Classical Mechanics Phys105A, Winter 2007

Transcription

1 Classical Mechanics Phys5A, Winter 7 Wim van Dam Room 59, Harold Frank Hall vandam@cs.ucsb.edu Phys5A, Winter 7, Wim van Dam, UCSB

2 Midterm New homework has been announced last Friday. The questions are the same as the Midterm It is due this Friday. Regarding the Midterm: Future homework assignments will be more aligned with the kind of questions for the Final. Suggestions, as always, are welcome. Phys5A, Winter 7, Wim van Dam, UCSB

3 Chapter 5: Oscillations Phys5A, Winter 7, Wim van Dam, UCSB

4 Hooke s Law For a spring with force constant k (with units kg m/s ) Hooke s Law states F(x) = kx, such that the potential is U(x) = ½kx (the system is stable as long as k>). All conservative, d, stable systems at x=, can be approximated for small displacements x by such a parabolic U. In other words: d, oscillating, conservative systems can always be approximated by Hooke s law (provided the oscillations are small enough). Phys5A, Winter 7, Wim van Dam, UCSB

5 Simple Harmonic Motion The equation of motion is d x/dt = (k/m)x = ω x with the angular frequency ω= (k/m). The general solution is the superposition x(t) = C e iωt + C e iωt, which has period τ = π/ω = π (m/k) (with units s). The constants C and C are determined by the position and velocity at (say) t=. We know of course that e iωt = cos ωt + sin ωt, yet x(t) will typically be real valued. Hence the constants C and C will be such that the complex components cancel each other. Phys5A, Winter 7, Wim van Dam, UCSB

6 Solving the SHM Equivalently, we can say we have the simple harmonic motion (SHM): x(t) = B cos(ωt) + B sin (ωt), where the requirement x R equals B,B R. For initial (t=) postion x and velocity v, we get x(t) = x cos(ωt) + (v /ω) sin (ωt). For general B,B, there is a phase shift δ = tan (B /B ) with B cos(ωt) + B sin(ωt) = B + B cos(ωt δ ) Another way of visualizing all this is as the x-coordinate of a circular motion: Phys5A, Winter 7, Wim van Dam, UCSB

7 Energy Flow of a SHM From now on A = B + B The potential energy fluctuates as U = ½kx = ½kA cos (ωt δ) The kinetic energy goes like T = ½k(dx/dt) = ½kA sin (ωt δ) Hence the total energy we have E = T+U = ½kA. Phys5A, Winter 7, Wim van Dam, UCSB

8 Two Dimensional Oscillations For isotropic harmonic oscillators with F = kr we get the solution (picking t= appropriately): x(t) = A x cos(ωt) and y(t) = A y cos(ωt δ). Phys5A, Winter 7, Wim van Dam, UCSB

9 Anisotropic Oscillations If (more generally) F x = k x x and F y = k y y, then we have two independent oscillations, with solutions (again for right t=): x(t) = A x cos(ω x t) and y(t) = A y cos(ω y t δ). For such an anisotropic oscillator we have two angular frequencies ω x = (k x /m) and ω y = (k y /m). Three cases when ω x /ω y = ½: If ω x /ω y is irrational, the motion is quasiperiodic (see Taylor, page 7). Phys5A, Winter 7, Wim van Dam, UCSB

10 Damped Oscillations Often an oscillating system will undergo a resistive force f = bv that is linear in the velocity dx/dt (linear drag). Thus, for a one dimensional, x-coordinate system, the combined force on the particle equals kx bdx/dt such that md x/dt = kx bdx/dt, giving us the second order, linear, homogeneous differential equation: m && x + bx& + kx = with m the mass of the particle, bv the resistive force and kx the Hooke s law force. How to solve this damped oscillation? Phys5A, Winter 7, Wim van Dam, UCSB

11 Care versus Don t Care We are mainly interested in the properties of the system that hold regardless of the initial conditions. We care about: damping, frequencies, We care less about: specific velocities, angles, positions, and so on. Phys5A, Winter 7, Wim van Dam, UCSB

12 Differential Operators Solving the equations of damped oscillations becomes significantly easier with the use of the differential operator D = d/dt, such that we can rewrite the equation as md x + bdx + kx = (md + bd + k)x =, where D stands for D(D) = d /dt. To certain degree you can solve equations f(d)x= as if f(d) is scalar valued: if f(d)x= and g(d)x=, then we also have αf(d)g(d)x= and (αf(d)+βg(d))x=. An important exception occurs for D x=: besides the solution Dx= (hence x=c), it can also refer to the case of x being linear (x = at+c) such that Dx=a, but D x=. Phys5A, Winter 7, Wim van Dam, UCSB

13 Solving D Equations With D = d/dt, take the differential equations (D+4)x=. Rewrite it as Dx = 4x Observe that x = C e 4t is the general solution for x(t). Generally, (D a)x= has the solution x = C e at. For nd order equations f(d)x= with f(d) a quadratic polynomial in D, we solve the auxiliary equation f(d)= and use its solutions D=a and D=b to rewrite the equation as (D a)(d b)x=. As a result, we have (typically) the solutions x = C e at and x = C e bt. If a=b, then (D a) x= also gives: x = C t e at. Phys5A, Winter 7, Wim van Dam, UCSB

14 Damped Oscillations Often an oscillating system will undergo a resistive force f = bv that is linear in the velocity dx/dt (linear drag). Thus, for a one dimensional, x-coordinate system, the combined force on the particle equals kx bdx/dt such that md x/dt = kx bdx/dt, giving us the second order, linear, homogeneous differential equation: m && x + bx& + kx = with m the mass of the particle, bv the resistive force and kx the Hooke s law force. How to solve this damped oscillation? Phys5A, Winter 7, Wim van Dam, UCSB

15 Solving D Equations, Take With D = d/dt, (D a)x= has the solution x = C e at. For a nd order equation f(d)x= with f(d) a quadratic polynomial in D, we solve the auxiliary equation f(d)= and use its solutions D=a and D=b to rewrite the equation as (D a)(d b)x=. As a result, we have (typically) the general solution x = C e at + C e bt. If a=b, then (D a) x= also gives: x = C t e at, giving the general solution x = C e at + C t e bt What does this imply for the damped oscillation? Phys5A, Winter 7, Wim van Dam, UCSB

16 Solving the Equation We rewrite the damped oscillation equation by defining β=b/m and ω = (k/m) (both with frequency units /s) such that we have the equation (D +βd+ω )x=. Factorizing this gives: (D + β + β ω )(D + β ω There are thus three distinct scenarios: β<ω : underdamping, when the drag bv is small β>ω : overdamping, when the drag bv is large β=ω : critical damping β )x = Phys5A, Winter 7, Wim van Dam, UCSB

17 Weak Damping When β<ω we have for the differential equation ( D + β + ω )(D + β ω )x = with ω = (ω β ), such that the general solution is x(t) = = e βt A e (C βt cosω t C cos(ω t δ) + sinω t) The decay factor is β, and the evolution looks like: Note that for really small β we have ω ω Phys5A, Winter 7, Wim van Dam, UCSB

18 Strong Damping When β>ω we have for the differential equation ( D + β + ω )(D + β ω )x = such that the general solution is the sum of two decays: x(t) = C e (β ω (β+ The dominant decay factor is β (β ω ), and the evolution looks like:.4 β )t +. C e β ω )t Note that large β gives small decay factors Phys5A, Winter 7, Wim van Dam, UCSB

19 Critical Damping When β=ω we have for the differential equation ( D + β)(d + β)x = This time, the general solution is x(t) = C e βt βt + Ct e The decay factor is β, and the evolution looks like: or Phys5A, Winter 7, Wim van Dam, UCSB

20 Driven Damped Oscillations A damped oscillator (with m,b,k) driven by a time dependent force F(t) is described by the equation m && x + bx& + kx = F(t) Rewriting with β=b/m, ω = (k/m) and f(t) = F(t)/m gives (D + βd ω )x f(t) This is an inhomogeneous differential equation, for which we know how to solve the homogeneous part. We will describe a particular solution for f = f cos ωt, where ω is the driving frequency. + = Phys5A, Winter 7, Wim van Dam, UCSB

21 Solving the Driven Oscillator Solving the equation for the sinusoidal driving force gives ( D + βd + ω )x = f cosωt r t x (t) = A cos(ωt δ) + Ce + C r e t With A = (ω ω f ) + 4β ω δ = tan βω ω ω The C, C, r, r are determined by the homogeneous equation and do not matter in the limit t. Phys5A, Winter 7, Wim van Dam, UCSB