Damped harmonic oscillator

Transcription

1 Prof. O. B. Wright, Mechanics Notes Damped harmonic oscillator Differential equation Assume the mass on a spring is subject to a frictional drag force -'dx/dt. (This force always points in the opposite direction to the way the mass is moving.) This force is caused, for example, by the viscous medium in the damper. mx x x x x x Where ='/m and =/m. This equation is linear. It applies to many simple systems, such as to the motion of an object in a fluid in non-turbulent conditions. One method of solving this equation is to try a solution of the form cos( ) a t / x x t e Substitute and set the sin and cos terms to zero: this leads to the two equations -' +1/ a -/ a + = '/ a =' We find that this gives a solution for any value of provided that we choose Either a =/ ' = - /4. Note that '. For < 1

2 Prof. O. B. Wright, Autumn 7 Or '= a =/[( -4 ) 1/ ] For > Damped harmonic oscillations For <, the solution can also be written t / x e A1sin t B1cos t Initial conditions determine A 1 and B 1. If, at t=, x=x() then B 1 =x(). If, at t=, x x() then A1 x() x() / : t / 1 sint x e x() cos t x() x(). So x x x x 1 1 () () () 1/ This represents damped harmonic oscillations ' is the natural frequency of the damped oscillator. The amplitude x e -t/ decays exponentially (in the graph above / =.). After one cycle (period T') the amplitude

3 Prof. O. B. Wright, Autumn 7 decays by a factor e -T'/. If T ' is small compared to 1 (<<'), this factor is approximately equal to 1-T'/~1, and the oscillation is lightly damped. For example, if the oscillator is struck at rest at t=, x()=, then, where x() x() /. t / x x() e sin t Or, if the oscillator is released at t= from rest, x(), then t / sint x x() e cost Critical damping If then ' and sin't/' t. t x e x() ( x() x()) t 3

4 Prof. O. B. Wright, Autumn 7 This represents critical damping. For example, if the oscillator is struck at rest at t=, x()=, then x x() te t (always positive if x()>) Or, if the oscillator is released at t= from rest, x(), then t x x() e 1 t (always positive if x () >) Heavy Damping If > then ' and sin't/' t. x C e C e, t/ 1 t/ 1 where 1 =/[+( -4 ) 1/ ], =/[-( -4 ) 1/ ]. Putting in the initial conditions: e.g., at t=, x=x(). Then C 1 +C =x(). At t=, x x(), so C 1 / 1 +C / = x(). Constants can be found by elimination. For example, if the oscillator is struck at rest at t=, x()= then x x() e e t/ 1 t/ (always positive if x()>) 4

5 Prof. O. B. Wright, Autumn 7 Or, if the oscillator is released at t= from rest, x x(), x x() e e /( 4 ) t/ 1 t/ 1/ (always positive if x() ) (where ( -4 ) 1/ =1/ 1-1/ ) Solution with complex numbers x x x Try x=ce pt where C is complex This gives p +p+ = So p=[-( -4 ) 1/ ]/. If <, then p=-/i' (where '=( - /4) 1/ ). x=c 1 e (-/+i')t +C e (-/-i')t Since the two roots are complex conjugates, i.e. -/i' and-/i', and x must be real, then C 1 and C and must also be complex conjugates. This means that if C 1 =a+ib then C =a-ib. We use a star (*) to show the complex conjugate. 5

6 Prof. O. B. Wright, Autumn 7 x=c 1 e (-/+i')t +C 1 *e (-/-i')t. Then x=ae -t/ cos('t+). Let C 1 =Ae i Note that if < then the oscillations grow in amplitude. This can happen if an appropriate external energy source is present. If >, then p=-/( -4 ) 1/ /. p 1 =-1/, p =-1/ as above. t/ 1 t/ So x C1e Ce as before. C 1 and C must be real numbers to make x real. If =, then p=-/. So x 1 =Ae pt = Ae -t/ is one solution. But this cannot be the general solution because there is only one arbitary constant A. We need two constants for a nd order differential equation. To find another solution start with x Ae t u() t x in / 1, and try the substitution 1 x the differential equation x x x x ux ux, and 4. Using 1 1 x ux ux ux, the equation x x x 4 becomes ux u( x x ) u( x x 4 x ). But we know that x x 4 x, and also that x1x1. Therefore the equation becomes ux1. Therefore u. The solution is u c1t c. We can choose c = because the c part just reproduces the first solution x 1. Therefore the general solution is x x A c t e t / 1 ( ). With B=c we get x=(a+bt)e -/t where A and B must be real. Energy loss for very light damping 6

7 Prof. O. B. Wright, Autumn 7 Very light damping means <<. cos( ) t x x t e / (for simplicity put =. Analysis also applies for ) x x e sin t ( / )cos t x e sint t / t / Kinetic energy T=m x /(1/)m x e -t sin 't Potential energy V=x /=(1/)m x e -t cos 't So T+V(1/)m x e -t Or E=E e-t (1) The energy decays with time constant a /=1/. This is half the time constant for the amplitude decay. (Energy decays more rapidly that the amplitude.) de/dt=-e. The rate of loss of energy is therefore equal to E. Work done by the force + x on the viscous medium, i.e. the force pointing in the opposite direction to the frictional force on the mass x, since <<. In one cycle from time t Wf xdx x dt xe sin tdt t to t +/ (since ' ), / t t t sin t Wf m x e tdt m x e. 1 1 (Note: you can check this integral using the relation sin 1 sin.) This is the energy lost in one cycle (This is positive because it is the work done by the force from the moving part of the damper on the viscous medium.) We can also find the energy lost in one cycle using equation (1). Can you? Define the quality factor: Q= / 7

8 Prof. O. B. Wright, Autumn 7 With this definition, Q>1/ means light damping, Q=1/ means critical damping and Q<1/ means heavy damping. E E We can also write Q provided that << de de dt dt Stored energy Stored energy Then Q. Energy lost in one cyle Rate of energy loss Successive amplitudes vary as x, x e -, x e -, x e -3, where is known as the decrement. e - / =x(t+/ )/x(t)= exp e Therefore, =/ =/Q 8