Physics 203 College Physics I Fall 2012

1. Physics 203College Physics IFall 2012 S. A. Yost Chapter 11 Simple Harmonic Motion

2. 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.)

3. Clicker Question • Are you here? • A = Yes • B = No • C = Both • D = Neither • E = Can’t be determined

4. 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 kx2 x F = -k x

5. 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. This is called simple harmonic motion.

6. Circular Motion Analogy • Where else have we seen a linear restoring force? • Look at uniform circular motion in components. y v m r x

7. Circular Motion Analogy • The acceleration vector always points toward the center and has magnitude • a = v2/r • v = r w • a= w2r • In vectors: y → v → a m → r x →→ a = -w2r

8. →→ a = -w2r Circular Motion Analogy • The components of the acceleration vector • are • ax= - w2x • ay= - w2 y • The acceleration in each direction is proportional to the displacement. y → a m → r x

9. Circular Motion Analogy • Just focus on the x component: • a = - w2 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 • w =√k/m y → a m → r x ___

10. Circular Motion Analogy • Circular motion • x = rcosq = r coswt • The maximum value of x is called the amplitude. • In SHM, it is usually written as A: • x = A coswt y → a m → r q x

11. 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, w = √ k/m

12. Simple Harmonic Motion x = A coswtwith angular frequency w = √ k/m T= 2p/wis the period of the motion. f = 1/T = w/2pis the frequency of the motion. Units: 1/s = Hz. x A 0 t T -A

13. Circular Motion Analogy • The velocity vector in uniform circular motion has magnitude v = rwand is perpendicular to the radius vector: • vx= -r wsinq • vy = -r wcosq, q = wt • For SHM, take the x component, with r →A: y v → q m → r q x v = - A wsinwt

14. Simple Harmonic Motion x= A coswt v = - A wsinwt Note that v = 0at the turning points x = ± A. The maximum speed is vmax = Aw, where x = 0. v = 0 x A 0 t vmax -A

15. Simple Harmonic Motion Acceleration: a = F/m = - k x/m = - w 2A coswt Note that a = 0at the equilibrium points x = 0. The maximum acceleration is amax= w2A a = -w2A x A 0 t a = 0 -A a = +w2A

16. Simple Harmonic Motion • Energy is conserved: • K = ½ mv2, U = ½ kx2. • Total energy: E = K + U = ½ kA2. • The potential energy is maximum at the • turning points. • The kinetic energy is maximum at the • equilibrium position. • Kmax = ½ mvmax2 = ½ kA2 → vmax = Aw.

17. 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? w = √k/m = √(280 N/m) / 3.0 m = 9.66 s-1 f = w/2p = 1.5 Hz ___ ______________

18. 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 mv2, U = ½ kx2. Total energy: E = K + U = ½ kA2 = ½ mv2 +½ kx2 A2 = (m/k) v2 +x2 = (v/w)2+ x2 = [(0.55 m/s)/(9.66 s-1)]2 + (0.020 m)2 = 0.00364 m2 A = 0.060 m = 6.0 cm.

19. 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? vmax= Aw = (0.060 m) (9.66 s-1) = 0.58 m/s