Periodic Motion or Oscillations. Physics 232 Lecture 01 1

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1 Periodic Motion or Oscillations Physics 3 Lecture 01 1

2 Periodic Motion Periodic Motion is motion that repeats about a point of stable equilibrium Stable Equilibrium Unstable Equilibrium A necessary requirement for periodic motion is a Restoring Force Physics 3 Lecture 01

3 Characteristics T A A Amplitude - A The maximum displacement from the equilibrium position Period - T The time to complete one cycle of motion, peak to peak or valley to valley Frequency - f The number of cycles per unit time Physics 3 Lecture 01 3 f 1 T

4 Simple Harmonic Motion Consider a mass m attached to a horizontal spring having a spring constant k with the spring unstretched We now pull the mass to the right and then let the mass go What is the subsequent motion of the mass? The restoring force is given by Hooke s Law F ma k x Physics 3 Lecture 01 4

5 Simple Harmonic Motion This is a second order differential equation d x m k x dt Since this is a second order differential equation, there are two constants of integration and the general solution is x Acos t ( ω +φ ) where A is the maximum displacement, φ is a phase angle, and ω is the angular velocity and is given by ω A and φ are determined from the initial, boundary, conditions k m Physics 3 Lecture 01 5

6 Phase Angle φ Physics 3 Lecture 01 6

7 Simple Harmonic Motion The period of the motion is related to ω by T π π ω Note that the period of the motion is independent of the displacement! The velocity of the particle is given by v m k dx ω Asin t + dt with the maximum velocity being ( ω φ ) ω A Physics 3 Lecture 01 7

8 Simple Harmonic Motion We have so far that dx dt x Acos( ω t +φ ) and v ω Asin( ω t + φ ) The constants A and φ are determined from the initial conditions, that is what are x and v at a specified time or some other suitable initial conditions The maximum displacement occurs when the velocity is zero, which occurs at the extreme of motion (two locations: x ±A) The maximum velocity occurs when the displacement from the equilibrium position is zero (two values: v ±ωα) Physics 3 Lecture 01 8

9 SHM Energy Considerations We assume that the system is isolated and frictionless With the force being given by F k It can be shown that there is a potential energy in the system that is U 1 k x x The total energy is then given by E Total 1 KE m v + U + 1 k x Physics 3 Lecture 01 9

10 SHM Energy Considerations If we substitute for x and v we find that ETotal The total energy is constant 1 k A The energy shifts back and forth between the kinetic energy and the potential energy, but with the sum of the two being constant Physics 3 Lecture 01 10

11 SHM Energy Considerations The relationship between kinetic energy, potential energy, displacement, velocity and acceleration can be seen in the following diagram Physics 3 Lecture 01 11

12 Choice of Oscillatory Function Note that we used the cosine function in our development of Simple Harmonic Motion Acos t ( ω +φ ) We could have also used the sine function for our description of SHM The only difference is in the phase angle x ( ω + ) x Asin t φ' Physics 3 Lecture 01 1

13 Simple Harmonic Motion Simple Harmonic Motion can be used to describe motion in many situations under appropriate approximations Any potential energy function that under appropriate circumstances can be approximated by a parabolic function will exhibit SHM Physics 3 Lecture 01 13

14 Simple Pendulum Consider a mass m suspended from a massless, unstretchable string of length L The forces acting on the mass are as shown The restoring force is the one perpendicular to the string F restoring mg sinθ But this is a nonlinear function of θ However for small angles We then have that F restoring mgθ sinθ θ Physics 3 Lecture 01 14

15 From before we have that We also have that This then yields that If we then let Simple Pendulum k θ mg L F restoring mgθ this is then becomes the equation of SHM x L F restoring mg L The frequency of oscillation is given by x f 1 π which is independent of the mass attached to the string g L Physics 3 Lecture 01 15

16 Damped Motion Real life situations have dissipative forces The fact that there is a dissipative force, leads to motion that is damped, that is the amplitude decreases with time The dissipative force is often related to the velocity that is in the motion F diss bv The minus sign indicates that this force is in the opposite direction to the velocity The total force is then the sum of the restoring force and this dissipative force F net k x bv Physics 3 Lecture 01 16

17 Damped Motion This net force leads to a slightly more complicated second order differential equation d x m dt dx + b + kx dt The exact solution to this equation depends upon the damping constant 0 There are three possible solutions Underdamped: b < km Critically Damped: Overdamped: b b > Physics 3 Lecture km km

18 Damped Motion - Underdamped b < km The general solution for this situation is given by x Ae ( b / m) t cos( ω ' t +φ ) with ω ' k m b 4m This solution looks like The envelope is the the term describing the decreasing amplitude Envelope Ae ( b / m)t Physics 3 Lecture 01 18

19 Damped Motion- Critically Damped b km The general solution for this situation is given by x b b t m A 1 + t e m This solution looks like Physics 3 Lecture 01 19

20 b > Damped Motion - Overdamped km The general solution for this situation is given by ( b / m) t ( ωt ω t A e + A e ) x e 1 with ω b 4 m k m This solution looks like Physics 3 Lecture 01 0

21 Damped Motion The most efficient damping, as far getting back to zero amplitude, is the critically damped case Note that the overdamped case may yield some interesting behavior depending on the relative values of the parameters Physics 3 Lecture 01 1

22 Forced Oscillation It is also possible to drive a system such as an oscillator with an external force that is also time varying The general differential equation is of the form d x m dt + k x + bv F max cosωd where the term on the right hand side is the driving force This solution to this equation involves two functions, a complementary function and a particular solution t Physics 3 Lecture 01

23 The solution is given by x( t) Forced Oscillation F max cos t d + b ω ( k mω ) d d ( ω δ ) with δ tan 1 ω d k / m ( b / m) ω d δ represents the phase difference between the driving force and the resulting motion There can be a delay between the action of the driving force and the response of the system Physics 3 Lecture 01 3

24 Resonance The term in front of the cosine function represents the amplitude A F ( ) k mω + b ω Physics 3 Lecture 01 4 max If we vary the angular velocity of the driving force, we find that the system has its maximum amplitude when ω d This phenomenon of the amplitude peaking at a driving frequency that is near the natural frequency of the system is known as resonance d k m d

25 Resonance The strength of the amplitude depends upon the magnitude of the damping coefficient The smaller the value of b the more pronounced the peak The larger the damping the peak becomes broader, less sharp, and shifts to lower frequencies If m b k > the peak disappears completely Physics 3 Lecture 01 5

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