1 \) Therefore, \[ \pi_2 = h b^{-1} = \frac{h}{b} \] --- #### **Third Pi Term (\( \pi_3 \)) — Involving \( k \):** Let: \[ \pi_3 = k b^a V^b \rho^c \] - \( k \): \( ML^2T^{-2} \) \[ [ML^2T^{-2}][L]^a[LT^{-1}]^b[M L^{-3}]^c = M^{1+c} L^{2+a+b-3c} T^{-2-b} \] Set exponents to zero: - \( M \): \( 1 + c = \Rightarrow c = -1 \) - \( L \): \( 2 + a + b - 3c = \) - \( T \): \( -2 - b = \Rightarrow b = -2 \) So, - \( c = -1 \) - \( b = -2 \) - \( 2 + a - 2 + 3 = \Rightarrow 2 + a + (-2) - 3(-1) = \Rightarrow 2 + a -2 +3 = \rightarrow a + 3 = \rightarrow a = -3 \) So, \[ \pi_3 = k b^{-3} V^{-2} \rho^{-1} = \frac{k}{\rho V^2 b^3} \] --- ### **Step 7: Write the Final Functional Relationship** The original relationship: \[ f(\omega, b, h, V, \rho, k) = \] Using the Pi terms: \[ F\left(\frac{\omega b}{V}, \frac{h}{b}, \frac{k}{\rho V^2 b^3}\right) = \] Or, equivalently: \[ \frac{\omega b}{V} = f\left(\frac{h}{b}, \frac{k}{\rho V^2 b^3}\right) \] --- ### **Step 8: Box the Final Set of Pi Terms** **The suitable set of dimensionless Pi terms is:** \[ \boxed{ \pi_1 = \frac{\omega b}{V} \qquad \pi_2 = \frac{h}{b} \qquad \pi_3 = \frac{k}{\rho V^2 b^3} } \] --- ## **Summary Table** | Pi Term | Expression | Physical Meaning | |---------|----------------------------------|------------------------------------| | \( \pi_1 \) | \( \dfrac{\omega b}{V} \) | Oscillation frequency ratio | | \( \pi_2 \) | \( \dfrac{h}{b} \) | Aspect ratio of the sign | | \( \pi_3 \) | \( \dfrac{k}{\rho V^2 b^3} \) | Elastic-to-inertia ratio | --- ## **Final Answer** > **A suitable set of dimensionless Pi terms for this problem is:** > > \[ > \boxed{ > \pi_1 = \frac{\omega b}{V}, \quad > \pi_2 = \frac{h}{b}, \quad > \pi_3 = \frac{k}{\rho V^2 b^3} > } > \] Let me know if you'd like any step explained in more detail!