Question

A single degree of freedom system is represented as a \(2 \mathrm{~kg}\) mass attached to a spring possessing a stiffness of \(4 \mathrm{~N} / \mathrm{m}\). Determine the response of the vertically configured system if the mass is displaced 1 meter downward and released from rest. What is the amplitude, period and phase lag for the motion? Sketch and label the response history of the system.

Short answer

Based on the given problem and solution, the amplitude of the motion is 1 m, the period is \(\frac{2 \pi}{\sqrt{2}} \mathrm{~s}\), and the phase lag is 0. The response history is represented by a cosine wave oscillating between +1 m and -1 m with an oscillation frequency of \(\sqrt{2} \mathrm{~rad/s}\).

Step-by-step solution

Calculate the natural frequency

The natural frequency of the system can be found using the formula: $$\omega_n = \sqrt{\frac{k}{m}}$$Where \(\omega_n\) is the natural frequency, \(k = 4 \mathrm{~N} / \mathrm{m}\) is the stiffness of the spring, and \(m = 2 \mathrm{~kg}\) is the mass. Using the given values, the natural frequency can be calculated as:$$\omega_n = \sqrt{\frac{4}{2}} = \sqrt{2} \mathrm{~rad/s}$$

Determine the amplitude and phase

Since the system is initially displaced 1 meter downward and released from rest, the initial displacement \(x(0) = 1 \mathrm{~m}\) and initial velocity \(\dot{x}(0) = 0 \mathrm{~m/s}\). The general solution for the displacement of a single degree of freedom system can be written as:$$x(t) = A \cos(\omega_n t + \phi)$$Where \(A\) is the amplitude of the motion, \(\omega_n\) is the natural frequency, \(t\) is the time, and \(\phi\) is the phase angle. Using the initial conditions, we get:$$1 = A \cos(\phi)$$And:$$0 = -A \sqrt{2} \sin(\phi)$$The first equation suggests that the amplitude \(A=1 \mathrm{~m}\), and since the system is released from rest, the phase angle \(\phi=0\).

Calculate the period of the motion

The period of the motion can be found using the formula:$$T = \frac{2 \pi}{\omega_n}$$Using the calculated natural frequency, the period of the motion is:$$T = \frac{2 \pi}{\sqrt{2}} = \frac{2 \pi}{\sqrt{2}} \mathrm{~s}$$

Sketch the response history

The response history can be sketched by plotting the displacement formula, \(x(t) = \cos(\sqrt{2} t)\), over the time period \(T\). Since the amplitude is 1 m and the period is \(\frac{2 \pi}{\sqrt{2}} \mathrm{~s}\), the curve will oscillate between +1 m and -1 m with an overall oscillation frequency of \(\sqrt{2} \mathrm{~rad/s}\). Note that the phase lag is 0, so the curve will start at the maximum displacement. To summarize, the amplitude of the motion is 1 m, the period is \(\frac{2 \pi}{\sqrt{2}} \mathrm{~s}\), and the phase lag is 0. The response history of the system can be sketched as a cosine wave with these characteristics.

Key concepts

Natural Frequency

In a single degree of freedom system, the natural frequency is a fundamental concept. Imagine it as the frequency at which the system vibrates when disturbed from its equilibrium position without any external force acting upon it. To calculate it, we use the formula: \[\omega_n = \sqrt{\frac{k}{m}}\]Here, \(k\) represents the stiffness of the spring, and \(m\) is the mass. It's crucial because it dictates how quickly the system will oscillate back and forth. From the given exercise, with a spring stiffness \(k\) of \(4 \mathrm{~N/m}\) and a mass \(m\) of \(2 \mathrm{~kg}\), the natural frequency comes out to be \(\sqrt{2} \mathrm{~rad/s}\). Understanding this helps us predict the behavior of the system under given conditions.

Whenever you're dealing with such oscillatory motion, identifying the natural frequency helps in foreseeing potential resonance issues, where a system might suffer excessive oscillations at specific frequencies.

Amplitude of Motion

The amplitude of motion in a vibratory system signifies the maximum displacement the mass undergoes from its mean position. It's the peak value of motion and is essential to understanding how far the system will move when disturbed. In our case, since the mass is displaced 1 meter downward and released from rest, the initial position itself becomes the amplitude. Hence, the amplitude \(A\) is \(1 \mathrm{~m}\).

This means that in a cycle of motion, the mass will oscillate between \(+1 \mathrm{~m}\) and \(-1 \mathrm{~m}\) around the equilibrium position. Recognizing the amplitude is crucial, especially in engineering applications, as it could determine design constraints like maximum allowable displacement without failure.

Period of Motion

The period of motion represents the time it takes for the system to complete one full cycle of oscillation. Let's think of it as how long it takes for the mass to return to the same point in its path. Calculating the period can be done using the natural frequency:\[T = \frac{2 \pi}{\omega_n}\]With the natural frequency \(\omega_n = \sqrt{2} \mathrm{~rad/s}\), the period comes out to \(\frac{2 \pi}{\sqrt{2}} \mathrm{~s}\). This value is vital in scenarios where synchronization and timing of oscillations matter, such as in tuning musical instruments or designing systems to avoid unwanted vibrations.

Knowing the period helps in visualizing how frequently the mass cycles through the paths, which is incredibly valuable for both maintenance and the efficient functionality of vibratory systems.

Phase Angle

The phase angle \(\phi\) describes how much the oscillation is shifted horizontally along the time axis. It's a crucial part of understanding the initial conditions of the motion. For systems released from rest, as in this exercise, the phase angle helps determine when within the oscillatory cycle the motion starts. The general displacement equation,\[x(t) = A \cos(\omega_n t + \phi)\]gives insight into this. Since the initial velocity is zero and displacement is maximum, the phase angle \(\phi\) is \(0\). This indicates that the motion begins at the peak displacement, simplifying our calculation, as it starts at this maximum point in the cycle.

Understanding the phase angle is like understanding the timing of an event within a repeating sequence, making it valuable for phase-sensitive applications, where alignment and synchronization are imperative.