# Mass-Spring-Damper System Design Given: - Mass \( m = 8\, \text{kg} \) - Settling time \( t_s = 50\, \text{s} \) (to steady state) - Maximum overshoot \( M_p = 30\% \) - Find: Spring constant \( k \) and damping coefficient \( c \) --- ## Step 1: System Model A mass-spring-damper system is described by: \[ m \ddot{x} + c \dot{x} + kx = \] The corresponding standard second-order system: \[ \ddot{x} + 2\zeta\omega_n \dot{x} + \omega_n^2 x = \] where: - \( \omega_n = \) natural frequency - \( \zeta = \) damping ratio Comparing: - \( 2\zeta\omega_n = \frac{c}{m} \) - \( \omega_n^2 = \frac{k}{m} \) --- ## Step 2: Using Overshoot to Find Damping Ratio Max overshoot formula: \[ M_p = e^{-\frac{\pi \zeta}{\sqrt{1-\zeta^2}}} \] Given \( M_p = .3 \): Set up the equation: \[ .3 = e^{-\frac{\pi \zeta}{\sqrt{1-\zeta^2}}} \] Take natural log: \[ \ln(.3) = -\frac{\pi \zeta}{\sqrt{1-\zeta^2}} \] \[ -1.204 = -\frac{\pi \zeta}{\sqrt{1-\zeta^2}} \] \[ 1.204 = \frac{\pi \zeta}{\sqrt{1-\zeta^2}} \] \[ \sqrt{1-\zeta^2} = \frac{\pi \zeta}{1.204} \] \[ 1 - \zeta^2 = \left(\frac{\pi \zeta}{1.204}\right)^2 \] \[ 1 - \zeta^2 = \frac{\pi^2 \zeta^2}{1.204^2} \] \[ 1 = \zeta^2 \left(1 + \frac{\pi^2}{1.204^2}\right) \] Calculate: \[ \frac{\pi^2}{1.204^2} = \frac{9.870}{1.450} \approx 6.806 \] So, \[ 1 = \zeta^2 (1 + 6.806) = \zeta^2 \times 7.806 \] \[ \zeta^2 = \frac{1}{7.806} \approx .1281 \] \[ \zeta \approx \sqrt{.1281} \approx .358 \] --- ## Step 3: Using Settling Time to Find Natural Frequency Settling time (for 2% criterion): \[ t_s = \frac{4}{\zeta \omega_n} \] Given \( t_s = 50 \): \[ 50 = \frac{4}{.358\, \omega_n} \] \[ .358\, \omega_n = \frac{4}{50} = .08 \] \[ \omega_n = \frac{.08}{.358} \approx .2235\, \text{rad/s} \] --- ## Step 4: Compute \( k \) and \( c \) Recall: \[ \omega_n^2 = \frac{k}{m} \implies k = m \omega_n^2 \] \[ 2\zeta\omega_n = \frac{c}{m} \implies c = 2m\zeta\omega_n \] Plug in values: - \( m = 8 \) - \( \omega_n = .2235 \) - \( \zeta = .358 \) ### Calculate \( k \): \[ k = 8 \times (.2235)^2 = 8 \times .04996 \approx .400\, \text{N/m} \] ### Calculate \( c \): \[ c = 2 \times 8 \times .358 \times .2235 \] \[ = 16 \times .358 \times .2235 \] \[ = 16 \times .07999 \approx 1.28\, \text{N}\cdot\text{s/m} \] --- ## Final Answers **Spring Constant:** \[ \boxed{k \approx .40\, \text{N/m}} \] **Damping Coefficient:** \[ \boxed{c \approx 1.28\, \text{N}\cdot\text{s/m}} \] --- ## Summary Table | Parameter | Value | |-----------|-------| | Mass (\(m\)) | \(8\, \text{kg}\) | | Spring constant (\(k\)) | \(.40\, \text{N/m}\) | | Damping coefficient (\(c\)) | \(1.28\, \text{N}\cdot\text{s/m}\) | | Damping ratio (\(\zeta\)) | \(.358\) | | Natural frequency (\(\omega_n\)) | \(.2235\, \text{rad/s}\) | --- **Note:** - All results are rounded to two decimal places. - The system is underdamped, as expected from the overshoot requirement.