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Chapter 2: Problem 4

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

Expert verified
Answer: The amplitude of the motion is approximately 3.67 m, the period is approximately 5.15 sec, and the phase lag is approximately 1.79 rad.

Step by step solution

01

Calculate the natural frequency of the system

The first step is to define the natural frequency of the mass-spring system using the formula: $$\omega_n = \sqrt{\frac{k}{m}}$$ where \(k = 6 \mathrm{~N} / \mathrm{m}\) is the stiffness of the spring, and \(m = 4 \mathrm{~kg}\) is the mass. $$\omega_n = \sqrt{\frac{6}{4}} = \sqrt{1.5} \mathrm{~rad} / \mathrm{sec}$$
02

Calculate the period of the motion

The next step is to calculate the period of the motion, which is the time it takes for the system to complete one full oscillation. The formula is: $$T = \frac{2\pi}{\omega_n}$$ Now, substituting the natural frequency, we get: $$T = \frac{2\pi}{\sqrt{1.5}} \approx 5.15 \mathrm{~sec}$$
03

Determine the initial conditions

The mass is initially displaced 2 meters to the right, which means \(x(0) = 2\mathrm{~m}\). The mass is released with a velocity of 4 m/s, which means \(v(0) = 4\mathrm{~m} / \mathrm{sec}\).
04

Determine the damping ratio and amplitude

Since the system is not damped, the damping ratio \(\zeta = 0\). Therefore, the amplitude of the motion, \(A\), can be calculated using the initial displacement and the initial velocity: $$A = \sqrt{x(0)^2 + \frac{v(0)^2}{\omega_n^2}}$$ Substituting the known values, we get: $$A = \sqrt{2^2 + \frac{4^2}{(\sqrt{1.5})^2}} = \sqrt{4 + \frac{16}{1.5}} \approx 3.67 \mathrm{~m}$$
05

Determine the phase lag

The phase lag 蠁 can be determined using the formula: $$\tan\phi = \frac{v(0)}{\omega_n x(0)}$$ Inserting the known values, we get: $$\tan\phi = \frac{4}{\sqrt{1.5} \cdot 2}$$ $$\phi = \tan^{-1} \left(\frac{4}{\sqrt{1.5} \cdot 2}\right) \approx 1.79 \mathrm{~rad}$$
06

Sketch and label the response history of the system

To sketch the response history of the system, plot the displacement as a function of time using the equation: $$x(t) = A \cos(\omega_n t + \phi)$$ Substitute the values of amplitude, natural frequency, and phase lag: $$x(t) = 3.67 \cos(\sqrt{1.5}t + 1.79)$$ The resulting graph would be a sinusoidal wave with an amplitude of 3.67 m, a period of approximately 5.15 sec, and a phase shift of approximately 1.79 rad. The wave will oscillate between maximum and minimum values with a sinusoidal motion, starting at the initial displacement of 2 m and moving with the initial velocity of 4 m/s.

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