lecture_5

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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 = -
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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.

Page1 / 8

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online