Question

A single degree of freedom system is represented as a \(4 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(6 \mathrm{~N} / \mathrm{m}\). Determine the response of the horizontally configured system if the mass is displaced 2 meters to the right and released with a velocity of \(4 \mathrm{~m} / \mathrm{sec}\). What is the amplitude, period and phase lag for the motion? Sketch and label the response history of the system.

Short answer

To create a short answer question based on this solution, we can ask: **Question:** Consider a single-degree-of-freedom system with mass \(4\mathrm{~kg}\) and spring stiffness \(6\mathrm{~N/m}\). At the beginning, the displacement is \(2\mathrm{~m}\) and the velocity is \(4\mathrm{~m/s}\). Calculate the natural frequency, amplitude, period, and phase lag (angle in degrees) of the motion. **Answer:** Natural frequency: \(\omega_n = \sqrt{1.5}\mathrm{~rad/sec}\) Amplitude: \(A = \frac{2}{\cos\left( \tan^{-1} \left( \frac{-2}{\sqrt{1.5}}\right) \right)}\) Period: \(T = \frac{2\pi}{\sqrt{1.5}}\mathrm{~s}\) Phase lag: \(\phi = \tan^{-1} \left( \frac{-2}{\sqrt{1.5}}\right)\) in degrees

Step-by-step solution

Find the natural frequency

We can find the natural frequency (\(\omega_n\)) of the system using the mass (m) and stiffness (k) of the spring. The formula for natural frequency is: \(\omega_n = \sqrt{\frac{k}{m}}\) In our case, the mass is \(4\mathrm{~kg}\) and the stiffness is \(6\mathrm{~N/m}\). Plugging in these values, we get: \(\omega_n = \sqrt{\frac{6}{4}} = \sqrt{1.5}\mathrm{~rad/sec}\)

Find the amplitude and phase lag

To find the amplitude (A) and phase lag (\(\phi\)), we'll use the initial conditions of the system: - Initial displacement: \(x_0 = 2\mathrm{~m}\) (2 meters to the right) - Initial velocity: \(v_0 = 4\mathrm{~m/s}\) Since the system is oscillating with a sinusoidal function, we can represent the displacement, x(t), as: \(x(t) = A\cos(\omega_n t + \phi)\) The velocity, v(t), is the derivative of x(t) with respect to time: \(v(t) = -A\omega_n\sin(\omega_n t + \phi)\) At t = 0, the initial displacement and velocity are: \(x(0) = A\cos(\phi) = 2\mathrm{~m}\) \(v(0) = -A\omega_n\sin(\phi) = 4\mathrm{~m/s}\) We need to solve this system of equations for A and \(\phi\). Divide the second equation by the first equation: \(\frac{-A\omega_n\sin(\phi)}{A\cos(\phi)} = \frac{4}{2}\) \(- \omega_n \tan(\phi) = 2\) Now, plug in the value of \(\omega_n\) from Step 1: \(- \sqrt{1.5} \tan(\phi) = 2\) Solving for \(\phi\), we get: \(\phi = \tan^{-1} \left( \frac{-2}{\sqrt{1.5}}\right)\) To find A, plug in the value of \(\phi\) into the first equation: \(A\cos\left( \tan^{-1} \left( \frac{-2}{\sqrt{1.5}}\right) \right)= 2\) Now, solve for A: \(A = \frac{2}{\cos\left( \tan^{-1} \left( \frac{-2}{\sqrt{1.5}}\right) \right)}\)

Calculate the period of oscillation

The period (T) of oscillation is the time it takes for the oscillation to complete one full cycle. The relation between the period and the natural frequency is given by: \(T = \frac{2\pi}{\omega_n}\) Using the value of \(\omega_n\) we calculated in Step 1, we get: \(T = \frac{2\pi}{\sqrt{1.5}}\mathrm{~s}\)

Sketch and label the response history

Now that we have the amplitude, period, and phase lag, we can sketch the response history for the single-degree-of-freedom system. Plot the curve of x(t) using the values we found for A, \(\omega_n\), and \(\phi\). The period T represents how wide the cycle is, while the amplitude A gives its height. The phase lag \(\phi\) will determine the phase shift of the oscillation.

Key concepts

Natural Frequency

Natural frequency plays a crucial role in the dynamics of a single degree of freedom (SDOF) system. It is defined as the frequency at which a system oscillates in the absence of any damping force or external driving force. For the problem at hand, the natural frequency, denoted as \(\omega_n\), can be calculated using the formula \(\omega_n = \sqrt{\frac{k}{m}}\) where \(k\) is the stiffness of the spring and \(m\) is the mass attached to it.

When the mass of \(4 \text{kg}\) is connected to a spring with stiffness \(6 \text{N/m}\), the natural frequency is found by squaring the quotient of the stiffness and the mass. This gives us \(\omega_n = \sqrt{1.5}\) rad/sec, which is a measure of how quickly the system tends to oscillate when disturbed.

Oscillation Amplitude

The oscillation amplitude, referred to as \(A\), indicates the maximum displacement of an oscillating system from its equilibrium position. It's a measure of the energy in the oscillation—the larger the amplitude, the more energy the oscillation has. To determine the amplitude in our example, we used the initial conditions of the system and sinusoidal function relationships. After applying initial conditions and solving the combined equations - including the phase lag - we found that \(A\) equals twice the cosine of the tangent inverse of \(\frac{-2}{\sqrt{1.5}}\). This provides us with a number that represents the maximum extent to which the mass will move away from its rest position during each oscillation.

System Damping

System damping is a force that reduces the amplitude of oscillations over time and is not covered in the given problem since it focuses on an undamped SDOF system. Damping is typically due to resistance factors like friction or air resistance and is critical in real-world applications. In engineering, we want to control system damping to prevent excessive vibrations and potential damage. For instance, in vehicle suspension, proper damping ensures comfort and stability. Even though the provided example assumes a system without damping, understanding damping is necessary for comprehensively analyzing the behavior of oscillatory systems.

Phase Lag

Phase lag, symbolized by \(\phi\), describes the shift in phase between the oscillation and a reference point in time. In oscillatory systems, it represents how much the waveform is shifted from its expected position in the absence of initial velocity. To find the phase lag in the given scenario, we looked at the initial conditions of velocity and displacement. By taking the ratio of initial velocity over initial displacement and equating this to the system's natural frequency times the tangent of the phase lag, we found \(\phi\). The phase lag also affects how the oscillation starts; in this case, the negative tangent inverse indicates that the mass begins its motion by moving to the left, even though it was displaced to the right initially.

Period of Oscillation

The period of oscillation, known as \(T\), is the time it takes for the mass to complete one full cycle of motion and return to its starting position. It's inversely proportional to the natural frequency; thus, a higher natural frequency means a shorter period. To calculate the period, we use the relation \(T = \frac{2\pi}{\omega_n}\). From the natural frequency calculated previously, we determine that the period for our \(4 \text{kg}\) mass attached to a \(6 \text{N/m}\) stiffness spring system is \(T = \frac{2\pi}{\sqrt{1.5}}\) seconds. This means that it takes that much time for the mass to oscillate back and forth once, which is a handy reference for predicting future positions and velocities of the system.