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Chapter 15

Chapter 15. Oscillatory Motion (GERAKAN BERAYUN). Periodic Motion (Gerakan Berkala). Periodic motion is motion of an object that regularly repeats The object returns to a given position after a fixed time interval

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Chapter 15

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  1. Chapter 15 Oscillatory Motion (GERAKAN BERAYUN)

  2. Periodic Motion (Gerakan Berkala) • Periodic motionis motion of an object that regularly repeats • The object returns to a given position after a fixed time interval • A special kind of periodic motion occurs in mechanical systems when the force acting on the object is proportional to the position of the object relative to some equilibrium position • If the force is always directed toward the equilibrium position, the motion is called simple harmonic motion (gerakan harmonik mudah)

  3. Motion of a Spring-Mass System (Gerakan sistem spring-jisim) • A block of mass m is attached to a spring, the block is free to move on a frictionless horizontal surface • When the spring is neither stretched nor compressed, the block is at the equilibrium position (Kedudukan Keseimbangan) • x = 0

  4. Hooke’s Law • Hooke’s Law states Fs = - kx • Fs is the restoring force (daya pemulih) • It is always directed toward the equilibrium position • Therefore, it is always opposite the displacement from equilibrium • k is the force (spring) constant • x is the displacement (sesaran)

  5. More About Restoring Force • The block is displaced to the right of x = 0 • The position is positive • The restoring force is directed to the left

  6. More About Restoring Force, 2 • The block is at the equilibrium position • x = 0 • The spring is neither stretched nor compressed • The force is 0

  7. More About Restoring Force, 3 • The block is displaced to the left of x = 0 • The position is negative • The restoring force is directed to the right

  8. Acceleration (Pecutan) • The force described by Hooke’s Law is the net force in Newton’s Second Law

  9. Acceleration, cont. • The acceleration is proportional to the displacement of the block • The direction of the acceleration is opposite the direction of the displacement from equilibrium • An object moves with simple harmonic motion whenever its acceleration is proportional to its position and is oppositely directed to the displacement from equilibrium

  10. Acceleration, final • The acceleration is not constant • Therefore, the kinematic equations cannot be applied • If the block is released from some position x = A, then the initial acceleration is –kA/m • When the block passes through the equilibrium position, a = 0 • The block continues to x = -A where its acceleration is +kA/m

  11. Motion of the Block • The block continues to oscillate between –A and +A • These are turning points of the motion • The force is conservative • In the absence of friction, the motion will continue forever • Real systems are generally subject to friction, so they do not actually oscillate forever

  12. Orientation of the Spring • When the block is hung from a vertical spring, its weight will cause the spring to stretch • If the resting position of the spring is defined as x = 0, the same analysis as was done with the horizontal spring will apply to the vertical spring-mass system

  13. Simple Harmonic Motion – Mathematical Representation • Model the block as a particle • Choose x as the axis along which the oscillation occurs • Acceleration • We let • Then a = -w2x

  14. Simple Harmonic Motion – Mathematical Representation, 2 • A function that satisfies the equation is needed • Need a function x(t) whose second derivative is the same as the original function with a negative sign and multiplied by w2 • The sine and cosine functions meet these requirements

  15. Simple Harmonic Motion – Graphical Representation • A solution is x(t) = A cos (wt + f) • A, w, f are all constants • A cosine curve can be used to give physical significance to these constants

  16. Simple Harmonic Motion – Definitions • A is the amplitude of the motion • This is the maximum position of the particle in either the positive or negative direction • w is called the angular frequency (frekuensi sudut) • Units are rad/s • f is the phase constant (pemalar fasa) or the initial phase angle (sudut fasa awal)

  17. Simple Harmonic Motion, cont • A and fare determined uniquely by the position and velocity of the particle at t = 0 • If the particle is at x = A at t = 0, then f = 0 • The phase of the motion is the quantity (wt + f) • x (t) is periodic and its value is the same each time wt increases by 2p radians

  18. An Experiment To Show SHM • This is an experimental apparatus for demonstrating simple harmonic motion • The pen attached to the oscillating object traces out a sinusoidal on the moving chart paper • This verifies the cosine curve previously determined

  19. Period (Kala atau tempoh ayunan) • The period, T, is the time interval required for the particle to go through one full cycle of its motion • The values of x and v for the particle at time t equal the values of x and v at t + T

  20. Frequency • The inverse of the period is called the frequency • The frequency represents the number of oscillations that the particle undergoes per unit time interval • Units are cycles per second = hertz (Hz)

  21. Summary Equations – Period and Frequency • The frequency and period equations can be rewritten to solve for w • The period and frequency can also be expressed as:

  22. Period and Frequency, cont • The frequency and the period depend only on the mass of the particle and the force constant of the spring • They do not depend on the parameters of motion • The frequency is larger for a stiffer spring (large values of k) and decreases with increasing mass of the particle

  23. Motion Equations for Simple Harmonic Motion • Remember, simple harmonic motion is not uniformly accelerated motion

  24. Maximum Values of v and a • Because the sine and cosine functions oscillate between ±1, we can easily find the maximum values of velocity and acceleration for an object in SHM

  25. Graphs • The graphs show: • (a) displacement as a function of time • (b) velocity as a function of time • (c ) acceleration as a function of time • The velocity is 90o out of phase with the displacement and the acceleration is 180o out of phase with the displacement

  26. SHM Example 1 • Initial conditions at t = 0 are • x (0)= A • v (0) = 0 • This means f = 0 • The acceleration reaches extremes of ± w2A • The velocity reaches extremes of ± wA

  27. SHM Example 2 • Initial conditions at t = 0 are • x (0)=0 • v (0) = vi • This means f = - p/2 • The graph is shifted one-quarter cycle to the right compared to the graph of x (0) = A

  28. Contoh 1 • Satu objek berayun mengikut (gerakan harmonik mudah) GHM pada paksi x. Kedudukannya berubah mengikut masa seperti persamaan x=(4.00m)cos(t + /4) t saat dan sudut radian. (a) Tentukan amplitud, frekuensi, dan kala ayunan. (b) Hitung halaju dan pecutan pada bila-bila masa t. (c) Tentukan kedudukan, halaju, dan pecutan pada masa t=1.00s. (d) Tentukan laju maksima dan pecutan maksima objek.

  29. Penyelesaian Contoh 1

  30. Penyelesaian Contoh 1 (contd)

  31. Contoh 2 • Blok berjisim 200g telah disambung pada satu spring ringan di mana pemalar dayanya adalah 5.00N/m. Sistem ini berayun mengufuk di atas satu permukaan licin. Blok itu telah disesar sebanyak 5.0cm dari keseimbangan dan dilepaskan dari keadaan rehat (rujuk Fig. 15.7). (a) Cari kala gerakan blok. (b) Tentukan laju maksima dan pecutan maksima blok. (c) Ungkapkan kedudukan, laju dan pecutan blok dlm sebutan masa.

  32. Penyelesaian Contoh 2

  33. Energy of the SHM Oscillator • Assume a spring-mass system is moving on a frictionless surface • This tells us the total energy is constant • The kinetic energy can be found by • K = ½ mv 2 = ½ mw2A2 sin2 (wt + f) • The elastic potential energy can be found by • U = ½ kx 2 = ½ kA2 cos2 (wt + f) • The total energy is K + U = ½ kA 2

  34. Energy of the SHM Oscillator, cont • The total mechanical energy is constant • The total mechanical energy is proportional to the square of the amplitude • Energy is continuously being transferred between potential energy stored in the spring and the kinetic energy of the block

  35. Energy of the SHM Oscillator, cont • As the motion continues, the exchange of energy also continues • Energy can be used to find the velocity

  36. Energy in SHM, summary

  37. Molecular Model of SHM • If the atoms in the molecule do not move too far, the force between them can be modeled as if there were springs between the atoms • The potential energy acts similar to that of the SHM oscillator

  38. Contoh 3 • Satu bongkah berjisim 0.5 disambung dgn satu spring ringin yg mempunyai pemalar daya 20.0N/m. Bongkah ini berayun mengufuk di atas satu landasan licin. (a) Hitung jumlah tenaga sistem dan laju maksima bongkah jika amplitud gerakan adalah 3.00cm. (b) Apakah halaju bongkah pada kedudukan 2.00cm Hitung tenaga kinetik dan keupayaan sistem di sini.

  39. Penyelesaian contoh 3

  40. Penyelesaian contoh 3 (contd)

  41. SHM and Circular Motion • This is an overhead view of a device that shows the relationship between SHM and circular motion • As the ball rotates with constant angular velocity, its shadow moves back and forth in simple harmonic motion

  42. SHM and Circular Motion, 2 • The circle is called a reference circle • Line OP makes an angle f with the x axis at t = 0 • Take P at t = 0 as the reference position

  43. SHM and Circular Motion, 3 • The particle moves along the circle with constant angular velocity w • OP makes an angle q with the x axis • At some time, the angle between OP and the x axis will be q = wt + f

  44. SHM and Circular Motion, 4 • The points P and Q always have the same x coordinate • x (t) = A cos (wt + f) • This shows that point Q moves with simple harmonic motion along the x axis • Point Q moves between the limits ±A

  45. SHM and Circular Motion, 5 • The x component of the velocity of P equals the velocity of Q • These velocities are • v = -wA sin (wt + f)

  46. SHM and Circular Motion, 6 • The acceleration of point P on the reference circle is directed radially inward • P ’s acceleration is a = w2A • The x component is –w2A cos (wt + f) • This is also the acceleration of point Q along the x axis

  47. SHM and Circular Motion, Summary • Simple Harmonic Motion along a straight line can be represented by the projection of uniform circular motion along the diameter of a reference circle • Uniform circular motion can be considered a combination of two simple harmonic motions • One along the x-axis • The other along the y-axis • The two differ in phase by 90o

  48. Contoh 4 • Satu zarah berputar dlm satu bulatan mengikut arah lawan jam dgn jejari 3.0m dan laju sudut malar 8.00rad/s. At t=0, zarah berada pada x=2.0m dan bergerak arah ke kanan. (a) Tentukan kordinat x mengikut masa. (b) Cari komponen x halaju dan pecutan zarah pada bila-bila masa.

  49. Penyelesaian Contoh 4

  50. Penyelesaian Contoh 4 (contd)

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