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Simple harmonic motion

Simple harmonic motion. SIMPLE HARMONIC MOTION. This unit is made up of the following: Reference circle (page 92 – 94). Graphs, Phasors & Equations (page 94 – 101). SHM of a spring (page 102 – 103). SHM of a pendulum (page 104). Energy of SHM. Forced SHM & Resonance (page 105 – 106).

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Simple harmonic motion

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  1. Simple harmonic motion

  2. SIMPLE HARMONIC MOTION • This unit is made up of the following: • Reference circle (page 92 – 94). • Graphs, Phasors & Equations (page 94 – 101). • SHM of a spring (page 102 – 103). • SHM of a pendulum (page 104). • Energy of SHM. • Forced SHM & Resonance (page 105 – 106).

  3. Definition of SHM Repeated motion where the acceleration is proportional to, and in the opposite direction to, the displacement This is an essential definition learn it by heart it appears somewhere in almost all NCEA L3 papers

  4. Oscillatory motion is motion that occurs periodically over and over again. • E.g. • rotation of the earth on its axis, • rise and fall of tides in the ocean; • vibration of a guitar string; • regular beating of the human heart. • Common type of oscillatory motion is Simple Harmonic Motion (SHM). Object moves back and forth regularly over the same path. • E.g. • Wave motion (seismic/sound/water) • Vibrations of molecules gases of the atmosphere • Movement of electrons in circuits.

  5. A Equilibrium point y y A DEFINITION OF SHM SHM is a particle motion with an acceleration (a) that is directly proportional to the particle’s displacement (y) from a fixed point (rest point), and this acceleration always points towards the fixed point. or The negative value denotes that the acceleration is back towards the equilibrium point. The ‘y’ value could be ‘x’ but for a spring displacement is up & down.

  6. + A - A Abody will undergo SIMPLE HARMONIC MOTION when the force that tries to restore the object to its REST (EQUILIBRIUM) POSITION is PROPORTIONAL TO the DISPLACEMENT of the object. A pendulum and a mass on a spring both undergo this type of motion which can be described by a SINE WAVEor a COSINE WAVE depending upon the start position. Displacement y Equilibrium Time t A = maximum amplitude possible from equilibrium point.

  7. Aspects of SHM Displacement Sin graph Velocity Cosine graph Upside down sine graph Acceleration

  8. Displacement y T time + A Frequency ( f ):The number or oscillations the object completes per unit time. Units Hz = s-1 . - A Angular Frequency ( ω ):The frequency in radians per second, 2πper cycle. Amplitude( A ):The maximum distance that an object moves from its rest position. y = A and y = - A . Period ( T ):The time that it takes to execute one complete cycle of its motion. Units seconds,

  9. Example 1: • During exercise, a runner’s heart beats at 60 times in 20 seconds. Later, while resting, her resting heart beats 36 times in 30 seconds. • Calculate the frequency and period: • During exercise • At rest • SOLUTION: • 60 beats in 20 seconds  60/20 beats per second = 3.0Hz • T = 1/f = 1/3.0 • T = 0.33s • 36 beats in 30 seconds 36/30 beats per second = 1.2 Hz • T = 1/f = 1/1.2 • T = 0.83s

  10. The Reference Circle Equilibrium Position Reference Circle The imaginary green point moves in a circle at constant speed (ω). The vertical component of the green point’s motion is SHM

  11. Rotational Motion and SHM • SHM can be analysed by using our knowledge of circular motion; as the vertical component of circular motion of the reference circle is the same as SHM

  12. 1  2  4 3 PHASOR OSCILLATING OBJECT • Always work anticlockwise from equilibrium point. • Object oscillates at a speed of . • Is moving down at max displacement ‘A’. • Has moved  rad of the circle at equilibrium point. • Max ‘A’ changing direction therefore max ‘a’. • Now moving upwards moved through 1½ +  rad

  13. Your turn • Read ESA pg145-147 • Use this reading to supplement your PowerPoint notes • Complete Activity 10A.

  14. Rotational equations:  =  / t • For 1 revolution • = 2 t = T  = 2 / T As f = 1/T  = 2f It stands to reason that: Maximum displacement is given by the radius of the circle, ‘A’. Maximum translational velocity, vmax = A Maximum translational acceleration, amax = A2

  15. a + A x - A EQUATION OF SHM Acceleration – Displacement graph Gradient = - ω2 MAXIMUM ACCELERATION = ±ω2 A

  16. Example 2: • The tides go up and down with SHM. The period of the tides is 12 hours. • Calculate the cyclic frequency, f , of the tides. • Calculate the angular frequency, , of the tides • A boat is tied up alongside a wharf at exactly low tide. How long will it take the boat to rise 1.7m up the side of the wharf if the total rise of the tide is 5.4m? SOLUTION: a. Period is 12 hours,  T = 12 x 60 x 60 s = 43 200s f = 1/T = 2.3 x 10-5 Hz

  17. b.  = 2f = 2 x  x 2.314815 x 10-5 = 1.5 x 10-4 rads-1 c. While the boat rises a distance of 1.7m the displacement phasor turns through an angle . Yellow triangle: cos = (A – 1.7) /A = (2.7 – 1.7)/2.7  = 1.19 rad But  = t t = 1.19/1.45 x 10-4 = 8191.3877s = 2.3 hours 5.4m A  A 1.7m 1.7m

  18. To find a proportion of the circle i.e.  or the height after a certain portion of the circle has been travelled, we need to produce a right angle triangle with the displacement, , the hypotenuse, ‘A’ and opposite, y. NB: if  & t given then  can be calculated. A Y  From this we are able to determine : 1. Displacement 2. Velocity 3. Acceleration equations at any point of the phasor or reference circle.

  19. Aspects of SHM Displacement Sin graph Velocity Cosine graph Upside down sine graph Acceleration

  20. DISPLACEMENT A A Y y     From this we are able to find any size of displacement ‘y’, using the triangle and the following equation: y = A sin BUT  = t y = A sint

  21. Example 3: • A pendulum swings with a period of 1.9s and the bob has amplitude 0.14m. Calculate the distance of the pendulum bob from the central position: • 0.35s after the timing starts • 1.1s after the timing starts SOLUTION: a. y = A sint = 0.14 x sin(2/1.9 x 0.35) [substituting  = 2/T] = 0.13m b. y = A sint = 0.14 x sin(2/1.9 x 1.1) = -0.067m  y = 0.067m in the opposite direction

  22. Velocity • The direction of the velocity of an object with circular motion is always changing. • The size can be found by v=r • The velocity of the SHM will be the component of the velocity in the direction of the SHM • The direction and length of the velocity vector will always be changing. • When the object is at the end position-no component in direction of SHM • When the object is at the equilibrium position-whole vector in direction of SHM

  23. VELOCITY A  v A A     v From this we are able to find any size of velocity ‘v’, using the triangle and the following equation: v = A cos BUT  = t v = A cost

  24. Example 4: • The pendulum in example 3, swings with a period of 1.9s and the bob has amplitude 0.14m. Calculate the velocity of the pendulum bob: • 0.35s after the timing starts • 1.1s after the timing starts SOLUTION: a. v = Acost = 0.14 x (2/1.9) x cos(2/1.9 x 0.35) [substituting  = 2/T] = 0.19ms-1 b. v = Acost = 0.14 x (2/1.9) x cos(2/1.9 x 1.1) = -0.41ms-1  v = 0.41ms-1 in the opposite direction

  25. Acceleration • Find acceleration using acceleration phasor. • Direction of acceleration during circular motion is towards the centre of the circle. • When object at end position-whole vector is the SHM component. • When object at equilibrium position-no vector in direction of SHM.

  26. ACCELERATION A2 A2 a   a   From this we are able to find any size of acceleration ‘a’, using the triangle and the following equation: a = - A2 sin a = v2/r & v = r Centripetal acc acts to centre thus: a = v2/A = (A)2/A a = A2 BUT  = t a = - A2 sint

  27. Example 5: • The pendulum in examples 3 & 4, swings with a period of 1.9s and the bob has amplitude 0.14m. Calculate the acceleration of the pendulum bob: • 0.35s after the timing starts • 1.1s after the timing starts SOLUTION: a. a = - A2sint = - 0.14 x (2/1.9)2 x sin(2/1.9 x 0.35) [substituting  = 2/T] = - 1.4ms-2 b. a = - A2sint = - 0.14 x (2/1.9)2 x sin(2/1.9 x 1.1) = - - 0.73ms-2  a = 0.73ms-2 in the upward direction

  28. Velocity = gradient of displacement- time graph Acceleration = gradient of velocity - time graph Displacement x t Note that in this graph the timing starts when y is maximum hence the cosine instead of sine Velocity v Maximum velocity in the middle of the motion t ZERO velocity at the end of the motion Acceleration a Maximum acceleration at the end of the motion – where the restoring force is greatest! t ZERO acceleration in the middle of the motion. Note direction opposite to displacement, going negative here.

  29. Note: The following equations: y = A sint v = A cost a = - A2 sint Only apply when the object is at equilibrium position and moving in a positive direction, anticlockwise when timing starts. If timing starts when y = A i.e. at max amplitude then the equations are used as follows: y = A cost v = A sint a = - A2 cost

  30. Velocity = gradient of displacement- time graph Acceleration = gradient of velocity - time graph Displacement x t Note that in this graph the timing starts when y is maximum hence the cosine instead of sine Velocity v Maximum velocity in the middle of the motion t ZERO velocity at the end of the motion Acceleration a Maximum acceleration at the end of the motion – where the restoring force is greatest! t ZERO acceleration in the middle of the motion. Note direction opposite to displacement, going negative here.

  31. COMPLETE EXERCISES PAGE 92 - 101 RUTTER

  32. MASS ON A SPRING y M M M A F = -ky Stretch & Release M k = the spring constant in N m-1 If two springs are attached to a mass ‘k’ is the combined spring constants for both springs. Equation is the same.

  33. Example 6: When a 0.40kg mass is attached to a spring suspended vertically, its length increases from 27.0cm to 35.0cm. Find the spring constant and period of oscillation. SOLUTION: The extension y = 35.0 – 27.0 = 8.0cm or -0.080m as the extension is opposite to the force. k = f/-y = 3.92/0.080 = 49Nm-1 T = 2(m/k) = 2(0.40/49) = 0.57s

  34. l THE PENDULUM The period, T, is the time for one complete cycle. Where l = length of string (m); g = gravitational acceleration (ms-2)

  35. Example 7: Find the length of a pendulum which will have a period of 1.0 seconds. SOLUTION: T = 2(L/g) T2 = 42L/g [squaring] L = T2g/42 = 1.02 x 9.8/42 = 0.25m

  36. COMPLETE EXERCISES PAGE 102 - 104 RUTTER

  37. M M M ENERGY IN SHM PENDULUM SPRING kinetic potential EP Potential EP Kinetic EK potential ETOTAL = Ep + EK

  38. ENERGY IN A SPRING 0 Ek when the mass moves though 0 at max speed Ep when the spring is at max compression & momentarily stopped. Ek again as the mass moves back through 0 at max speed. Ep as the spring is a max extension & the mass is momentarily stopped. Ek again as the mass moves back through 0 at max speed. At intermediate positions, the energy is partly Ep and Ek

  39. The total energy is the energy stored in the spring when it was initially extended by the amplitude of the SHM. Total E = ½kA2 At any instant, the kinetic energy is the total energy less the potential energy. Ek = ½kA2 - ½ky2

  40. Example 8: • An object of mass 0.35kg bounces vertically on the end of a spring that has a force constant of 110Nm-1. The amplitude of the SHM is 25cm (0.25m) • Calculate the total energy in the oscillating system • Calculate the maximum speed of the object. • Calculate the speed of the object when the displacement is 15cm. • SOLUTION: • Total E = ½kA2 • = 0.3 x 110 x 0.252 • = 3.4J • b. ½mv2 = 3.4375 • = 4.4ms-1 c. ½mv2 = 3.4375 - ½k x 0.152 v = 3.5ms-1

  41. ENERGY IN A PENDULUM In a pendulum as we saw earlier the main energies are Ep and Ek. Energy is applied to the bob by pulling it to one side gaining gravitational potential energy. h A • At end positions Ek = 0 & max potential energy. • At centre position Ep = 0 & max kinetic energy. • Between these positions the energy is partly Ek & Ep. Ep = mgh Ek = ½mv2

  42. Example 9: • A pendulum has a bob of mass 0.15kg. To set it in SHM, the bob is pulled aside until it has gained a height of 0.092m. • Calculate the total energy of the SHM. • Calculate the maximum speed of the bob. • SOLUTION: • Total E = mgh • = 0.15 x 9.8 x 0.092 • = 0.14J • b. ½mv2 = 0.13524 • v = 1.3ms-1

  43. ENERGY IN SHM = kinetic = potential energy = TOTAL ENERGY, E E E x -A +A 0 energy time Energy Change with POSITION Energy Change with TIME N.B. Both the kinetic and the potential energies reach a maximum TWICE in on cycle. T/2 T

  44. Example 10: • A pendulum of length 2.5m and mass 5.0kg swings with a amplitude of 0.40m. • Find • The period T. • The angular frequency, . • The total energy of the SHM. • The potential energy at the point where the mass is moving with a speed of 5.0ms-1. SOLUTION: a. T = 2(l/g) = 2(2.5/9.8) = 3.2s b.  = 2/T = 2/3.1734 = 2.0rads-1 c. Total E = ½mvmax2 = ½m(A)2 = 1.6J

  45. d. When the mass is moving at 0.50ms-1: Ek = ½ x 5.0 x 0.502 = 0.625J Ep = Ek TOT – Ek = 1.56800 – 0.625 = 0.94J

  46. INITIAL AMPLITUDE time DAMPING DISPLACEMENT The causes of damping often can be friction and air resistance which generate heat and thus loss of energy. Also if there is rubbing of particles again friction will result. THE AMPLITUDE DECAYS EXPONENTIALLY WITH TIME

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