Physics 203 – College Physics I Department of Physics – The Citadel Physics 203 College Physics I Fall 2012 S. A. Yost Chapter 11 Simple Harmonic Motion

Physics 203 – College Physics I Department of Physics – The Citadel Announcements Problem set 10B is due Thursday. Read sections 1 – 4 and 7 – 9 if you haven’t Topics: simple harmonic motion, intro to waves. Next Tuesday: Ch. 11, sec. 7 – 9 and 11 – 13 with some material from Ch. 12, sec. 1 – 4 & 7 mixed in. (The topics are related.)

Physics 203 – College Physics I Department of Physics – The Citadel Clicker Question Are you here? A = Yes B = No C = Both D = Neither E = Can’t be determined

Physics 203 – College Physics I Department of Physics – The Citadel Hooke’s Law From chapter 6: Hooke’s Law describes a linear restoring force when a spring is displaced from its equilibrium position. Elastic potential energy: U = ½ kx 2 x F = -k x

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion This is called simple harmonic motion. When a mass oscillates under a linear restoring force F =  kx, the acceleration is always opposite the displacement from equilibrium, but proportional to it. a = F/m =  (k/m) x.

Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion Analogy Where else have we seen a linear restoring force? Look at uniform circular motion in components. x y m r v

Physics 203 – College Physics I Department of Physics – The Citadel m Circular Motion Analogy The acceleration vector always points toward the center and has magnitude a = v 2 /r v = r  a=  2 r In vectors: x y r v a a =   2 r → → → →

Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion Analogy The components of the acceleration vector are a x =  2 x a y =  2 y The acceleration in each direction is proportional to the displacement. m r a x y a =   2 r → → →

Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion Analogy Just focus on the x component: a =  2 x This is simple harmonic motion, if we match the Hooke’s Law condition: a = F/m =  (k/m) x. Identify the angular velocity for SHM as  =  √ k/m m r a x y → → ___

Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion Analogy Circular motion x = r cos  = r cos  t The maximum value of x is called the amplitude. In SHM, it is usually written as A : x = A cos  t m r a x y → → 

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion General feature of simple harmonic motion: The frequency doesn’t depend on the amplitude. No matter how far you displace the object from equilibrium,  = √ k/m

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion x = A cos  t with angular frequency  = √ k/m A 0 T t x T = 2  /  is the period of the motion. f = 1/T =  /2  is the frequency of the motion. Units: 1/s = Hz. 

Physics 203 – College Physics I Department of Physics – The Citadel Circular Motion Analogy The velocity vector in uniform circular motion has magnitude v = r  and is perpendicular to the radius vector: v x =  r  sin  v y =  r  cos  t For SHM, take the x component, with r →A: m r x y v   → → v =  A  sin  t

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion A 0t x Note that v = 0 at the turning points x = ± A. The maximum speed is v max = A , where x = 0.  v max v = 0 v =  A  sin  t x = A cos  t

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion A 0t x Note that a = 0 at the equilibrium points x = 0. The maximum acceleration is a max =  2 A  a = 0 a =   2 A a =  2 A Acceleration: a = F/m =  k x/m =    A cos  t

Physics 203 – College Physics I Department of Physics – The Citadel Simple Harmonic Motion Energy is conserved: K = ½ mv 2, U = ½ kx 2. Total energy: E = K + U = ½ kA 2. The potential energy is maximum at the turning points. The kinetic energy is maximum at the equilibrium position. K max = ½ mv max 2 = ½ kA 2 → v max = A .

Physics 203 – College Physics I Department of Physics – The Citadel Example A 3.0 kg object is attached to a spring with k = 280 N/m and is oscillating in SHM. When it is 2.0 cm from equilibrium, it moves 0.55 m/s. What is the frequency of the motion?  = √k/m = √(280 N/m) / 3.0 m = 9.66 s -1 f =  Hz ___ ______________

Physics 203 – College Physics I Department of Physics – The Citadel Example A 3.0 kg object is attached to a spring with k = 280 N/m and is oscillating in SHM. When it is 2.0 cm from equilibrium, it moves 0.55 m/s. What is the amplitude of the motion? K = ½ mv 2, U = ½ kx 2. Total energy: E = K + U = ½ kA 2 = ½ mv 2 + ½ kx 2 A 2 = (m/k) v 2 + x 2 = (v/  ) 2 + x 2 = [(0.55 m/s)/(9.66 s -1 )] 2 + (0.020 m) 2 = m 2 A = m = 6.0 cm.

Physics 203 – College Physics I Department of Physics – The Citadel Example A 3.0 kg object is attached to a spring with k = 280 N/m and is oscillating in SHM. When it is 2.0 cm from equilibrium, it moves 0.55 m/s. What is the maximum force on the object? F = kA = (280 N/m) (0.060 m) = 17 N. What is its maximum speed? v max = A  = (0.060 m) (9.66 s -1 ) = 0.58 m/s