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lecture_12 - Lecture 12 0.1 Shock response One of the...

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Lecture 12 0.1 Shock response One of the typical vibration engineering design tasks involves protecting a device from a shock impulse, such as when one drops a cell phone on the ﬂoor. A more extreme example is that of a torpedo hitting a ship whose ﬁber-optics for communications should still survive - video of shock vibration testing for that case was shown in class. The basic physics that plays in shock response is that the maximum force that the device body will experience is proportional to its max- imum decceleration. Consider the standard spring-mass system with mass M and stiﬀness k . Assume we drop an object of mass m o from height h . We set the y coordinate to be 0 at the place we dropped the object from and h at the position where it hits the mass M of the mass-spring system. By integrating m o ¨ y = m o g we get ˙ y ( t ) = gt and y ( t ) = gt 2 / 2 . Thus, time of impact is t imp = p (2 h/g ) . Plugging this into the velocity equation, we get the velocity at impact is going to be v imp = v 0 = p 2 hg. Now we consider the continuing motion of the mass m 0 and the testing equipment of mass M . Thus the joint mass is m = m 0 + M . The diﬀerence in the total linear momentum of the system after the collision can be attributed to the impulse of all the forces acting on it during the collision. This leads to m o ˙ x (0 + δt ) + M ˙ x (0 + δt ) - m o v imp = Z δt 0 F s dt, 1

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M h y 0 k m x Figure 1: Testing the forces on an object falling from height h . where F s is the force in the spring. Now, since the impulse lasts for a very short time, we can take δt to zero and obtain m ˙ x (0 + ) = m o v imp , where, as before 0 + denotes the time right after the impact. Thus, the initial velocity of the mass-spring system is v 0 = m o m v imp . We take the initial position to be
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lecture_12 - Lecture 12 0.1 Shock response One of the...

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