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Simple Harmonic Motion

Simple Harmonic Motion. Periodic Motion :. any motion which repeats itself at regular, equal intervals of time. Link to SHM Applet. Simple Harmonic Motion. It is actually anything but simple!. Simple harmonic motion is a special case of periodic motion.

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Simple Harmonic Motion

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  1. Simple Harmonic Motion Periodic Motion: any motion which repeats itself at regular, equal intervals of time.

  2. Link to SHM Applet

  3. Simple Harmonic Motion It is actually anything but simple! Simple harmonic motion is a special case of periodic motion

  4. Characteristics of simple harmonic motion: • the displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), • Velocity is maximum when displacement is zero, • Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, • The period does not depend on the amplitude. • It is periodic oscillatory motion about a central equilibrium point, A Positive amplitude Equilibrium position O Negative amplitude B

  5. Characteristics of simple harmonic motion: • It is periodic oscillatory motion about a central equilibrium point, • The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), • Velocity is maximum when displacement is zero, • Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, • The period does not depend on the amplitude. A A Displacement O A Positive amplitude Time Equilibrium position O B B Negative amplitude B

  6. Characteristics of simple harmonic motion: • It is periodic oscillatory motion about a central equilibrium point, • The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), • Velocity is maximum when displacement is zero,(equilibrium point) • Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, • The period does not depend on the amplitude. O A B Velocity Time O A A Displacement O Time B B remember velocity is the gradient of displacement A Positive amplitude Equilibrium position O Negative amplitude B

  7. Characteristics of simple harmonic motion: • It is periodic oscillatory motion about a central equilibrium point, • The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), • Velocity is maximum when displacement is zero, • Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, • The period does not depend on the amplitude. O B A B Velocity Acceleration O Time Time O A A A Displacement O A Positive amplitude Time Equilibrium position O B B Negative amplitude B remember acceleration is the gradient of velocity

  8. Characteristics of simple harmonic motion: • It is periodic oscillatory motion about a central equilibrium point, • The displacement is a sinusoidal function of time, it ranges from zero to a maximum displacement (amplitude), • Velocity is maximum when displacement is zero, • Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction, • The period does not depend • on the amplitude. O B A B Velocity Acceleration O Time Time O A Displacement A Positive amplitude Equilibrium position O Negative amplitude B

  9. Simple Harmonic Motion • Equilibrium: the position at which no net force acts on the particle. • Displacement: The distance of the particle from its equilibrium position. Usually denoted as x(t) with x=0 as the equilibrium position. • Amplitude: the maximum value of the displacement with out regard to sign. Denoted as xmaxor A.

  10. What is the equation of this graph? Hint HIVO, HOVIS A cos graph takes 2 radians to complete a cycle

  11. Relation between Linear SHM and Circular Motion A A Displacement O Time A B O B B A B

  12. O B A B Velocity Acceleration O Time Time O A Relation between Linear SHM and Circular Motion A A Displacement O Time O A B B B A B

  13. T = periodic time = the time to complete 1 cycle O A B xo A A xo Displacement O Time A B B B At A, t = 0 the displacement is maximum positive x = xo

  14. T = periodic time = the time to complete 1 cycle A O B xo A A Displacement O Time A B B B At O, t = the displacement is zero x = 0

  15. T = periodic time = the time to complete 1 cycle A O B -xo xo A A Displacement O Time A B B B At B, t = the displacement is maximum negative x = -xo

  16. T = periodic time = the time to complete 1 cycle A O B xo A A Displacement O Time A B B B At O, t = the displacement is zero x = 0

  17. T = periodic time = the time to complete 1 cycle A O B xo A xo A Displacement O Time A B B B At A, t = T the displacement is maximum positive x = xo

  18. T = periodic time = the time to complete 1 cycle O A B xo A A xo Displacement O Time A B B B At A, t=0 the displacement is maximum positive x = xo θ = 0°

  19. A O B xo A A Displacement O Time A B B B At O, t = the displacement is zero x = 0 θ= 90°

  20. T = periodic time = the time to complete 1 cycle A O B -xo xo A A Displacement O Time A B B B At B, t = the displacement is maximum negative x = -xo θ = π 180°

  21. A O B xo A A Displacement O Time A B B B At O, t = the displacement is zero x = 0 θ= 270°

  22. T = periodic time = the time to complete 1 cycle B O A xo A xo A Displacement O Time B B B A At A, t = T the displacement is maximum positive x = xo θ = 2π 360°

  23. M2 reminder w = Angular velocity = rads per sec Units

  24. T = periodic time = the time to complete 1 cycle B M O A x xo A A Displacement O Time θ B B B A x At a random point M , the displacement is x andθ= θ° since ω =  θ = ω t cos (ω t) = cosθ =  Using trigonometry x = xocos (ω t)

  25. If timing begins when x = 0 at the centre of the SHM A B A Displacement O Time B

  26. T = periodic time = the time to complete 1 cycle B M O A x xo A A Displacement O Time θ B B B A x If however timing begins when x = 0 at the centre of the SHM then the displacement equation is : ω=  θ = ω t x = xosin (θ) So x = xosin (ω t)

  27. Displacement Formulae If x = x0 when t = 0 then use x = xocos (ω t) This is usually written as x = Acos (ω t) where A is the amplitude of the motion If x = 0 when t = 0 then use x = xosin (ω t) This is usually written as x = Asin (ω t) where A is the amplitude of the motion Memory aid Sin graphs start at the origin(centre) Cos graphs start at the top(end)

  28. 0.45 B O A M x Amp A A Displacement 0.45 O Time A B B B x What is the displacement of a ball from the end undergoing SHM between points A and B starting at A where the distance between them is 1 m and the angle is rads x = Acos (ω t) when t = 0 A = 0.5 i.e the Amplitude x = 0.5cos = 0.45 m so it is 0.05m from the end

  29. Periodic Time ω = So t = To complete one cycle t = T and q = 2p T = Tom is 2 pies over weight So the time to complete one cycle isT = =

  30. What is the displacement of a ball from the centre undergoing SHM between points A and B starting at the A, after 2secs where the distance between them is 1 m and the periodic time is 6 A A Amp Displacement O Time -0.25 B x = Acos (ω t) when t = 0 A = 0.5 i.e the Amplitude x = 0.5cos2ω Memory aid Tom is 2 pies overweight T= To find ω use the fact that ω= To complete one whole cycle then q = 2p and t = 6 So ω = = = M A O B x = 0.5cos2ω = 0.5cos2 = -0.25 x -0.25

  31. What is the displacement from the of a ball going through SHM between points A and B starting at the centre, after 3secs where the distance between them is 0.8m and the periodic time is 8s A xo Displacement O B B 0.4 i.e the Amplitude x = Asin (ω t) when t = 0 A = x = 0.4sin 3ω Memory aid Tom is 2 pies overweight T= To find ω use the fact that ω= To complete one whole cycle then q = 2p and t = 8 So ω = = = M A O B x = 0.4sin 3ω =0.4sin 3 = x

  32. Calculus Memory Aid s v a Integrate Differentiate Some Vehicles Accelerate Thanks to Chloe Barnes Starving Vultures Attack

  33. O A B Velocity Time O Velocity Equation. x = Acos (ω t) Displacement since v = • -ω Asin (ω t) v = Characteristics of simple harmonic motion: Velocity is maximum when displacement is zero,

  34. O B A B Velocity Acceleration O Time Time O A Acceleration Equation. v = -ωAsin (ω t) since a = • -ω2 Acos (ω t) a = • -ω2 x a = since x = Acos (ω t) Characteristics of simple harmonic motion: Acceleration is always directed toward the equilibrium point, and is proportional to the displacement but in the opposite direction,

  35. Equation for velocity where v depends on x rather than t • v = -ωAsin (ω t) v depends on t Velocity = rate of change of displacement = • = Acceleration = rate of change of velocity = Chain Rule for acceleration in terms of displacement So acceleration =

  36. Equation for velocity where v depends on x rather than t • v = -ωAsin (ω t) v depends on t Separate the variables = ) 0 = c =

  37. Equation for velocity where v depends on x rather than t • The max velocity occurs at the centre where x = 0 • Vmax = wAmp

  38. Summary x = Acos (ωt) if x = A when t = 0 x = Asin (ωt) if x = 0 when t = 0 v = -ωAsin (ωt) if x = A when t = 0 Differentiate the displacement equation ) • Vmax = ω A a = -ω2 x • T=Tom is 2 pies over weight

  39. Stop and look at Course notes

  40. Horizontal elastic string Consider a GENERAL point P where the extension is x . Resolving 0 – T = m but – = m so = Comparing with the S.H.M eqn= –w2xw2 = T l +ve x x T = for an elastic string

  41. Ex.1 String l = 0.8m l= 20N m = 0.5kg Bob pulled so that extension = 0.2m. Find time for which string taught b) Time for 1 oscillation c) Velocity when ext. = 0.1m T l +ve x x

  42. l = 0.8m l= 20N m = 0.5kg Horizontal elastic string Resolving 0 – T = mbut T = for an elastic string – = 0 so = Comparing with the S.H.M eqn= –w2x T l +ve x x w2 = 50 Hence w= 50 As the string goes slack at the natural length it then travels with constant velocity before the string goes taught again.

  43. This is the periodic time when the string is tight a) b) Time for which initially taught is a Tso t = As the initial extension is 0.2 m then Amp = 0.2m Velocity when string goes slack v = wAmpas x = 0 vslack = 500.2 = 1.414m/s

  44. So the distance travelled when string is slack is 40.8 So time for which slack So total time for 1 oscillation = + = 3.15secs c) Vel. when ext = 0.1 v2 = w2(Amp2 – x2) v2 = 50(0.22 – 0.12) = 1.5ms–1 v = 1.225ms–1

  45. Two Connected Horizontal Elastic Strings A and B are two fixed points on a smooth surface with AB = 2m. A string of length 1.6m and modulus 20N is stretched between A and B. A mass of 3kg is attached to a point way along the string from A and pulled sideways 9cm towards A. Show that the motion is S.H.M and find the max acceleration and max velocity. A B

  46. E N1 N2 0.2 0.8 0.8 T1 x T1 = 0.2 A B T2 As x is increasing towards A make  +vex is measured from the equilibrium position +vedirection T1– T2 = 3a - = 3

  47. E N1 N2 0.2 0.8 0.8 T1 x T1 = 0.2 A B T2 - = 3 -50x = 3 - x =

  48. E N1 N2 0.2 0.8 0.8 T1 x T1 = 0.2 A B T2 = - x 0.09 = 0.37ms-1 Max velocity = wAmp=

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