Harmonic Oscillator - Model Systems

Transcription

1 3_Model Systems HarmonicOscillators.nb Chapter 3 Harmonic Oscillator - Model Systems 3.1 Mass on a spring in a gravitation field a Force Method The two forces on the mass are due to the spring, F s kzz, and the gravitational field, F g mgz.

2 3_Model Systems HarmonicOscillators.nb 3 Out[47]= As shown, the displacement is negative and therefore the force is in the positive z direction. Evaluating the total force on the mass, F t m z F t g F s mgkz Defining a new variable, zmgk and noting that z t t then the equation becomes m t k (3.1) because mgk is a constant (3.) From our previous discussions, the general solution is of the form A t with k m and therefore the motion for z is again harmonic but with a displaced zero Energy Method The kinetic energy is only due to the motion of the mass whereas the potential energy is a combination of potential energy due to the spring and gravity. KE 1 m z t (3.3) PE 1 kz mgz (3.4) The equilibrium position of the mass is given by the minimum of the potential energy. Setting

3 4 3_Model Systems HarmonicOscillators.nb the derivative of the potential energy to zero and solving for z PE z gm kz 0 (3.5) or z equilibrium g m k (3.6) The effective spring constant is given by the second derivative of the potential energy. k effective PE z k (3.7) Because the effective mass is just m, the frequency of oscillation is just k m Pendulum 3..1 Force Method The forces on the mass of a pendulum are the gravitational force pointing in the z direction, mgz, and the tension in the pendulum rod that points along the rod towards the pivot point. A component of the gravitational force perpendicular to the rod, mgsin, provides a torque, mglsin, that results in a changing angular momentum of the body.

4 3_Model Systems HarmonicOscillators.nb 5 t 0. Show Forces Show Total Force Show Normal and Tangential F mg The torque equation then becomes EprintGFODL, t, eq,m, L, GFOD, t,, eq,m, g, L, Sin dl dt d ml mglsin d t (3.8) We see then that the resulting equation does not describe a simple harmonic oscillator except in the limit of small angle. From the Taylor expansion of the Sine, O7 (3.9) we see that when 3 6 or Sqrt[6].45R, this system will exhibit simple harmonic motion. The frequency of oscillation will be given by k eff m eff glm ml g L (3.10) which is independent of the mass of the pendulum and only depends on it length and the acceleration due to gravity. For larger amplitudes, the period of oscillation is no longer independent of the amplitude and increases. At very large amplitudes, the motion is no longer sinusoidal.

5 6 3_Model Systems HarmonicOscillators.nb observation time length of pendulum initial angle radians i ' i 0 Harmonic Motion Harmonic Motion release system initial angular velocity Energy Method The potential energy of the pendulum relative to its equilibrium position is given by U mgh mgl1 Cos

6 3_Model Systems HarmonicOscillators.nb 7 The equilibrium position is given by U glmsin 0 (3.11) or z equilibrium 0 (3.1) For small oscillations about the equilibrium position, expand U about the equilibrium position, 0, U 1 glm O 4 (3.13) The kinetic energy is 1 mv but v L so that KE 1 ml d dt (3.14) or equivalently KE 1 I d where I is the moment of inertia. The total energy becomes dt total 1 L m d dt 1 glm (3.15) Comparing this to the general form of the energy of a harmonic oscillator,

7 8 3_Model Systems HarmonicOscillators.nb E 1 m dx eff dt 1 k eff x, we see that the oscillation frequency becomes k eff m eff glm ml g L (3.16) Capacitor and Inductor Circuit For a parallel circuit of an inductor and capacitor, the energy oscillate between the magnetic field of the inductor and the electric field of the capacitor Energy Method The total energy is given by the energy of the capactitor, 1 CV, and the energy of the inductor, 1 LI or noting that the voltage of the capacitor is V Q C where I d Q dt total 1 L d Q dt 1 1 C Q (3.17) Comparing this to the general form of the energy of a harmonic oscillator, E total 1 m dx eff d t 1 k eff x, we see that the oscillation frequency becomes k eff m eff 1 LC (3.18) 3.3. Circuit Equation

8 3_Model Systems HarmonicOscillators.nb 9 One could also have solved this problem using the circuit equations. The voltage of the capacitor is equal to that of the inductor, V C V L or Q C L d I. Differentiating once we have that dt L d 1 dt C dq dt 1 C (3.19) or d 1 dt L C (3.0) Gas Piston The ideal gas law states that PV Nk B T. For an isothermal process, the product of the pressure and volume will be a constant, i.e. PV P 0 V 0. Area A L mg Volume V AL The forces on the piston will be the downwards gravitational force and the upwards force due to the pressure F PA mg (3.1) In general the pressure will depend upon the volume of the gas or the length of the chamber, P P 0 V 0 V P 0 V 0. The force equation then becomes A L F PA mg P 0 V 0 L mg (3.) In equilibrium the force on the piston will be zero and so the equilibrium length of the chamber

9 10 3_Model Systems HarmonicOscillators.nb will be P 0 V 0 mg. For deviations about this equilibrium length, L y, the force equation becomes F P 0 V 0 y mg (3.3) If y is small, we can expand in a Taylor s series. Note that the 0th order terms cancel. The first order terms gives up a force proportional to the displacement. F gm P 0 V 0 yp 0 V 0 P 0 V 0 y P 0 V 0 y 3 Oy 3 (3.4) Equating this force to the acceleration of the piston, we see that we will give harmonic motion with a frequency of g P 0 V 0 m. But P 0 V 0 which resembles the frequency of a pendulum. mg and so the frequency can also be written as Orbital Mechanics For a mass, m, orbiting a larger mass, M, the force of gravity is F mmg r r which corresponds to a potential energy of the form Vr m M G. The kinetic energy of the smaller mass r is simply 1 mv. For a central force the angular momentum, L mvr is conserved and is a constant. For a circular orbit, the kinetic energy can be expressed in terms of this angular momentum. KE mv 1 m L mr L mr (3.5) The total energy then becomes L GmM mr r (3.6) Graphing the total energy,

10 3_Model Systems HarmonicOscillators.nb 11 1 r r 0 1 r At a given angular momentum, L, the object will have a specific orbit at r 0 given by the minimum of the energy d dr d L mr GmM r dr L mr 3 GmM r 0 (3.7) Solving for r 0 r 0 L Gm M (3.8) Alternatively Eq..7 is just a statement that the gravitational force is equal to the centripetal force GmM r 0 L mv mr3 0 r 0 (3.9) Now what happens if you nudge teh object in the radial direction with an impulse. Because the impulse is in the radial direction, the angular momentum is unchanged. The object will oscillate about hte point r 0 as it continues in its orbit. The effective spring constant is given by the second derivative of the energy k eff d 3 L GmM dr rr mr 4 0 G4 m 7 M 4 L 6 r 3 rr 0 (3.30) or substituting from Eq..9 G L m Mr 0

11 1 3_Model Systems HarmonicOscillators.nb k eff L mr 0 4 (3.31) We can reexpress k eff in terms of the frequency of the orbits. From L mvr 0 mr 0 0 we have that k eff G4 m 7 M 4 L 6 m 0 (3.3) Therefore the frequency of the radial oscillations becomes osc k eff m 0 (3.33) i.e. the period of the radial oscillations is the same as the circular motion. Plotting the orbit then eps nth The orbit is the form of an ellipse. A complete analysis in mechanics shows that a general orbit is given by an ellipse with eccentricity,

12 3_Model Systems HarmonicOscillators.nb 13 r Cos (3.34) For small we have that r Cos Cos Cos O 3 (3.35) or to lowest order r Cos 4 (3.36) which is exactly the orbit that we have found....