Question

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\) and a viscous damper whose coefficient is \(2 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\). (a) Determine the response of the horizontally configured system if the mass is displaced 1 meter to the right and released from rest. Plot and label the response history of the system. (b) Determine the response and plot its history if the damping coefficient is \(8 \mathrm{~N}-\mathrm{sec} / \mathrm{m}\).

Short answer

Question: Calculate and compare the response of a single degree of freedom system with an initial displacement of 1 m, mass of 2 kg, and spring stiffness of 4 N/m when subjected to (a) a viscous damper coefficient of 2 N·s/m and (b) a viscous damper coefficient of 8 N·s/m. Briefly describe the differences in the response histories. Answer: For case (a) with a viscous damper coefficient of 2 N·s/m, the system is underdamped, and its response is given by the equation: $$x(t) = e^{-\frac{\sqrt{2}}{4}t}( \cos\left(\frac{\sqrt{15}}{2}t\right) + \frac{\sqrt{2}}{\sqrt{15}}\sin\left(\frac{\sqrt{15}}{2} t\right))$$ For case (b) with a viscous damper coefficient of 8 N·s/m, the system is critically damped, and its response is given by the equation: $$x(t) = (1 + \sqrt{2}t)e^{-\sqrt{2}t}$$ The underdamped system (case a) exhibits oscillatory behavior, with the amplitude of oscillations decaying over time due to damping, whereas the critically damped system (case b) returns to the equilibrium position as quickly as possible without any oscillations.

Step-by-step solution

Obtain the governing equation of motion

The governing equation of motion of the system can be represented by the following differential equation: $$m \ddot{x} + c \dot{x} + kx = 0$$ where m is the mass, c is the viscous damping coefficient, k is the spring stiffness, and x is the displacement of the mass. In our case, we have: - m = 2 kg - k = 4 N/m - Case (a): c = 2 N·s/m - Case (b): c = 8 N·s/m

Analyze the type of damping

To determine the type of damping, we need to calculate the damping ratio \(\zeta\). The damping ratio is given by: $$\zeta = \frac{c}{2\sqrt{mk}}$$ For case (a), the damping ratio is: $$\zeta_a = \frac{2}{2\sqrt{2\cdot4}} = \frac{1}{4}$$ For case (b), the damping ratio is: $$\zeta_b = \frac{8}{2\sqrt{2\cdot4}} = 1$$ Since \(\zeta_a < 1\) for case (a), we have an underdamped system. For case (b), \(\zeta_b = 1\), which indicates a critically damped system.

Calculate the natural frequency and damped frequency

The natural frequency \(\omega_n\) is given by: $$\omega_n = \sqrt{\frac{k}{m}}$$ So, \(\omega_n = \sqrt{\frac{4}{2}} = \sqrt{2} ~\mathrm{rad/s}\) For the underdamped case (a), we need to calculate the damped frequency \(\omega_d\). The damped frequency is given by: $$\omega_d = \omega_n\sqrt{1-\zeta^2}$$ $$\omega_d = \sqrt{2}\sqrt{1-\left(\frac{1}{4}\right)^2} = \frac{\sqrt{15}}{2} ~\mathrm{rad/s}$$

Find the response of the system

For the underdamped case (a), the response is given by: $$x(t) = e^{-\zeta \omega_n t}(A\cos(\omega_d t) + B\sin(\omega_d t))$$ We also have the initial conditions: - Initial displacement: \(x(0) = 1 ~\mathrm{m}\) - Initial velocity: \(\dot{x}(0) = 0 ~\mathrm{m/s}\) Applying the initial conditions, we get: $$A = 1$$ and $$B = \frac{\omega_n \zeta}{\omega_d}$$ The response of the system for case (a) is: $$x(t) = e^{-\frac{\sqrt{2}}{4}t}( \cos\left(\frac{\sqrt{15}}{2}t\right) + \frac{\sqrt{2}}{\sqrt{15}}\sin\left(\frac{\sqrt{15}}{2} t\right))$$ For the critically damped case (b), the response is given by: $$x(t) = (A + Bt)e^{-\omega_n t}$$ Applying the initial conditions as before, we get: $$A = 1$$ and $$B = \omega_n$$ The response of the system for case (b) is: $$x(t) = (1 + \sqrt{2}t)e^{-\sqrt{2}t}$$

Plot the response history

To plot the response history, generate time values (e.g., from 0 to 10 seconds with 0.1-second increments) and calculate corresponding displacement values using the response equations we found for each case: - Case (a): \(x(t) = e^{-\frac{\sqrt{2}}{4}t}( \cos\left(\frac{\sqrt{15}}{2}t\right) + \frac{\sqrt{2}}{\sqrt{15}}\sin\left(\frac{\sqrt{15}}{2} t\right))\) - Case (b): \(x(t) = (1 + \sqrt{2}t)e^{-\sqrt{2}t}\) Then, plot the displacement against time for both cases and label the response histories accordingly.