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

Expert verified
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

01

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:$$\omega_n = \sqrt{\frac{4}{2}} = \sqrt{2} \mathrm{~rad/s}$$
02

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 \(\phi=0\).
03

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:$$T = \frac{2 \pi}{\sqrt{2}} = \frac{2 \pi}{\sqrt{2}} \mathrm{~s}$$
04

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.

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