A damped single degree of freedom mass–spring system has mass m = 10 kg, spring coefficient k = 4000 N/m, and damping coefficient c = 40 N·s/m. The amplitude of the forcing function F0 = 60 N, and the forcing frequency ωf = 40 rad/s. Determine the displacement of the mass as a function of time, and

Question

A damped single degree of freedom mass–spring system has mass m = 10 kg, spring coefficient k = 4000 N/m, and damping coefficient c = 40 N·s/m. The amplitude of the forcing function [latex]F_{0}[/latex] = 60 N, and the forcing frequency [latex]ω_{f}[/latex] = 40 rad/s. Determine the displacement of the mass as a function of time, and determine also the transmissibility and the amplitude of the force transmitted to the support.

Accepted Answer

The circular frequency of the system is
ω = [latex]\sqrt{\frac{k}{m}} = \sqrt{\frac{4000}{10}}[/latex] = 20 rad/s

The frequency ratio r is given by
r = [latex]\frac{ω_{f}}{ω} = \frac{40}{20}[/latex] = 2
The critical damping coefficient [latex]C_{c}[/latex] is defined as

[latex]C_{c}[/latex] = 2mω = 2(10)(20) = 400 N · s/m
The damping factor ξ is then given by
ξ = [latex]\frac{c}{C_{c}} = \frac{40}{400}[/latex] = 0.1

which is the case of an underdamped system, and the complete solution can be written in the following form
x(t) = [latex]x_{h} + x_{p} = X e^{−ξωt}[/latex] sin([latex]ω_{d}[/latex]t + Φ) + [latex]X_{0}[/latex]β sin([latex]ω_{f}[/latex]t − ψ)
where [latex]ω_{d}[/latex] is the damped circular frequency

[latex]ω_{d} = ω \sqrt{1 − ξ²} = 20 \sqrt{1 − (0.1)²}[/latex] = 19.8997 rad/s
The constants [latex]X_{0}[/latex], β, and ψ are

[latex]X_{0} = \frac{F_{0}}{k} = \frac{60}{4000}[/latex] = 0.015 m

β = [latex]\frac{1}{\sqrt{(1 − r²)² + (2rξ)²}} = \frac{1}{\sqrt{(1 − (2)²)² + (2 × 2 × 0.1)²}}[/latex] = 0.3304

ψ = [latex]tan^{−1} (\frac{2rξ}{1 − r²}) = tan^{−1}\frac{2(2)(0.1)}{1 − (2)²}[/latex] = −0.13255 rad
The displacement can then be written as a function of time as

x(t) = X[latex]e^{−2t}[/latex] sin(19.8997t + Φ) + 0.004956 sin(40t + 0.13255)

The constants X and Φ can be determined using the initial conditions. The transmissibility [latex]β_{t}[/latex] is defined by Eq. 4.58 as

[latex]β_{t} = β \sqrt{1 + (2rξ)²} = \frac{\sqrt{1 + (2rξ)²}}{\sqrt{(1 − r²)² + (2rξ)²}}[/latex] (4.58)

[latex]β_{t} = \frac{\sqrt{1 + (2rξ)²}}{\sqrt{(1 − r²)² + (2rξ)²}}[/latex] = [latex]\frac{\sqrt{1 + (2 × 2 × 0.1)²}}{\sqrt{[1 − (2)²]² + (2 × 2 × 0.1)²}}[/latex] = 0.35585

The amplitude of the force transmitted is given by
[latex]|F_{t}| = F_{0}β_{t}[/latex] = (60)(0.35585) = 21.351 N

Question 4.4: A damped single degree of freedom mass–spring system has mas...

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