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lecture_5 - Lecture 5 Example 1 Consider a car with 1/4...

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Lecture 5 Example 1 Consider a car, with 1 / 4 mass m = 500 kg, whose sus- pension for a single tire has k = 200000 N/m . The damping coeﬃcient is c = 2000 Nm/s . A car goes over a bump on the road that gives the suspension system a 6 cm displacement. In how many periods will the car reach less than 0 . 06 cm, i.e. less than 1 / 100 of the initial value of the displacement? The logarithmic decrement for the n -th period reads ln x 1 x n = ln x 1 x 2 · x 2 x n 3 · ... · x n x n - 1 = ( n - 1) ln x 1 x 2 . Thus, n = 1 + ln(100) p (1 - ζ 2 ) 2 πζ . since ζ = 0 . 1 , n = 9 . Note that c c = 20000 Nm/s , f = 3 Hz . Visualization of viscous damping The three types of viscous damping that we studied in the last lecture have a distinct visual diﬀerences associated with them. In ﬁgure 1 we plot for reference the case of free, undamped vibrations with mass m = 1 kg, the stiﬀness k = 50 N/m, giving the natural frequency f n = 1 . 1254 Hz (the angular natural frequency ω n = 7 . 0711. The top of that ﬁgure shows the position of the mass starting from the equilibrium x (0) = 0 with velocity ˙ x (0) = 1. The amplitude of the periodic oscillation is ˙ x (0) n = 0 . 1414. The graphic in the middle represents velocity vs. time for the same free oscillation. Starting from velocity of ˙ x (0) = 1, the curve of velocity is also periodic with the same period as the position-time curve. The graphic at the bottom is the representation of motion that we will talk about more later and that will be very useful for you when you study automatic control later. It is the velocity vs. position curve. Let’s take a moment and describe it: look at the vertical axis at position x = 0, where the velocity is ˙ x = 1. That’s the situation at the beginning of the motion. Looking from that point to the right, the motion proceeds so that the velocity is reduced and the position increases, until the curve hits ˙ x = 0. There, the system reaches the maximum positive x with zero velocity. From 1

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that point, the curve turns to the left, position decreases towards zero and velocity goes negative and decreases to its maximal negative value ˙ x = - 1 reached at x = 0. After that, the velocity increases towards zero and becomes positive again after the system reaches the maximum negative position x = -
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This note was uploaded on 03/23/2009 for the course ME 163 taught by Professor Mezic,i during the Winter '08 term at UCSB.

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lecture_5 - Lecture 5 Example 1 Consider a car with 1/4...

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