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- 1. 1 CHAPTER 2 SIMPLE HARMONIC MOTION 2.1 Definitionof SimpleHarmonicMotion Simple harmonicmotion, occurswhenthe accelerationisproportionaltodisplacementbutthey are inopposite directions. Simple HarmonicMotion(SHM).The motionthatoccurs whenan objectisacceleratedtowards a mid-point.The size of the accelerationisdependentuponthe distance of the objectfromthe mid-point.Verycommontype of motion,eg.seawaves,pendulums,spring. Simple harmonicmotionoccurswhenthe force Facting onan objectisdirectlyproportional to the displacementx of the object,butinthe opposite direction. Mathematical statementF= - kx The force iscalleda restoringforce because italwaysactson the objectto returnit to its equilibriumposition. 2.2 Descriptive terms a. The amplitude A isthe maximumdisplacementfromthe equilibriumposition. b. The periodT isthe time forone complete oscillation.Aftertime Tthe motionrepeatsitself.In general x(t) =x (t + T). c. The frequencyf isthe numberof oscillationspersecond.The frequencyequalsthe reciprocal of the period. f = 1/T. d. Althoughsimple harmonicmotionisnotmotioninacircle,it isconvenienttouse angular frequency bydefining= 2pf = 2p/T 2.3 Simple HarmonicMotion A bodyat simple harmonicmotionif : a) The accelerationalwaysdirectedtowardafixedpointsontheirpath. b) The accelerationisproportional toitsdisplacementfromafixedpointsandalwaysdirectedtoward that points.
- 2. 2 Example forsimple harmonicmotion a) When a mass hanging from the spring is deflected it will move with simple harmonic motion. b) Motion of the piston in a cylinder is close simple harmonic motion. c) Weight of the pendulum moves with simple harmonic motion if the angle is small.
- 3. 3 2.4 Simple HarmonicMotionDiagram Q θ V Vt Vo θ θ x a P o o P V a X ω Q iii) space diagramii) vectordiagrami Figure 2.1
- 4. 4 Figure 2.1 (i) shows a point P rotating with constant velocity , in a circular path of radius , a Linear velocity , V = a From figure 2.1 (ii) , the horizontal component of velocity V , the velocity of point Q Vq = V sin = a sin From space diagram, PQ = (a2 – x2) sin = (a2 - x2) a Vq = (a) (a2 - x2) a = () (a2 - x2) P V O , Q P , Q X P V O, Q i) Vq maksimum=a ii) Vq minimum=0 Figure : 2.2
- 5. 5 Whenx = 0 , Vq ismaximum, Vq maks = (a2 – 0) = a (figure 2.2 i) Whenx = a , Vq is minimum Vq min= (a2 – a2 ) = 0 (figure 2.2 ii) If P rotate with constant angular speed , it also has a central acceleration f , goes to center of rotation O. f = a 2 o P V a X ω Q P f = a2 f q Q 0 (ii) Vectordiagram P Q0 a x (i) Figure : 2.3 (iii) space diagram
- 6. 6 (fromfigure 2.3 ii : vectordiagram ) central acceleration of the horizontal component f , the acceleration of point Q fq = a 2 kos fromfigure 2.3 iii , kos = x a fq = (a 2 ) x a = x 2 Whenx = 0 , fq isminimum fq min = 02 = 0 Whenx = a , fq is maximum fq maks = a 2 2.5 PeriodicTime ,FrequencyandAmplitud Periodic time or period , T is the time taken by the point Q to make a swing back and forth to complete . And that time isequal to the time taken by the OP toturn a rotation 2 radians with angularspeed rad/s . Periodictime ,T= 2 But , fq = x2 dan = fq X T = 2 x = 2 distance fq acceleration
- 7. 7 Frequency, n is the number of a complete cycle of oscillation in the penetration by the point Q in the second. The unit is the Hertz ( Hz), ie one cycle per second . Frequency , n = Hz 2 n = 1/T =1______________ Hz 2 distance/acceleration Amplitude , a is the maximum displacement of point Q from a fixed point O. The distance , 2a traveled by the point Q is known as a stroke or swing . EXAMPLE 1 A point moves with simple harmonic motion with the acceleration of 9 m/s2 and velocity of 0.92 m/s when it is 65 mm from the center of travel. Find : i. amplitude , ii. time of the periodic motion
- 8. 8 EXAMPLE 2 A particle moving with simple harmonic motion has a periodic time of 0.4 s and it was back and forth between two points is 1.22 m. Determine : i. The frequency and amplitude of the oscillation . ii. Velocity and acceleration of the particle when it is 400 mm from the center of oscillation . iii.Velocity and maximum acceleration of the movement . EXAMPLE 3 A mass of body 1.5 kg moving with simple harmonic motion is towards to the end of the swing . At the time he was at A, 760 mm from the center of oscillation , velocity and acceleration is 9 m/s and 10 m/s2 , respectively. Find : a) frequency and amplitude of the oscillation , b) the maximum acceleration and the inertia of the body when it is at the end of the swing , c) the time has elapsed for it to go and back to A.
- 9. 9 2.6 Elastic system - mass and spring. mg d Stiffnessof spring,S Ked.tak tegang Ked.tegang dan pegun Ked.keseim bangan,o Ked.terpesong Daya spring x Mf f mg
- 10. 10 The figure shows a body of mass M kg supported by a spring of stiffness S N/m . Static spring deflection is d meters , then: Mg = Sd If the body is in the pull- down x meters from the equilibrium position O ( a fixed point) and then released , so Body weight + inertia force = total spring force Mg + Mf = Sd + Sx Where, Mg = Sd Mf = Sx f = (S/M) x The reference S/M is constant for a system under consideration. So , the acceleration f is proportional to the distance x from the equilibrium position O . This indicates that the body is moving with simple harmonic motion . From SHM, f = x2 = √ 𝑆 𝑀
- 11. 11 So, time periodic , T = 2 ∴ 𝑇 = 2𝜋√ 𝑀 𝑆 Mg = Sd dan 𝑀 𝑆 = 𝑑 𝑔 ∴ 𝑇 = 2𝜋 √( 𝑑 𝑔 ) EXAMPLE 4 A body of mass 14 kg is suspended by springs up from the end attached to a rigid support . The body produces a static deflection of 25 mm . It is in the pull down as far as 23 mm and then released . Find : i. Initial acceleration of the body , ii. the periodic time oscillations , iii. the maximum spring force , iv. the velocity and acceleration of the body when it is 12 mm from the equilibrium position .
- 12. 12 EXAMPLE 5 Two body , each type per 6.5 kg , suspended on a vertical spring with stiffness 2.45 kN/m. A body is removed and this causes the system to oscillate. Find : i. The maximum extension spring , ii. the periodic time and amplitude of the oscillation , iii. the velocity and acceleration of the mass when it is at the center of the amplitude , iv. the total energy of the oscillations . 2.7 Simple pendulum , Mff mg mf P
- 13. 13 The diagram shows a simple pendulum of length cord, L and weight B with mass M. Amplitude of the oscillation is small, not exceeding 120 , the angular displacement and the three forces , the heavy weight , cord tension and inertia forces , is in equilibrium( figure ii) From the force triangle . sin 𝜃 = 𝑀𝑓 𝑀𝑔 = 𝑓 𝑔 sin θ ≅ θ , sebab θ adalah kecil 𝜃 = 𝑓 𝑔 length of arc , x = L (s = r) = 𝑥 𝐿 ∴ 𝑓 𝑔 = 𝑥 𝐿 𝑓 = 𝑥 ( 𝑔 𝐿 ) for pendulum f = x2 for SHM
- 14. 14 This shows that the oscillations of a simple pendulum is simple harmonic. 2 = 𝑔 𝐿 and = √( 𝑔 𝐿 ) 𝑇 = 2𝜋 𝜔 for SHM ∴ 𝑇 = √( 𝐿 𝑔 ) for pendulum Periodic time and frequency of a simple pendulum is independent of the mass of the pendulum and the angle oscillation (if this angle exceeds 120) . If is the angular speed line generating SHM . angular speed of the pendulum = (2 - 2 ) and maximum = angular acceleration of the pendulum = 2 and maximum = 2 Periodic time , T = 2 𝑎𝑛𝑔𝑙𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑎𝑛𝑔𝑙𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 2 (( 𝜃 𝛼 )
- 15. 15 EXAMPLE 6 The amplitude of a simple pendulum is 7 0 and the periodic time is 5s , find: i. maximum linear velocity of the pendulum weight and the maximum angular speed of the pendulum cord , ii. maximum linear acceleration of the pendulum weight and maximum angular acceleration of the pendulum cord . EXAMPLE 7 A pendulum clock required rhythm second, with periodic time 2 s , was found late 80 s a day. The pendulum is shortened so that it rhythm seconds exactly. Find the difference in the length of the pendulum clock . 2.8 SIMPLE CONE
- 16. 16 figure shows a body B of mass M kg of rotate with rad/s on the vertical axis A-A and the angle is assumed small. The forces acting on the body is shown in the figure. Tan = Mr2 = r2 Mg g But, tan ≅ 𝜃 (Because 𝜃 is too small) = r2 g A P A
- 17. 17 And from space diagram, Sin = 𝑟 𝐿 = 𝑟 𝐿 ∴ r2 = r and 𝜔 = √( 𝑔 𝐿 ) g L This is the minimum value of for a conical pendulum , and the value is equal to the value for a simple pendulum and the pendulum is in equilibrium. From the force triangle Tan = Mr2 = r2 Mg g From the space diagram Tan = r / h ∴ r2 = r and = (g/h) g h
- 18. 18 Periodic time , T = 2𝜋 𝜔 for ∴ 𝑇 = 2𝜋√( ℎ 𝑔 ) to get the tensile cord P, from triangle of forces, sin = 𝑀𝑟𝜔 𝑃 2 and from spacediagram, sin = 𝑟 𝐿 ∴ 𝑟 𝐿 = 𝑀𝑟𝜔 𝑃 2 P = ML2 = 𝑀𝑔𝐿 ℎ EXAMPLE 8 Cord length a conical pendulum is 200 mm and weight is 2.4 kg. Find the rotation of the pendulum at the moment leads to upward from its equilibrium position. If the cone pendulum is rotating with a 88.8 rpm, determine: i. Periodic time , ii. high-pendulum, iii. the tension of the cord. EXAMPLE 9
- 19. 19 Maximum permissible tension of the cord used for a conical pendulum is 7 times heavier weight . If the cord length is 1.2 m , find: i. angular speed in revolutions per minute , ii. high- pendulum, iii. the change of high pendulum , when the angular speed dropped to 20% .

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