The block of weight W is connected in a rigid frame between a linear spring and a viscous damper. The frame is subjected to the time dependent vertical displacement y(t) = Y sin ωt. The displacement x of the block is measured from its static equilibrium position (with support stationary at y = 0).

Question

The block of weight W is connected in a rigid frame between a linear spring and a viscous damper. The frame is subjected to the time dependent vertical displacement y(t) = Y sin ωt. The displacement x of the block is measured from its static equilibrium position (with support stationary at y = 0). Determine the steadystate solution for (1) the relative displacement z = x − y; and (2) the absolute displacement x. Use Y = 40 mm, ω = 400 rad/s, M = 3 kg, k = 2.63 × 10[latex]^5[/latex] N/m, and c = 585 N · s/m.

Accepted Answer

Part 1

Because the given system is equivalent to that shown in Fig. 20.10, Eqs. (20.30)–(20.32) can be used to determine the relative displacement z. The parameters appearing in these equations are

z = Z sin(ωt − Φ) (20.30)

[latex]\phi = tan^{−1} [\frac{2ζω/p}{1 − (ω/p)^2}][/latex] (20.32)

[latex]p = \sqrt{\frac{k}{m}} =\sqrt{\frac{2.63 × 10^5}{3}}[/latex] = 296 rad/s

[latex]ζ = \frac{c}{2mp} = \frac{585}{2(3)(296)}[/latex] = 0.329 (underdamped)

[latex]\frac{ω}{p} = \frac{400}{296}[/latex] = 1.351

The relative amplitude of the steady-state solution is now obtained from Eq. (20.31)

[latex]Z = \frac{mY ω^2/k}{\sqrt{[1 − (ω/p)^2]^2 + (2ζω/p)^2}}[/latex]

[latex]= Y \frac{(ω/p)^2}{\sqrt{[1 − (ω/p)^2]^2 + (2ζω/p)^2}}[/latex] (20.31)

[latex]Z = Y \frac{(ω/p)^2}{\sqrt{[1 − (ω/p)^2]^2 + (2ζω/p)^2}}[/latex]

[latex] = 0.04\frac{(1.287)^2}{\sqrt{[1 − (1.351)^2]^2 + [2(0.329)(1.351)]^2}}[/latex]

= 0.07 m

and the phase angle can be computed from Eq. (20.32)

[latex]\phi = tan^{−1} [\frac{2ζω/p}{1 − (ω/p)^2}] = tan^{−1} [\frac{2(0.329)(1.351)}{1 − (1.351)^2}][/latex]

= −0.8226

The relative displacement thus becomes

z(t) = 2.249 sin(400t + 0.8226) m

Note that z(t) leads y(t) by 0.8226 rad (47.1°).

Part 2

Using the trigonometric identity sin(a + b) = sin a cos b + cos a sin b, the expression for z(t) in Part 1 can be written as

z(t) = 0.07(cos 0.8226 sin 400t + sin 0.8226 cos 400t)

= 0.048 sin 400t + 0.0512 cos 400t m

Therefore, the expression for the absolute displacement of the mass becomes

x = z + y
= (0.048 sin 400t + 0.0512 cos 400t) + 0.04 sin 400t
= 0.088 sin 400t + 0.0512 cos 400t m

Question 20.6: The block of weight W is connected in a rigid frame between ...

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