# Given Information - Two degree of freedom spring-mass system is shown. - Masses: \( m \) (top), \( 2m \) (bottom) - Spring constants: All springs have constant \( k \) - Displacements: \( x_1 \) (top mass), \( x_2 \) (bottom mass) - External forces: \( F_1 \) on \( m \), \( F_2 \) on \( 2m \) # What Have to Find - Differential equations of motion for the system. - Natural frequencies of the system. # Definition or Concept Used - **Newton’s Second Law**: For each mass, sum of forces = mass × acceleration. - **Free Vibration (No Forcing)**: Set \( F_1 = \), \( F_2 = \) for natural frequencies. - **Characteristic Equation**: Used to compute natural frequencies from equations of motion. # Step by Step Solution **Equation of motion for mass \( m \) (top mass):** - Upward forces: Spring attached to ceiling (\( -kx_1 \)) - Downward forces: Spring between \( m \) and \( 2m \) (\( -k(x_1-x_2) \)) - Equation: \[ m\ddot{x}_1 = -k x_1 - k(x_1 - x_2) + F_1 \] \[ m\ddot{x}_1 + 2k x_1 - k x_2 = F_1 \] **Equation of motion for mass \( 2m \) (bottom mass):** - Upward force: Spring between \( m \) and \( 2m \) (\( k(x_1-x_2) \)) - Downward forces: Two parallel springs at bottom (\( -2k x_2 \)) - Equation: \[ 2m\ddot{x}_2 = k(x_1 - x_2) - 2k x_2 + F_2 \] \[ 2m\ddot{x}_2 + 3k x_2 - k x_1 = F_2 \] **In matrix form (when \( F_1 = F_2 = \)):** \[ \begin{bmatrix} m & \\ & 2m \end{bmatrix} \begin{bmatrix} \ddot{x}_1 \\ \ddot{x}_2 \end{bmatrix} + \begin{bmatrix} 2k & -k \\ -k & 3k \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} \\ \end{bmatrix} \] **For natural frequencies:** Assume solution: \( x_1 = X_1 e^{i\omega t},\ x_2 = X_2 e^{i\omega t} \) \[ \left| \begin{matrix} 2k - m\omega^2 & -k \\ -k & 3k - 2m\omega^2 \end{matrix} \right| = \] \[ (2k - m\omega^2)(3k - 2m\omega^2) - k^2 = \] \[ 6k^2 - 4km\omega^2 - 3k m\omega^2 + 2m^2\omega^4 - k^2 = \] \[ 5k^2 - 7km\omega^2 + 2m^2\omega^4 = \] \[ 2m^2\omega^4 - 7km\omega^2 + 5k^2 = \] Quadratic in \( \omega^2 \): \[ \omega^2 = \frac{7km \pm \sqrt{(7km)^2 - 4 \cdot 2m^2 \cdot 5k^2}}{2 \cdot 2m^2} \] \[ \omega^2 = \frac{7km \pm \sqrt{49k^2m^2 - 40k^2m^2}}{4m^2} \] \[ \omega^2 = \frac{7km \pm 3km}{4m^2} \] \[ \omega^2_1 = \frac{10km}{4m^2} = \frac{5k}{2m} \] \[ \omega^2_2 = \frac{4km}{4m^2} = \frac{k}{m} \] \[ \omega_1 = \sqrt{\frac{5k}{2m}},\quad \omega_2 = \sqrt{\frac{k}{m}} \] # Final Answers (Summary) - Differential equations of motion: \[ m\ddot{x}_1 + 2k x_1 - k x_2 = F_1 \] \[ 2m\ddot{x}_2 + 3k x_2 - k x_1 = F_2 \] - Natural frequencies: \[ \omega_1 = \sqrt{\frac{5k}{2m}},\quad \omega_2 = \sqrt{\frac{k}{m}} \]

# Part (a): Analysis Using Bernoulli's Theorem ## Given Information - **Fluid**: Real and viscous fluid - **Device**: Inclined Venturi meter connected to piezometers - **Factors to analyze**: Elevation change, velocity variation, pressure head differences ## What to Find - Effect of elevation change, velocity variation, and pressure head differences on the fluid flow. - Justification for the applicability of Bernoulli's equation despite viscosity. ## Definition or Concept Used - **Bernoulli's Equation**: For steady, incompressible, non-viscous flow, expressed as: \[ \frac{P}{\rho g} + z + \frac{v^2}{2g} = \text{constant} \] ## Analysis - **Elevation Change**: The elevation difference between piezometer points alters potential energy. This can be expressed as: \[ z_1 - z_2 = \Delta z \] - **Velocity Variation**: Velocity varies between the inlet and throat due to the conservation of mass (continuity equation). If \( A_1 v_1 = A_2 v_2 \), and assuming \( v_2 = 2v_1 \), then: \[ A_1 v_1 = 2 A_2 v_1 \Rightarrow A_1 = 2A_2 \] - **Pressure Head Differences**: The pressure difference between the inlet and throat can be represented as: \[ P_1 - P_2 = \rho g(z_2 - z_1) + \frac{1}{2} \rho (v_2^2 - v_1^2) \] ### Justification for Bernoulli's Equation Despite viscosity, Bernoulli's equation can be applied in a real fluid flow for the following reasons: - **Average Flow**: The Venturi meter operates under conditions where the effects of viscosity are minimal at the macroscopic level. The average flow can be approximated as inviscid due to high Reynolds numbers. - **Minor Losses**: While viscous losses exist, they can be considered as minor losses relative to the total energy head, allowing the application of Bernoulli's principles over short distances. # Part (b): Worked Example ## Given Information - Elevation difference between inlet and throat: \( \Delta z = 0.8 \, \text{m} \) - Inlet piezometric head: \( P_1 = 2.4 \, \text{m} \) - Throat piezometric head: \( P_2 = 1.2 \, \text{m} \) - Throat velocity is twice the inlet velocity: \( v_2 = 2v_1 \) ## What to Find - Velocity at inlet (\( v_1 \)) and throat (\( v_2 \)). ## Step by Step Solution 1. **Apply Bernoulli's Equation** between the inlet and the throat: \[ P_1 + \rho g z_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g z_2 + \frac{1}{2} \rho v_2^2 \] 2. **Substituting known values**: - Set \( z_1 = 0 \) and \( z_2 = -0.8 \, \text{m} \) (since throat is lower). - Replace \( P_1 \) and \( P_2 \): \[ 2.4 + 0 + \frac{1}{2} \rho v_1^2 = 1.2 - 0.8g + \frac{1}{2} \rho (2v_1)^2 \] 3. **Simplify the equation**: - Since \( g \approx 9.81 \, \text{m/s}^2 \): \[ 2.4 + \frac{1}{2} \rho v_1^2 = 1.2 - 0.8(9.81) + 2\rho v_1^2 \] - Calculate \( 0.8g \): \[ 0.8 \times 9.81 = 7.848 \, \text{m} \] - The equation becomes: \[ 2.4 + \frac{1}{2} \rho v_1^2 = 1.2 - 7.848 + 2\rho v_1^2 \] \[ 2.4 + \frac{1}{2} \rho v_1^2 = -6.648 + 2\rho v_1^2 \] 4. **Rearrange and solve for \( v_1 \)**: \[ 2.4 + 6.648 = (2 - \frac{1}{2}) \rho v_1^2 \] \[ 9.048 = \frac{3}{2} \rho v_1^2 \] \[ v_1^2 = \frac{9.048}{\frac{3}{2} \rho} = \frac{9.048 \times 2}{3\rho} \] 5. **Substitute a value for \( \rho \)** (assume \( \rho = 1000 \, \text{kg/m}^3 \)): \[ v_1^2 = \frac{18.096}{3000} \implies v_1 = \sqrt{\frac{18.096}{3000}} \approx 0.0773 \, \text{m/s} \] - Calculate \( v_2 \): \[ v_2 = 2v_1 = 0.1546 \, \text{m/s} \] # Final Answers (Summary) - Inlet velocity: \[ v_1 \approx 0.0773 \, \text{m/s} \] - Throat velocity: \[ v_2 \approx 0.1546 \, \text{m/s} \]

# Part (a): Analysis Using Bernoulli's Theorem ## Given Information - **Fluid**: Real and viscous fluid - **Device**: Inclined Venturi meter connected to piezometers - **Factors to analyze**: Elevation change, velocity variation, pressure head differences ## What to Find - Effect of elevation change, velocity variation, and pressure head differences on the fluid flow. - Justification for the applicability of Bernoulli's equation despite viscosity. ## Definition or Concept Used - **Bernoulli's Equation**: For steady, incompressible, non-viscous flow, expressed as: \[ \frac{P}{\rho g} + z + \frac{v^2}{2g} = \text{constant} \] ## Analysis - **Elevation Change**: The elevation difference between piezometer points affects potential energy, expressed as: \[ z_1 - z_2 = \Delta z \] - **Velocity Variation**: Velocity varies between inlet and throat due to the conservation of mass (continuity equation). If \( A_1 v_1 = A_2 v_2 \), and given \( v_2 = 2v_1 \): \[ A_1 v_1 = 2 A_2 v_1 \Rightarrow A_1 = 2A_2 \] - **Pressure Head Differences**: Represented as: \[ P_1 - P_2 = \rho g(z_2 - z_1) + \frac{1}{2} \rho (v_2^2 - v_1^2) \] ### Justification for Bernoulli's Equation Despite viscosity, Bernoulli's equation can be applied because: - **Average Flow**: The Venturi meter operates under conditions where the effects of viscosity are minimal at the macroscopic level, allowing approximation as inviscid flow. - **Minor Losses**: While viscous losses exist, they are relatively small and can be neglected for short distances, allowing Bernoulli's principles to apply. # Part (b): Worked Example ## Given Information - Elevation difference between inlet and throat: \( \Delta z = 0.8 \, \text{m} \) - Inlet piezometric head: \( P_1 = 2.4 \, \text{m} \) - Throat piezometric head: \( P_2 = 1.2 \, \text{m} \) - Throat velocity is twice the inlet velocity: \( v_2 = 2v_1 \) ## What to Find - Velocity at inlet (\( v_1 \)) and throat (\( v_2 \)). ## Step by Step Solution **Apply Bernoulli's Equation** between the inlet and the throat: \[ P_1 + \rho g z_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g z_2 + \frac{1}{2} \rho v_2^2 \] **Substituting known values**: - Set \( z_1 = 0 \) and \( z_2 = -0.8 \, \text{m} \) (throat is lower). - Replace \( P_1 \) and \( P_2 \): \[ 2.4 + 0 + \frac{1}{2} \rho v_1^2 = 1.2 - 0.8g + \frac{1}{2} \rho (2v_1)^2 \] **Calculate \( 0.8g \)** (using \( g \approx 9.81 \, \text{m/s}^2 \)): \[ 0.8 \times 9.81 = 7.848 \, \text{m} \] - The equation becomes: \[ 2.4 + \frac{1}{2} \rho v_1^2 = 1.2 - 7.848 + 2\rho v_1^2 \] **Rearranging**: \[ 2.4 + \frac{1}{2} \rho v_1^2 = -6.648 + 2\rho v_1^2 \] \[ 2.4 + 6.648 = (2 - \frac{1}{2}) \rho v_1^2 \] \[ 9.048 = \frac{3}{2} \rho v_1^2 \] **Solve for \( v_1^2 \)**: \[ v_1^2 = \frac{9.048}{\frac{3}{2} \rho} = \frac{9.048 \times 2}{3\rho} \] **Substitute a value for \( \rho \)** (assume \( \rho = 1000 \, \text{kg/m}^3 \)): \[ v_1^2 = \frac{18.096}{3000} \implies v_1 = \sqrt{\frac{18.096}{3000}} \approx 0.0773 \, \text{m/s} \] - Calculate \( v_2 \): \[ v_2 = 2v_1 = 0.1546 \, \text{m/s} \] # Final Answers (Summary) - Inlet velocity: \[ v_1 \approx 0.0773 \, \text{m/s} \] - Throat velocity: \[ v_2 \approx 0.1546 \, \text{m/s} \]

# Part (a): Analysis Using Bernoulli's Theorem ## Given Information - **Fluid**: Real and viscous - **Device**: Inclined Venturi meter with piezometers - **Factors to Analyze**: Elevation change, velocity variation, pressure head differences ## What to Find - Effects of elevation change, velocity variation, and pressure differences on flow. - Justification for applying Bernoulli's equation despite viscosity. ## Definition or Concept Used - **Bernoulli's Equation** for steady, incompressible flow: \[ \frac{p_1}{\gamma} + \frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} + \frac{V_2^2}{2g} + z_2 + h_L \] where \( h_L \) is head loss due to viscosity. ## Analysis ### Effect of Elevation Change - The elevation difference affects potential energy: \[ z_1 \neq z_2 \] - If \( z_2 > z_1 \), pressure head decreases further; if \( z_2 < z_1 \), pressure drop is partly balanced by elevation gain. ### Effect of Velocity Variation - From the continuity equation: \[ A_1 V_1 = A_2 V_2 \quad \text{and } V_2 > V_1 \] - Increased velocity head results in decreased pressure head at the throat. ### Pressure Head Differences from Piezometers - The difference in piezometric heads: \[ \left( \frac{p_1}{\gamma} + z_1 \right) - \left( \frac{p_2}{\gamma} + z_2 \right) = \frac{V_2^2}{2g} - \frac{V_1^2}{2g} + h_L \] ### Justification for Bernoulli's Equation - Despite viscosity, Bernoulli's equation applies due to streamlined flow and minor losses in the Venturi meter. - Correction for losses can be incorporated: \[ \text{Ideal Bernoulli} + h_L \] # Part (b): Worked Example ## Given Information - Elevation difference: \( z_1 - z_2 = 0.8 \, \text{m} \) - Piezometric heads: \( H_1 = 2.4 \, \text{m}, H_2 = 1.2 \, \text{m} \) - Relationship: \( V_2 = 2V_1 \) ## What to Find - Velocities at inlet (\( V_1 \)) and throat (\( V_2 \)). ## Step by Step Solution **Apply Bernoulli's Equation:** \[ H_1 + \frac{V_1^2}{2g} = H_2 + \frac{V_2^2}{2g} \] **Substituting known values:** \[ 2.4 + \frac{V_1^2}{2g} = 1.2 + \frac{(2V_1)^2}{2g} \] **Calculate \( V_2 \):** \[ 2.4 + \frac{V_1^2}{2g} = 1.2 + 2\frac{V_1^2}{g} \] **Rearranging:** \[ 2.4 - 1.2 = 2\frac{V_1^2}{g} - \frac{V_1^2}{2g} \] \[ 1.2 = \frac{4V_1^2 - V_1^2}{2g} \] \[ 1.2 = \frac{3V_1^2}{2g} \] **Solving for \( V_1^2 \):** \[ V_1^2 = \frac{2g \cdot 1.2}{3} \] **Assuming \( g \approx 9.81 \, \text{m/s}^2 \):** \[ V_1^2 = \frac{2 \cdot 9.81 \cdot 1.2}{3} \approx 7.848 \] \[ V_1 \approx \sqrt{7.848} \approx 2.8 \, \text{m/s} \] **Calculating \( V_2 \):** \[ V_2 = 2V_1 = 2 \cdot 2.8 \approx 5.6 \, \text{m/s} \] # Final Answers (Summary) - Velocity at inlet: \[ V_1 \approx 2.8 \, \text{m/s} \] - Velocity at throat: \[ V_2 \approx 5.6 \, \text{m/s} \]

# Given Information - Judge's initial certainty of guilt: \( P(G) = 0.63 \) - Probability that the suspect is left-handed: \( P(L) = 0.22 \) - Probability that the criminal is left-handed given the judge's conviction: \( P(L | G) = 0.81 \) # What to Find - The updated probability that the suspect is guilty given they are left-handed: \( P(G | L) \) # Definition or Concept Used - **Bayes' Theorem**: \[ P(G | L) = \frac{P(L | G) \cdot P(G)}{P(L)} \] - **Total Probability** for \( P(L) \): \[ P(L) = P(L | G) \cdot P(G) + P(L | \neg G) \cdot P(\neg G) \] # Step by Step Solution **Step 1: Calculate \( P(\neg G) \)** \[ P(\neg G) = 1 - P(G) = 1 - 0.63 = 0.37 \] **Step 2: Use Bayes' Theorem to find \( P(G | L) \)** 1. **Calculate \( P(L | \neg G) \)**: - Since 22% of the population is left-handed, we can assume this probability is the same for those not guilty: \[ P(L | \neg G) = 0.22 \] 2. **Calculate \( P(L) \)** using the Law of Total Probability: \[ P(L) = P(L | G) \cdot P(G) + P(L | \neg G) \cdot P(\neg G) \] \[ P(L) = (0.81 \cdot 0.63) + (0.22 \cdot 0.37) \] \[ P(L) = 0.5103 + 0.0814 = 0.5917 \] 3. **Now apply Bayes' Theorem**: \[ P(G | L) = \frac{P(L | G) \cdot P(G)}{P(L)} \] \[ P(G | L) = \frac{0.81 \cdot 0.63}{0.5917} = \frac{0.5103}{0.5917} \approx 0.8614 \] # Final Answers (Summary) - The updated certainty of the judge regarding the guilt of the suspect is approximately: \[ P(G | L) \approx 0.8614 \quad \text{(or 86.14\%)} \]