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Erich Poppitz

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Lecture 9: Damping + Restoring Forces ▯ relevant reading: Morin 4.3 ▯ dynamics when you have restoring forces and damping Damped Oscillations ▯ We know that all real oscillators experience some sort of dissipative damping, and that if there is nothing driving the motion, the oscilla- tions will eventually stop. ▯ Our simplest model for a resistive force like friction or air drag is F = ▯bx_; drag where b is a constant and the minus sign indicates the force is in the direction opposite to the motion. ▯ The total net force will be the restoring force plus the drag force. For example, if the restoring force is due to a spring (so Hooke’s law), then the net force is F net= ▯kx ▯ bv and Newton’s law tells us that mx = ▯kx ▯ b_ or b k  + x_ + x = 0 (1) m m ▯ This is the ODE we need to solve for damped oscillation problems. Notice it is of the form  = F(x;x_) and therefore cannot be solved using separation of variables. ▯ We are now going to rename the parameters in front _ and x to make 2 some of the algebra simpler later on. Let 2 = b=m and !0= k=m. Then the equation becomes: 1 2  + 2 x_ + !0x = 0; ▯ The 2 in the above equation will make sense later. Notice that ! is0 the frequency of oscillations for a mass on a spring when there is NO damping. We will call this the \natural frequency" of the system. ▯ Since this is a HOMOGENEOUS, linear, 2nd order ODE, and we know we need x such that a combination of x;x _; and  has to give 0, the only function that can do that is an exponential. So just like in our previous undamped examples, we will suppose we have solutions of the ▯t form x = Ae . Taking the ▯rst and second derivatives and plugging them into the equation gives: 2 ▯t ▯t 2 ▯t ▯ Ae + 2 ▯Ae + ! 0e = 0 Dividing out the common Ae ▯tfrom each term gives: 2 2 ▯ + 2 ▯ + ! = 0 (2) ▯ This equation is known as the \characteristic equation" for this problem. Just like in last week’s lecture (oscillations with no damping), it allows us to solve for the ▯ in the solution but not the A. Solving this quadratic equation gives: q ▯2 ▯ (2 ) ▯ 4! 0 q ▯ = = ▯ ▯ ▯ ! 0 2 (notice how the "2" in our de▯nition of b=m = 2 made the algebra simpler). ▯ Just like last week, we get 2 possible solutions (one with the plus sign, one with the minus). Again, this is good because the mathematicians tell us that 2nd order linear homogeneous ODEs have 2 linearly inde- pendent solutions. ▯ To write the general solution we write a linear combination of the 2 solutions: 2 ▯ ▯ ▯ ▯ p 2 2 p 2 2 ▯ + ▯!0 t ▯ ▯ ▯!0 t x(t) = Ae + Be where A and B are constants determined by the initial conditions. ▯ t ▯ I can simplify the equation q little. If I pull out the e from both 2 2 expressions and de▯ne ▯ ▯ ! ,0I get: h i x(t) = e▯ t Ae t + Be ▯ t ▯ Now, is real so the exponential in front is purely real and cannot represent oscillations. may or may not be real depending on the sce- nario so this will determine whether we get oscillations. Let’s consider all the possible cases for : 1. < ! :0In this case, is imaginary. We can then write: q q q 2 2 2 2 2 2 = ▯ ! =0 (▯1)(! 0 ) = i (! ▯ 0 ) q ▯ De▯ne !~ ▯ !0▯ . Plugging this into our solution gives: ▯ th ~t ▯i~t x(t) = e Ae + Be ▯ We see that the term in square brackets represents oscillations with angular frequency !~. However, now the oscillations are mul- ▯ t tiplied by a negative exponential e . As time increases, this term will decrease exponentially. Since it multiplies the oscilla- tory term in brackets, the amplitude of the oscillations must also decay exponentially. ▯ Remember that we can write the complex exponential term in a variety of ways. For example, another way to write the above ▯ t solution is: x(t) = Ce cos(~t + ▯): ▯ Run with omega0=5, gamma=1. This shows the nat- ural undamped oscillation, the envelope and the damped solution. ▯ This type of damped oscillation is known as being \Underdamped". The \under" is because the damping force is not strong enough to overpower the restoring force and so there are some oscillations. Essentially there is time to get in a few good oscillations before the damping has a signi▯cant e▯ect. 3 ▯ Some useful jargon: Logarithmic decrement: After every pe- riod T = 2▯=! ~, the ampltiude x(t +
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