The damped single degree of freedom mass–spring system shown in Fig. 4.15 has a mass m = 25 kg, and a spring stiffness coefficient k = 2500 N/m. Determine the damping coefficient of the system knowing that the mass exhibits an amplitude of 0.01m when the support oscillates at the natural

Question

The damped single degree of freedom mass–spring system shown in Fig. 4.15 has a mass m = 25 kg, and a spring stiffness coefficient k = 2500 N/m.
Determine the damping coefficient of the system knowing that the mass exhibits an amplitude of 0.01m when the support oscillates at the natural frequency of the system with amplitude [latex]Y_0[/latex] = 0.005 m. Determine also the amplitude of dynamic force carried by the support and the displacement of the mass relative to the support.

Accepted Answer

From Eq. 4.98, the amplitude of the oscillation of the mass is given
by
[latex]X_{p} = Y_{0}β_{b} = Y_{0} \frac{\sqrt{1 + (2rξ)^{2}}}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}[/latex]
Since in this example, ωf = ω, that is, r = 1, the above equation reduces to

[latex]X_{p} = Y_{0} \frac{\sqrt{(1 + 2ξ)^2}}{2ξ}[/latex]
From which the damping factor ξ is

ξ = [latex]\frac{1}{2} \frac{1}{\sqrt{\left(\frac{X_p}{Y_0}\right)^{2}− 1}} = \frac{1}{2} \frac{1}{\sqrt{\left(\frac{0.01}{0.005}\right)^{2}− 1}}[/latex] = 0.28868

The critical damping coefficient [latex]C_c[/latex] of the system is
[latex]C_c = 2mω = 2 \sqrt{km} = 2 \sqrt{(2500)(25)}[/latex] = 500N · s/m
The damping coefficient c is
[latex]c = ξC_c[/latex] = 0.28868(500) = 144.34N · s/m

From Eq. 4.101, the amplitude of the force transmitted is
[latex](F_1)_{max} = Y_{0}kr^{2}β_{b} = Y_{0}k \frac{r^{2}\sqrt{1 + (2rξ)^{2}}}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}[/latex]
At resonance r = 1, therefore,

[latex](F_1)_{max} = Y_{0}k \frac{\sqrt{1 + (2rξ)^{2}}}{2ξ} = (0.005)(2500) \frac{\sqrt{1 + (2 × 0.28868)^2}}{2(0.28868)}\[/latex] = 24.9997N
The amplitude of the displacement of the mass relative to the support can be obtained using Eq. 4.105 as
Z = [latex]\frac{Y_0r^2}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}[/latex]

At resonance
Z = [latex]\frac{Y_0}{2ξ} = \frac{0.005}{2(0.28868)} = 8.66 × 10^{−3} [/latex] m
Observe that Z ≠ [latex]X_{p} − Y_{0}[/latex] because of the phase shift angles ψ and [latex]ψ_{b}[/latex].

Question 4.6: The damped single degree of freedom mass–spring system shown...

Related Answered Questions

Figure 4.16 depicts a simple pendulum with a support that has a specified motion y = Y0 sin ωft . Assuming small oscillations, determine the differential equation of motion and the steady state solution. ...

Verified Answer:

The uniform slender bar shown in Fig. 4.4, which has length l and mass m, is pinned at point O and connected to a linear spring which has a stiffness coefficient k. If the bar is subjected to the moment M = M0 sin ωft , obtain the steady state response of this system. ...

Verified Answer:

A vibrometer is used to measure the amplitude of a vibrating machine. It was observed that the machine is vibrating at a frequency of 16 Hz. The natural frequency and damping ratio of the vibrometer are, respectively, 8Hz and 0.7. The reading of the vibrometer indicates that the amplitude of ...

Verified Answer:

Verified Answer:

For the single degree of freedom mass–spring system shown in Fig. 4.1, m = 10 kg, k = 4000 N/m, F0 = 40 N, and ωf = 20 rad/s. The initial conditions are x0 = 0.02m and x0 = 0. Determine the displacement of the mass after t = 0.5 s and t = 1 s. ...

Verified Answer:

It was observed that a machine with a mass M = 600 kg has an amplitude of vibration of 0.01m when the operating speed is 10 Hz. If the machine iscritically damped with an equivalent stiffness coefficient k of 10 × 10³ N/m, determine the amount of unbalance me. ...

Verified Answer:

For the undamped single degree of freedom mass–spring system shown in Fig. 4.1, let m = 5 kg, k = 2500 N/m, F0 = 20 N, and the force frequency ωf = 18 rad/s. The mass has the initial conditions x0 = 0.01m and x0 = 1m/s. Determine the displacement of the mass at t = 1 s. ...

Verified Answer: