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lecture_3 - ME 163 Vibrations Lecture 3 In the second...

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ME 163 Vibrations Lecture 3 In the second lecture we emphasized the idea that the canonical equa- tion for free undamped vibrations is m ˙ x + kx = 0 where m is the mass and k the stiffness of the spring-mass system. We introduced various changes of coordinates that bring the vibrating system into that canonical form. We also discussed the method of linearization which we use to deal with nonlinear systems. When they vibrate around equilibrium x eq , we expand the force into a Taylor series around that equilibrium: f ( x ) = f ( x eq )+ df dx ( x eq ) · ( x - x eq )+ 1 2 d 2 f dx 2 ( x eq ) · ( x - x eq ) 2 + O ( x - x eq ) 3 , where O ( x - x eq ) 3 means we are neglecting terms of third order in dis- tance from equilibrium x - x eq . We plug this expansion into m ¨ x = f ( x ) and then neglect all the terms except the ones that have linear depen- dence on x - x eq . In practice, it is best to always change coordinates so that your x eq = 0 as we have shown in the last lecture. Now, we will often be able to bring our system to canonical form, but with some constants in front of ¨ x and x that have the units of mass and stiffness, but are actually the modified mass and modified stiffness. We call these constants ”effective mass” and ”effective stiffness”. We already saw examples of effective stiffness in the previous lecture, when we combined springs in parallel and series. In this lecture we will see that, in more complicated, but still single degrees of freedom vibrating system we can have an effective mass coefficient, too. In order to look at some examples of this, we will first talk about the energy of a vibrating system. Energy method for determining natural frequency: An easy way of obtaining the natural frequency of a vibrating system of a single degree of freedom is to write down its energy. If the expression for energy has kinetic and potential part as follows: E = 1 2 m ˙ x 2 + 1 2 kx 2 , (1) 1
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where T = 1 2 m ˙ x 2 is the kinetic energy and U = 1 2 kx 2 is the potential energy of the spring. The natural frequency is then obtained as ω 2 n = k m , i.e. as the ratio of the coefficients in front of x 2 and ˙ x 2 in the expression for energy. Let’s consider the example of the mathematical pendulum we stud- ied in the last lecture. Figure 1: Mathematical pendulum of length l and mass m . The pendulum has velocity only in the angular direction, v = v θ e θ , 2
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where e θ is the unit vector in the angular direction (see figure 1). The kinetic energy is then one half mass times velocity squared: T = 1 2 m ( l ˙ θ ) 2 , while the potential energy is that of gravity force U = mgz = mg ( l - lcosθ ) where z is the height measured from the straight down position of the mass. Thus, the energy reads E = 1 2 m ( l ˙ θ ) 2 + mg ( l - l cos θ ) Now we notice that for θ small (i.e. θ eq = 0, and θ - θ eq small) the Taylor expansion of cos θ = 1 - θ 2 / 2 + O ( θ 3 ) leads to E ( 1 2 ml 2 ) ˙ θ 2 + mgl θ 2 2 (2) Note that we kept the
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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_3 - ME 163 Vibrations Lecture 3 In the second...

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