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lecture_2 - ME 163 Vibrations Lecture 2 We are going to...

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Unformatted text preview: ME 163 Vibrations Lecture 2 We are going to consider a couple of examples where the equations of motion-these are second order ordinary differential equations that we derive using Newtonian mechanics - can be reduced to the above analyzed canonical form. Each example serves to introduce another physical concept. In the first one we emphasize that it is useful to define the coordinates that describe degrees of freedom starting from the unstretched length of the spring. Example 1 Consider the free vibrations of the system shown in figure 1 where the unstretched length of the spring is labeled l . Let y be the coordinate measuring the distance of the mass from the stationary wall on the left. The force in the spring is equal to F ( y ) =- k ( y- l ) . The minus sign comes about because if y > l i.e. the mass is to the right of the unstretched length of the spring, the spring is extended and the force pushes back on the mass, to the left. The sign of the force in that case is negative, so multiplying the positive y- l with negative- k gives it the correct sign. In the same way, if the mass is to the left of the unstretched length position y = l , the sign of y- l is negative, while the spring is compressed and the spring force pushes it to the right. We have Figure 1: m y =- k ( y- l ) Now we use what we will use a lot in this class: a change of coordinates. We define the new variable x = y- l . This new variable x measures the distance from the unstretched length. If it is positive, the spring is 1 extended. If it is negative, it is compressed. Since x = y and x = y , we have m x =- kx We have reduced the equation of motion to its canonical form. In the next example, we will study the effect of gravity on the spring- mass system. Example 2 Consider the vibrating system shown in figure 2. Deploying Figure 2: the free body diagram technique, as shown in the figure we obtain m y =- k ( y- l ) + w (1) where w = mg is the force of gravity, acting straight down. Consider the position of the mass y eq with zero velocity and zero acceleration: y = 0 and y = 0 Then- k ( y eq- l ) + w = 0 y eq- l = = w k , where is the static displacement under the force of gravity. Perform the change of coordinates x = y- y eq , 2 Figure 3: where y eq = l + to obtain m x = m y =- k ( y- l ) + w =- k ( y + - y eq ) + w...
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lecture_2 - ME 163 Vibrations Lecture 2 We are going to...

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