Question

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\) and a viscous damper whose coefficient is \(1 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (a) Determine the response of the horizontally configured system if the mass is displaced 2 meters to the right and released with a velocity of 4 \(\mathrm{m} / \mathrm{sec}\). Plot and label the response history of the system. (b) Determine the response and plot its history if the damping coefficient is \(5 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (c) Determine the response and plot its history if the damping coefficient is \(10 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\).

Short answer

Answer: The response function of a single degree of freedom system depends on the damping ratio (ζ). For different values of ζ, the response function can be divided into three scenarios: 1. Underdamped (\(\zeta < 1\)): \(x(t)=e^{-\zeta\omega_nt}\left[x_0\cos(\omega_d t)+\left(\frac{v_0+\zeta\omega_nx_0}{\omega_d}\right)\sin(\omega_d t)\right]\), where \(\omega_d=\omega_n\sqrt{1-\zeta^2}\). 2. Critically damped (\(\zeta = 1\)): \(x(t)=e^{-\omega_nt}\left[x_0+(v_0+\omega_n x_0)t\right]\) 3. Overdamped (\(\zeta > 1\)): \(x(t)=e^{-\zeta\omega_nt}\left[C_1e^{\omega_n\sqrt{\zeta^2-1}t}+C_2 e^{-\omega_n \sqrt{\zeta^2-1}t}\right]\), where \(C_1\) and \(C_2\) are coefficients determined from the initial conditions.

Step-by-step solution

Setup the equation of motion

For a single degree of freedom system with mass (\(m\)), spring stiffness (\(k\)) and damping coefficient (\(c\)), the equation of motion can be written as: \(m \ddot{x} + c \dot{x} + k x = 0\) We are given that \(m = 4\text{ kg}\), \(k = 6\text{ N/m}\), and the initial displacement \(x_0 = 2\text{ m}\) and velocity \(v_0 = 4\text{ m/s}\).

Determine the natural frequency and damping ratio

Compute the natural frequency (\(\omega_n\)) and damping ratio (\(\zeta\)) for the given system. They are calculated as follows: \(\omega_n = \sqrt{\frac{k}{m}}\) \(\zeta = \frac{c}{2 m \omega_n}\)

Solve the equation of motion for different damping coefficients

For each damping coefficient, we can now solve the equation of motion to compute the response of the system. Since we have 3 different damping coefficients: \(1\text{ Ns/m}\), \(5\text{ Ns/m}\), \(10\text{ Ns/m}\), we need to compute the response for each case.

Determine the response function

Depending on the value of the damping ratio \(\zeta\), the response function \(x(t)\) will have different forms: 1. For \(\zeta < 1\), the system is underdamped, and the response function is: \(x(t)=e^{-\zeta\omega_nt}\left[x_0\cos(\omega_d t)+\left(\frac{v_0+\zeta\omega_nx_0}{\omega_d}\right)\sin(\omega_d t)\right]\) where \(\omega_d=\omega_n\sqrt{1-\zeta^2}\) is the damped natural frequency. 2. For \(\zeta = 1\), the system is critically damped, and the response function is: \(x(t)=e^{-\omega_nt}\left[x_0+(v_0+\omega_n x_0)t\right]\) 3. For \(\zeta > 1\), the system is overdamped, and the response function is: \(x(t)=e^{-\zeta\omega_nt}\left[C_1e^{\omega_n\sqrt{\zeta^2-1}t}+C_2 e^{-\omega_n \sqrt{\zeta^2-1}t}\right]\) where \(C_1\) and \(C_2\) are coefficients determined from the initial conditions.

Calculate the response and plot the history

For each damping coefficient, compute the response function using the appropriate formula from Step 4. Then, plot the response history for each case (a), (b), and (c). For (a) \(c=1\text{ Ns/m}\), compute the response and plot the history using the appropriate formula. For (b) \(c=5\text{ Ns/m}\), compute the response and plot the history using the appropriate formula. For (c) \(c=10\text{ Ns/m}\), compute the response and plot the history using the appropriate formula.