simpleharmonicmotion-171003080504 1.pptx

for jee students most importANT

1. Simple Harmonic Motion

2. What is Simple Harmonic Motion? Is a type of periodic motion or oscillation motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement.

3. Periodic Motion This is the motion that repeats itself after a regular interval of time. Examples are; the motion of the earth around the sun whose period is 1 year, The motion of the moon around the earth is also a periodic motion whose time period is 27.3 days, etc.

4. Oscillatory or Vibratory Motion Is when a body is in a periodic motion and moves along the same path to and fro about a definite point (equilibrium). Examples are; A pendulum swing back and forth, the vibration of a guitar string, a mass vibrating at the end of a spring etc. N.B it may be noted that all oscillatory motions are periodic motions but all periodic motions are not oscillatory.

5. Why S.H.M is important? The study and control of oscillation are two of the primary goals of both physics and engineering.

6. S.H.M Consider a particle executing S.H.M about point O with an amplitude a . The particle is said to have completed 1 vibration (or oscillation) if starting from t = up to t = T . The vibration of the displacement x with t is a sine relation given by; o 𝑥 = 𝑎 𝑠𝑖𝑛𝜔𝑡

7. Sinusoidal motion of a particle executing S.H.M Time (s) Displacement ( x) Period T The displacement x is defined as the distance the body moves from its equilibrium Period T

8. Sine Function: Mathematically x y 2π π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π - 1 1 y=sin(x) y=cos(x)

9. Sin e function: employed for oscillations x y π/2 π 3π/2 2π 5π/2 3π 7π/2 4π 9π/2 5π 1 y=sin(x) Time t (s) T/2 T 2T - 1 Displacement y (m) 𝑎 - 𝑎 y= 𝑎 sin(ωt)

10. Sine function: employed for oscillations Time t (s) Displacement y (m) T/2 T 2T - 𝑎 𝑎 y= a sin(ωt) What do we need ? 1. Maximum displacement a ωT = 2π Initial condition

11. Amplitude of the motion ( a ) Amplitude Is the greatest displacement from the mean equilibrium position .

12. Time Period of the Motion ( T ) Time taken period is the time to complete 1 vibration.

13. Frequency of Oscillation ( f ) Frequency Is the number of oscillations that are completed each second . That is; SI unit of f is hertz 𝟏𝑯𝒆𝒓𝒕𝒛 = 𝟏𝑯𝒛 = 𝟏𝒐𝒔𝒄𝒊𝒍𝒂𝒕𝒊𝒐𝒏 𝒑𝒆𝒓 𝒔𝒆𝒄𝒐𝒏𝒅 = 𝟏𝒔 −𝟏 𝑓 = 1 / 𝑇

14. Angular frequency of S.H.M ( 𝝎 ) Angular Frequency is a scalar measure of rotation rate. Angular frequency ω T erms angular speed , (also referred to by the r adial frequency , circular frequency , orbital frequency , radian frequency ).

15. What is 𝝎 measured in? 𝝎 is measure in radians per second One revolution is equal to 2π radians , hence ; 𝜔 = 2𝜋 𝑇 = 2𝜋𝑓 In SI units , 𝝎 is normally presented in radians per second , even when it does not express a rotational value.

16. Phase Phase is the position of a point in time (an instant) on a waveform cycle. Phase difference is the difference, expressed in degrees or time, between two waves having the same frequency and referenced to the same point in time.

17. In phase and Out of phase Two oscillators that have the same frequency and no phase difference are said to be in phase . Two oscillators that have the same frequency and different phases have a phase difference, and the oscillators are said to be out of phase with each other.

18. In phase vis Out of Phase

19. Analysis of one- dimensional S.H.M in terms of Uniform Circular Motion It is possible to analyze S.H.M in terms of a uniform circular motion. Suppose the particle is moving with a uniform circular speed 𝝎 along a circumference of a circle of radius a and centre O .

20. Analysis… M moves to and fro along the straight line 𝐴𝐴 ′ ; 𝑂 is the equilibrium position. i.e. the motion of M is the S.H.M. To test the statement above , let us find the displacement of M (=y).

21. Analysis… At this instant, the expression of displacement of M is as shown below; sin 𝜃 = 𝑂𝑀 𝑂𝑃 𝑂𝑀 = 𝑂𝑃 sin 𝜃 But OM = y; OP= a; 𝜃 = 𝜔 ∴ 𝑦 = 𝑎sin 𝑤𝑡

22. General Initial Conditions for one dimensional S.H.M Suppose the initial time ( 𝑡 = ) is when the object is at P , with an initial angular position 𝜙 . At time t it is at s , having turned through an additional angle wt. From the basic trigonometry, an expression for the displacement x at time t in terms of 𝒂 , wt and 𝜙 is; 𝜙 is sometime called Phase constan t of the motion. 𝑡 ∴ 𝑥 = 𝑎 sin 𝑇 2𝜋𝑡 + 𝜙 = 𝑎 sin 𝜔𝑡 + 𝜙

23. V elocity in S.H.M If the displacement 𝑥 at time t is given by the function 𝑥(𝑡) then the gradient of the displacement- time graph at the time t 𝑑𝑥(𝑡) / 𝑑𝑡 is given by and this is also the instantaneous velocity 𝑣 𝑥 (𝑡) of the object at that time . 𝑑𝑡 𝑑𝑥 ∴ 𝑣 𝑥 𝑡 = (𝑡) …..(i)

24. V elocity Since, 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , substitute this equation into equation (i); 𝑑 𝑡 𝑑 𝑣 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 ∴ 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙

25. The speed of the particle executing S.H.M as a function displacement. From the displacement equation, 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , we have; 𝑥 2 = 𝑠𝑖𝑛 2 𝜔𝑡 + 𝜙 ……...(i) 𝑎 2 Also, from velocity equation, 𝒗 𝒙 𝒕 = 𝒂𝝎 𝒄𝒐𝒔 𝝎𝒕 + 𝝓 , we have; 𝑣 2 𝑎 2 𝜔 2 = 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙 ……(ii)

26. The speed of the particle … Sub stitute equation ( i ) and (ii) into trigonometric identity; 𝑠𝑖𝑛 2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1; then we have; 𝑥 2 𝑣 2 𝑎 2 + 𝑎 2 𝜔 2 = 1 ∴ 𝑣 = 𝜔 𝑎 2 − 𝑥 2

27. Acceleration in S.H.M Note that; the gradient of the velocity-time graph at time t is given by 𝑑𝑣 𝑥 (𝑡) / 𝑑𝑡 and is also the instantaneous acceleration 𝐴 𝑥 (𝑡) of the object at time t . 𝑥 ∴ 𝐴 t = 𝑑𝑣 𝑥 𝑑𝑡 𝑑𝑡 2 𝑑 2 𝑥 𝑡 = (𝑡) …..(ii)

28. Acceleration… Since, 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙 , substitute this equation into equation (ii); 𝑑 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙 𝑑𝑡 𝐴 𝑥 𝑡 𝐴 𝑥 𝑡 = −𝑎𝜔 2 sin 𝜔𝑡 + 𝜙 but 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 ∴ 𝐴 𝑥 𝑡 = −𝜔 2 𝑥

29. 𝒕 we have by the sine In our derivation of equation of 𝒗 𝒙 𝒕 and 𝑨 𝒙 arbitrary chosen to express the displacement function . If we could use cosine function representation, we could have produced the following equivalent expression; 𝑥 𝑡 = a cos(𝜔𝑡 + 𝜙) 𝑣 𝑥 𝑡 = −𝑎𝜔 sin(𝜔𝑡 + 𝜙 ) 𝐴 𝑥 𝑡 = −𝑎𝜔 2 cos 𝜔𝑡 + 𝜙 = −𝜔 2 𝑥

30. Graph Representation of S.H.M Graph representation of the displacement, velocity and acceleration for an object in one dimensional S.H.M as represented by equations 𝑥 𝑡 = 𝑎 sin 𝜔𝑡 + 𝜙 , 𝑣 𝑥 𝑡 = 𝑎𝜔 cos 𝜔𝑡 + 𝜙 and 𝐴 𝑥 𝑡 = −𝜔 2 𝑥 respectively. ∴ For the case 𝜙 =

31. Important Properties of a particle executing S.H.M 1. The displacement , velocity and acceleration all vary sinusoidal with time but are not in phase.

32. Important Properties… 2. The acceleration of the particle is proportional to the displacement but in the opposite direction.

33. Important Properties… The Frequency and period of the motion are independent of the amplitude of the motion. The particle moves to and fro about the mean (equilibrium) position in a straight line

34. Example 1 A simple harmonic oscillation is represented by; 𝑥 = 0.34 𝑐𝑜𝑠(3000𝑡 + 0.74) Where x and t are in mm and sec respectively. Determine; The amplitude The frequency and the angular frequency The time period.

35. Solution i. Amplitude; ∴ 𝑎 = 0.34 𝑚𝑚 ii. A. Angular frequency; ∴ 𝜔 = 3000 𝐻𝑧 ii. B. Frequency; 𝑓 = 𝜔 = 3000 2𝜋 2𝜋 𝜋 ∴ 𝑓 = 1500 Hz iii. Time period; 𝑇 = 1 𝑓 ∴ 𝑇 = 𝜋 1500 sec

36. Example 2 A particle executes S.H.M of amplitude 25 cm and time period 3 s . What is the minimum time required for the particle to move between two points 12.5 cm on either side of the mean position?

37. Solution 𝑇 𝑦 = 𝑎 sin 2𝜋 𝑡 Here 𝑦 = 12.5 𝑐𝑚 ; 𝑎 = 25 𝑐𝑚 ; 𝑇 = 3 𝑠 12.5 = 25 sin 2𝜋 3 𝑡 ∴ 𝒕 = 𝟏𝟒. 𝟑𝟐 𝒔

38. Example 3 The amplitude of a particle executing S.H.M with a frequency of 60 Hz is 0.01 m. Determine the maximum value of he acceleration of the particle.

39. Solution The maximum acceleration of a particle executing S.H.M is given by; 𝐴 𝑚𝑎𝑥 = 𝜔 2 𝑎 𝐴 𝑚𝑎𝑥 = (2𝜋𝑓) 2 𝑎 𝐴 𝑚𝑎𝑥 = (2𝜋 𝑥 60) 2 𝑥 0.01 ∴ 𝑨 𝒎𝒂𝒙 = 𝟏𝟒𝟐𝟏. 𝟐𝟐𝟑 𝒎 𝒔 −𝟐

40. Example 4 A particle executing S.H.M has a maximum displacement of 4 cm and its acceleration at a distance of 1 cm from its mean position is 3 𝒄𝒎 𝒔 −𝟐 . What will be its velocity when it is at a distance of 2 cm from its mean position?

41. Solution 𝐴 𝑥 𝑡 = −𝜔 2 𝑥 𝜔 = 𝑥 𝐴 = 3 1 𝝎 = 𝟏. 𝟕𝟑𝟐 𝒓𝒂𝒅 𝒔 −𝟏 𝑣 = 𝜔 𝑎 2 − 𝑥 2 𝑣 = 1.732 𝑥 4 2 − 2 2 ∴ 𝒗 = 𝟔 𝒎 𝒔 −𝟏

42. Forces acting in S.H.M of the particle From Newton’s second law of motion, 𝑭 = 𝒎𝑨 The acceleration expressed in terms of displacement executing S.H.M is; 𝑨 𝒙 𝒕 = −𝝎 𝟐 𝒙 Substitute the acceleration equation into the equation force then we have; ∴ 𝑭 = −𝒎𝝎 𝟐 𝒙

43. Forces acting in S.H.M In S.H.M . the magnitude of the force acting at any time is linearly proportional to the distance from the equilibrium at that time and the direction of the force is opposite to that of the displacement. Thus, the force causing S.H.M always acts in a direction that tends to reduce the displacement and for this reason it is usually called restoring force .

44. Hooke’s Law Law: The force exerted on a body by an external source is directly proportional to the mean displacement the object has displaced from its mean position. 𝐹 𝛼 − 𝑥 𝐹 = −𝑘𝑥

45. Force Law for S.H.M For a spring constant here being; 𝑘 = 𝑚𝜔 2 The angular frequency 𝜔 of the S.H.M of the block is related to 𝑘 and m of the block which yield; 𝜔 = 𝑘 𝑚

46. The period of the S.H.M of the block is also related to 𝑘 and m of the block which is; 𝑇 = 2𝜋 𝑚 𝑘

47. Example A block whose mass m is 680 g is fastened to a spring whose spring constant k is 65 N/m. The block is pulled a distance x = 11 cm from its equilibrium position at x = on a frictionless surface and released from rest at t = 0. What are the angular frequency, the frequency, and the period of the resulting motion? What is the maximum acceleration of the oscillating block.

48. Solution (a) The angular frequency is; 𝜔 = 𝑘 𝑚 𝜔 = 65 N/m 0.68 𝐾𝑔 ∴ 𝝎 = 𝟗. 𝟕𝟖 𝒓𝒂𝒅/𝒔 (a) The period is; 𝑇 = 2𝜋 𝑚 𝑘 𝑇 = 2𝜋 0.68 𝐾𝑔 65 N/m ∴ 𝑻 = 𝟎. 𝟔𝟒𝟐𝟕𝒔

49. Solution… (a) The frequency is; 𝑓 = = 𝜔 1 2𝜋 𝑇 𝑓 = 1 0.6427𝑠 𝒇 = 𝟏. 𝟓𝟓𝟔 𝑯𝒛 (b) Maximum acceleration; 𝑎 𝑚𝑎𝑥 = 𝜔 2 𝑥 𝑚𝑎𝑥 𝑎 𝑚𝑎𝑥 = 9.78 𝑟𝑎𝑑/𝑠 2 𝑥 0.11 𝑚 𝒂 𝒎𝒂𝒙 = 𝟏𝟎. 𝟓𝟐𝟏 𝒎𝒔 −𝟐

50. Springs in Parallel The 𝐹 𝑡𝑜𝑡 pul l ing the mass m at distance x is distributed among the three springs as shown in the equation below; 𝐹 𝑡𝑜𝑡 = 𝐹 1 + 𝐹 2 + 𝐹 3 i.e. displacement of each spring is equal to x 𝐹 𝑡𝑜𝑡 = 𝑘 1 𝑥 + 𝑘 2 𝑥 + 𝑘 3 𝑥

51. Spring in parallel… An equivalent spring constant 𝑘 𝐸 for springs connected in parallel is given by; 𝑘 𝐸 = 𝑘 1 + 𝑘 2 + 𝑘 3 Hooke’s law for the combination of springs is given by; 𝐹 𝑡𝑜𝑡 = 𝑘 𝐸 𝑥 The will springs execute in parallel a S.H.M whose period is given by; 𝑇 = 2𝜋 𝑚 𝑘 𝐸

52. Example Three springs with force constants 𝑘 1 = 10.0 N/m , 𝑘 2 = 12.5 N/m , and 𝑘 3 = 15.0 N/m are connected in parallel to a mass of 0.500 kg . The mass is then pulled to the right and released. Find the period of the motion.

53. Solution The period of the motion is; 𝑇 = 2𝜋 𝑚 𝑘 𝐸 𝑇 = 2𝜋 𝑚 𝑘 1 + 𝑘 2 + 𝑘 3 𝑇 = 2𝜋 0.5 10 + 12.5 + 15 ∴ 𝑻 = 𝟎. 𝟕𝟐𝟓𝟓 𝒔

54. Springs in Series The total displacement x of the spring connected in series is the sum of the displacements of each spring that is; 𝒙 = 𝒙 𝟏 + 𝒙 𝟐 + 𝒙 𝟑 i.e. Force in each spring is transmitted equally; 𝐹 𝑡𝑜𝑡 = 𝐹 1 = 𝐹 2 = 𝐹 3

55. Spring Series…. Since, 𝒙 = 𝒙 𝟏 + 𝒙 𝟐 + 𝒙 𝟑 Then, 𝑥 = 𝐹 𝑡𝑜𝑡 1 + 1 + 1 𝑘 1 𝑘 2 𝑘 3 Therefore; equivalent spring constant for spring connected in series is; 𝑘 𝐸 = 1 1 + 1 + 1 𝑘 1 𝑘 2 𝑘 3 The springs in series will execute a S.H.M whose period is given by; 𝑇 = 2𝜋 𝑚 𝑘 𝑘 𝑘 1 2 3 𝑘 2 𝑘 3 + 𝑘 1 𝑘 3 + 𝑘 1 𝑘 2

56. Example Three springs with force constants 𝑘 1 = 10.0 N/m , 𝑘 2 = 12.5 N/m , and 𝑘 3 = 15.0 N/m are connected in series to a mass of 0.500 kg . The mass is then pulled to the right and released. Find the period of the motion.

57. Solution The period of the motion is 𝑇 = 2𝜋 𝑚 𝑘 1 𝑘 2 𝑘 3 𝑘 2 𝑘 3 + 𝑘 1 𝑘 3 + 𝑘 1 𝑘 2 𝑇 = 2𝜋 0.5 10𝑥12.5𝑥15 12.5𝑥15 + 10𝑥15 + 10𝑥12.5 ∴ 𝑇 = 2.207 𝑠

58. Conservation of Energy The vibrating spring system can also be described in terms of the law of conservation of energy . When the spring is stretched to its maximum displacement a, work is done on the spring, and hence the spring contains potential energy . The kinetic energy is equal to zero at this point because v = at the maximum displacement. The total energy of the system is thus; 𝐸 𝑡𝑜𝑡 = 𝑃𝐸 + 𝐾𝐸

59. Kinetic Energy ( KE ) From KE equation, 𝐾𝐸 = 1 / 2 𝑚𝑣 2 ……….(i) Then, substitute equation; 𝑣 = 𝑎𝜔 cos(𝜔𝑡 + 𝜙) into equation (i), then we have; 𝐾𝐸 = 1 / 2 𝑎 2 𝑚𝜔 2 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙 But 𝐹 = −𝑘𝑥 = 𝑚𝑎 = −𝑚𝜔 2 𝑥 ; 𝝎 𝟐 = 𝒌 / 𝒎 ∴ 𝐾𝐸 = 1 / 2 𝑎 2 𝑘 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙

60. Potential Energy ( PE ) PE of the system is given by the amount of work required to move system from position 𝑡𝑜 𝑥 under the position of applied force ( 𝑘𝑥). Work done of the particle executing S.H.M; 𝑊 = −𝐹𝑑𝑥 𝑊 = 𝑘𝑥𝑑𝑥 𝑤 𝑥 ∫ 𝑊 = 𝑘 ∫ 𝑥 𝑑𝑥 𝑊 = 1 / 2 𝑘𝑥 2

61. PE… From equation ; 𝑥 = 𝑎 sin 𝜔𝑡 + 𝜙 𝑊 = 1 / 2 𝑘𝑥 2 substitute into equation 𝑃𝐸 = 𝑊 = 1 / 2 𝑘𝑎 2 𝑠𝑖𝑛 2 (𝜔𝑡 + 𝜙) ∴ 𝑃𝐸 = 1 / 2 𝑘𝑎 2 𝑠𝑖𝑛 2 (𝜔𝑡 + 𝜙)

62. Total Mechanical Energy Total mechanical energy of the system; 𝐸 𝑡𝑜𝑡 = 𝑃𝐸 + 𝐾𝐸 But, 𝐾𝐸 = 1 / 2 𝑎 2 𝑘 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙 𝑃𝐸 = 1 / 2 𝑘𝑎 2 𝑠𝑖𝑛 2 (𝜔𝑡 + 𝜙) Then; 𝐸 𝑡𝑜𝑡 = 1 / 2 𝑘𝑎 2 𝑠𝑖𝑛 2 𝜔𝑡 + 𝜙 + 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙

63. Total Mechanical Energy of the System… From trigonometric identity; 𝑠𝑖𝑛 2 𝜃 + 𝑐𝑜𝑠 2 𝜃 = 1 Then; 𝐸 𝑡𝑜𝑡 = 1 / 2 𝑘𝑎 2 𝑠𝑖𝑛 2 𝜔𝑡 + 𝜙 + 𝑐𝑜𝑠 2 𝜔𝑡 + 𝜙 𝐸 𝑡𝑜𝑡 = 1 / 2 𝑘𝑎 2 = 1

64. Total Mechanical Energy of the System… Thus 𝐸 𝑡𝑜𝑡 of the oscillator remains constant as 𝑥 is regained after every half cycle. If no energy is dissipated then all the PE becomes KE and vice versa. Figure on the right side shows the variation of KE and PE of harmonic oscillator with time where phase φ is set to zero for simplicity.

65. Example Conservation of ene r gy applied to a spring. A horizontal spring has a spring con s tant of 29.4 N/m. A m ass of 300 g is attached to t h e spring and displaced 5.00 cm. The m ass is then rele a sed. Find the total energy of the system, the maximum velocity of the system, and the potential energy and kinetic energy for x = 2.00 cm.

66. Solution… (a) The total Energy of the system( 𝑬 𝒕𝒐𝒕 ) is; 𝐸 𝑡𝑜𝑡 = 1 / 2 𝑘𝑎 2 𝐸 𝑡𝑜𝑡 = 1 / 2 𝑥 29.4 𝑥 (0.05) 2 ∴ 𝑬 𝒕𝒐𝒕 = 𝟑. 𝟔𝟕𝟓 𝒙 𝟏𝟎 −𝟐 𝑱

67. Solution… (b). The maximum velocity occurs when x = and the potential energy is zero; 𝐸 𝑡𝑜𝑡 = 𝐾𝐸 𝑚𝑎𝑥 = 1 / 2 𝑚𝑣 𝑚𝑎𝑥 2 𝑣 𝑚𝑎𝑥 = 2𝐸 𝑡𝑜𝑡 𝑚 1 / 2 𝑣 𝑚𝑎𝑥 = 2 𝑥 3.675 𝑥 10 −2 0.3 1 / 2 ∴ 𝑣 𝑚𝑎𝑥 = 0.495 𝑚𝑠 −1

68. Solution… 2 (c)The potential energy at 2.00 cm is; 𝑃𝐸 = 1 / 2 𝑘𝑥 2 = 1 / 2 𝑥 29.4 𝑥 0.02 ∴ 𝑃𝐸 𝑎𝑡 2.0𝑐𝑚 = 5.88 x 10 −3 𝐽 The kinetic energy at 2.00 cm is; K 𝐸 = 𝐸 𝑡𝑜𝑡 − 𝑃𝐸 𝑎𝑡 2.00𝑐𝑚 K 𝐸 = 3.675 𝑥 10 −2 𝐽 − 5.88 𝑥 10 −3 𝐽 ∴ 𝐾𝐸 𝑎𝑡 2.0𝑐𝑚 = 0.03087 𝐽