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Simple harmonic motionSIMPLE HARMONIC MOTIONThis 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).Definition of SHMRepeated 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 papersOscillatory 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.SHM is a particle motion with an acceleration (a) that is directly proportional to the particles displacement (y) from a fixed point (rest point), and this acceleration always points towards the fixed point.Equilibrium pointyAAy

or

DEFINITION OF SHMThe 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.W is angular frequency.5A pendulum and a mass on a spring both undergo this type of motion which can be described by a SINE WAVE or a COSINE WAVE depending upon the start position.

Displacement yTime t+ A- AEquilibriumA = maximum amplitude possible from equilibrium point.A body 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.

Aspects of SHM

DisplacementVelocityAccelerationSin graphCosine graphUpside down sine graph

Displacement ytimeAmplitude ( A ): The maximum distance that an object moves from its rest position. y = A and y = - A . + A- APeriod ( T ): The time that it takes to execute one complete cycle of its motion. Units seconds, T

Frequency ( f ): The number or oscillations the object completes per unit time. Units Hz = s-1 .

Angular Frequency ( ): The frequency in radians per second, 2 per cycle.

Example 1:During exercise, a runners 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 restSOLUTION:60 beats in 20 seconds 60/20 beats per second = 3.0HzT = 1/f = 1/3.0 T = 0.33s36 beats in 30 seconds 36/30 beats per second = 1.2 HzT = 1/f = 1/1.2T = 0.83sThe Reference Circle

Reference CircleThe imaginary green point moves in a circle at constant speed (). The vertical component of the green points motion is SHMEquilibrium PositionRotational Motion and SHMSHM 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

1234Always 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 1 + radPHASOROSCILLATING OBJECTYour turnRead ESA pg145-147Use this reading to supplement your PowerPoint notesComplete Activity 10A.

Rotational equations: = / t = 2 / T = 2f As f = 1/TFor 1 revolution = 2t = TIt stands to reason that:Maximum displacement is given by the radius of the circle, A.Maximum translational velocity, vmax = AMaximum translational acceleration, amax = A2

EQUATION OF SHMxaAcceleration Displacement graph

Gradient = - 2+ A- AMAXIMUM ACCELERATION = 2 A

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 Hzb. = 2f = 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 .5.4m1.7m1.7mAAYellow triangle:cos = (A 1.7) /A = (2.7 1.7)/2.7 = 1.19 radBut = t t = 1.19/1.45 x 10-4 = 8191.3877s = 2.3 hoursTo 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.YANB: if & t given then can be calculated.From this we are able to determine :1. Displacement2. Velocity3. Accelerationequations at any point of the phasor or reference circle.Aspects of SHM

DisplacementVelocityAccelerationSin graphCosine graphUpside down sine graphDISPLACEMENTAyYAFrom this we are able to find any size of displacement y, using the triangle and the following equation:y = A sintBUT = ty = A sinExample 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 startsSOLUTION:a. y = A sint = 0.14 x sin(2/1.9 x 0.35)[substituting = 2/T] = 0.13mb. y = A sint = 0.14 x sin(2/1.9 x 1.1) = -0.067m y = 0.067m in the opposite directionVelocityThe 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 SHMThe direction and length of the velocity vector will always be changing.When the object is at the end position-no component in direction of SHMWhen the object is at the equilibrium position-whole vector in direction of SHMVELOCITYAvAFrom this we are able to find any size of velocity v, using the triangle and the following equation:v = A cosvABUT = tv = A costExample 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 startsSOLUTION:a. v = A cost = 0.14 x (2/1.9) x cos(2/1.9 x 0.35)[substituting = 2/T] = 0.19ms-1b. v = A cost = 0.14 x (2/1.9) x cos(2/1.9 x 1.1) = -0.41ms-1 v = 0.41ms-1 in the opposite directionAccelerationFind 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.

ACCELERATIONaA2From this we are able to find any size of acceleration a, using the triangle and the following equation:a = - A2 sinaA2BUT = ta = - A2 sinta = v2/r & v = rCentripetal acc acts to centre thus:a = v2/A = (A)2/Aa = A2Example 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 startsSOLUTION:a. a = - A2 sint = - 0.14 x (2/1.9)2 x sin(2/1.9 x 0.35)[substituting = 2/T] = - 1.4ms-2b. a = - A2 sint = - 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

Displacement xVelocity vAcceleration a

ttt

Velocity = gradient of displacement- time graphMaximum velocity in the middle of the motionZERO velocity at the end of the motion

Acceleration = gradient of velocity - time graphMaximum acceleration at the end of the motion where the restoring force is greatest!ZERO acceleration in the middle of the motion.Note direction opposite to displacement, going negative here.Note that in this graph the timing starts when y is maximum hence the cosine instead of sineNote:The following equations:v = A costy = A sinta = - A2 sintOnly 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 costv = A sinta = - A2 cost

Displacement xVelocity vAcceleration a

ttt

Velocity = gradient of displacement- time graphMaximum velocity in the middle of the motionZERO velocity at the end of the motion

Acceleration = gradient of velocity - time graphMaximum acceleration at the end of the motion where the restoring force is greatest!ZERO acceleration in the middle of the motion.Note direction opposite to displacement, going negative here.Note that in this graph the timing starts when y is maximum hence the cosine instead of sineCOMPLETE EXERCISESPAGE 92 - 101RUTTERMASS ON A SPRINGMF = -kyyMStretch & ReleaseAMM

k = the spring constant in N m-1If two springs are attached to a mass k is the combined spring constants for both springs. Equation is the same.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-1T = 2(m/k) = 2(0.40/49) = 0.57sTHE PENDULUM

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

Example 7:Find the length of a pendulum which will have a period of 1.0 seconds.SOLUTION:T = 2(L/g)T2 = 42L/g[squaring]L = T2g/42 = 1.02 x 9.8/42 = 0.25mCOMPLETE EXERCISESPAGE 102 - 104RUTTERENERGY IN SHMpotentialEPKineticEKPotentialEPPENDULUMSPRINGMMMpotentialkineticETOTAL = Ep + EK ENERGY IN A SPRINGEk when the mass moves though 0 at max speedEp when the spring is at max compression & momentarily stopped.0Ek 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 EkThe total energy is the energy stored in the spring when it was initially extended by the amplitude of the SHM.Total E = kA2At any instant, the kinetic energy is the total energy less the potential energy.Ek = kA2 - ky2Example 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 = kA2 = 0.3 x 110 x 0.252 = 3.4Jb. mv2 = 3.4375 = 4.4ms-1c. mv2 = 3.4375 - k x 0.152 v = 3.5ms-1ENERGY IN A PENDULUMIn 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. hA 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 = mghEk = mv2Example 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 = 0.14Jb. mv2 = 0.13524 v = 1.3ms-1ENERGY IN SHMEnergy Change with POSITION= kinetic= potential= TOTAL ENERGY, EEnergy Change with TIME

x-A+A0energyE

energytimeTT/2 N.B. Both the kinetic and the potential energies reach a maximum TWICE in on cycle.EExample 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) = 3.2sb. = 2/T = 2/3.1734 = 2.0rads-1c. Total E = mvmax2 = m(A)2 = 1.6Jd. When the mass is moving at 0.50ms-1: Ek = x 5.0 x 0.502 = 0.625J Ep = Ek TOT Ek = 1.56800 0.625 = 0.94J

timeDAMPINGDISPLACEMENTINITIAL AMPLITUDETHE AMPLITUDE DECAYS EXPONENTIALLY WITH TIMEThe 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.RESONANCE:This is where a driving force causes an object to oscillate. This matches the natural frequency of the object causing a larger amplitude. This can produce some interesting effects: Musical instruments often produce sound by resonating such as strings or columns of air. Radio receivers work because a resonant electric circuit can be tuned to radio waves of a particular frequency. Buildings and bridges have a natural frequency of oscillation. Engineers must design them to make sure that seismic vibrations, winds or basic walking do not cause damaging resonance. E.g. 1940 USA, Tacoma Narrows suspension bridge was destroyed by high winds causing the bridge to shake itself to pieces & the Millennium bridge in London, UK.

COMPLETE EXERCISESPAGE 105 - 106RUTTER