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# CLASS17 - 1 CHAPTER 17 VIBRATING SYSTEMS 17.1 Introduction...

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1 CHAPTER 17 VIBRATING SYSTEMS 17.1 Introduction A mass m is attached to an elastic spring of force constant k , the other end of which is attached to a fixed point. The spring is supposed to obey Hooke’s law, namely that, when it is extended (or compressed) by a distance x from its natural length, the tension (or thrust) in the spring is kx , and the equation of motion is . kx x m = & & This is simple harmonic motion of period 2 π/ω , where ω 2 = k / m . Most readers will have no difficulty with that problem. But now suppose that, instead of one end of the spring being attached to a fixed point, we have two masses, m 1 and m 2 , one at either end of the spring. A diatomic molecule is much the same thing. Can you calculate the period of simple harmonic oscillations? It looks like an easy problem, but it somehow seems difficult to get a hand on it by conventional newtonian methods. In fact it can be done quite readily by newtonian methods, but this problem, as well as more complicated problems where you have several masses connected by several springs and several possible modes of vibration, is particularly suitable by lagrangian methods, and this chapter will give several examples of vibrating systems tackled by lagrangian methods. 17.2 The Diatomic Molecule Two particles, of masses m 1 and m 2 are connected by an elastic spring of force constant k . What is the period of oscillation? Let’s suppose that the equilibrium separation of the masses – i.e. the natural, unstretched, uncompressed length of the spring – is a . At some time suppose that the x -coordinates of the two masses are x 1 and x 2 . The extension q of the spring from its natural length at that FIGURE XVII.1 m 1 m 2 x 1 x 2 k

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2 moment is . 1 2 a x x q = We’ll also suppose that the velocities of the two masses at that instant are 1 x & and . 2 x & We know from chapter 13 how to start any calculation in lagrangian mechanics. We don’t have to think about it. We always start with T = . .. and V = ...: , 2 2 2 2 1 2 1 1 2 1 x m x m T & & + = 17.2.1 . 2 2 1 kq V = 17.2.2 We want to be able to express the equations in terms of the internal coordinate q . V is already expressed in terms of q . Now we need to express T (and therefore 1 x & and 2 x & ) in terms of q . Since , 1 2 a x x q = we have, by differentiation with respect to time, . 1 2 x x q & & & = 17.2.3 We need one more equation. The linear momentum is constant and there is no loss in generality in choosing a coordinate system such that the linear momentum is zero: . 0 2 2 1 1 x m x m & & + = 17.2.4 From these two equations, we find that . and 2 1 1 2 2 1 2 1 q m m m x q m m m x & & & & + = + = 17.2.5a,b Thus we obtain 2 2 1 q m T & = 17.2.6 and , 2 2 1 kq V = 17.2.2 where . 2 1 2 1 m m m m m + = 17.2.7 Now apply Lagrange’s equation . j j j q V q T q T dt d = & 13.4.13 to the single coordinate q in the fashion to which we became accustomed in Chapter 13, and the equation of motion becomes
3 , kq q m = & & 17.2.8 which is simple harmonic motion of period , / 2 k m π where m is given by equation 17.2.7. The frequency is the reciprocal of this, and the “angular frequency” ω , also

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CLASS17 - 1 CHAPTER 17 VIBRATING SYSTEMS 17.1 Introduction...

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