2. Chapter 11: Vibrations & Waves First half of Chapter: Vibrations Second half: Waves Chapter 12: Sound waves.

3. Sect. 11-1: Simple Harmonic Motion Vibration  Oscillation = back & forth motion of an object Periodic motion: Vibration or Oscillation that regularly repeats itself. Simplest form of periodic motion: Mass m attached to ideal spring, spring constant k (Hooke’s “Law”: F = -kx) moving in one dimension (x). –Contains most features of more complicated systems  Use the mass-spring system as a prototype for periodic oscillating systems.

4. Mass-Spring System Mass-spring (no friction): Simplest form of periodic motion. Prototype for ALL vibrating systems Restoring force obeys Hooke’s “Law”: F = -kx A “simple harmonic oscillator” (SHO) Undergoing “simple harmonic motion” (SHM) Spring constant k: Depends on spring.

5. Terminology & Notation Displacement = x(t) Velocity = v(t) Acceleration = a(t)  Constant!!!!!!!  Constant acceleration equations from Ch. 2 DO NOT APPLY!!!!!!!!!!! Amplitude = A (= maximum displacement) One Cycle = One complete round trip Period = T = Time for one round trip Frequency = f = 1/T = # Cycles per second Measured in Hertz (Hz) A and T are independent!!!

6. Repeat: Mass-spring system contains most features of more complicated systems  Use mass-spring system as a prototype for periodic oscillating systems. Results we get are valid for many systems besides this prototype system. ANY vibrating system with restoring force proportional to displacement (F = -kx) will exhibit simple harmonic motion (SHM) and thus is a simple harmonic oscillator (SHO). (whether or not it is a mass-spring system!)

7. Mass-Spring System

8. Vertical Mass-Spring System

9. Sect. 11-2: Energy in the SHO A Ch. 6 result: Compressing or stretching an ideal spring a distance x gives elastic PE: PE = (½)kx 2  Total mechanical energy of SHO (mass- spring system) is conserved (a constant!) E = KE + PE = (½)mv 2 + (½)kx 2 = const –Assuming that friction is neglected!

10. Energy conservation E = (½)mv 2 + (½)kx 2 = constant –The same throughout the motion Mass is moving back & forth.  Clearly x = x(t) The mass is speeding up & slowing down.  Clearly v = v(t)  E = (½)m[v(t)] 2 + (½)k[x(t)] 2 But E = const  E(t) (independent of t!)

11. E = (½)mv 2 + (½)kx 2 = constant Conservation of KE + PE gives some properties: At maximum x: At x = A, v = 0.  E x=A = (½)k(A) 2 At maximum v: At x = 0, v = v 0  E x=0 = (½)m(v 0 ) 2 Somewhere in between (any x: -A  x  A). E x = (½)mv 2 + (½)kx 2 But E = constant  E x=A = E x=0 = E x = E

12. E = (½)mv 2 + (½)kx 2 = constant Using results that E x=A = E x=0 = E x = E : (½)k(A) 2 = (½)m(v 0 ) 2 = (½)mv 2 + (½) kx 2 Relation between maximum x (x=A) & maximum v (v = v 0 ): (½)k(A) 2 = (½)m(v 0 ) 2  v 0 =  (k/m) ½ A Velocity at any x: (½)k(A) 2 = (½)mv 2 + (½)kx 2  v =  v 0 [1 - (x 2 /A 2 )] ½ Position at any v: (½)m(v 0 ) 2 = (½)mv 2 + (½)kx 2  (using above for v 0 ): x =  A[1 - (v 2 /v 0 2 )] ½

13. Energy in Mass-Spring System

14. Examples 11-4 & 11-5 Everything holds also for a vertical mass-spring system!

15. Sect. 11-3: Period & Nature of SHM Experiment: Period T of SHO does NOT depend on amplitude A! –T depends only on m & k Derivation: Use motion in a circle & compare SHM to object rotating in circle. Will give us T and expression for x(t).

16. Circular Motion Comparison

17. Similar triangles  (v/v 0 ) = [A 2 - x 2 ] ½ /(A) or v = v 0 [1 - (x 2 /A 2 )] ½ Same as for SHM!  Projection on x-axis of motion of mass moving in a circle is the same as for a mass on a spring!

18. Period T for one revolution of circle  Period T for SHO! Circle, radius A, circumference 2πA, velocity v 0. Mass goes once around (distance 2πA) in time T  2πA = v 0 T. Or, T = (2πA)/(v 0 ) (1) From energy discussion: (½)k(A) 2 = (½)m(v 0 ) 2  v 0 = (k/m) ½ A (2) Combining (1) & (2) gives: T = 2π(m/k) ½  Period of SHO

19. Period of SHO: T = 2π(m/k) ½ –Agrees with experiment! –Independent of amplitude A Frequency f = (1/T) f = [1/(2π)](k/m) ½ Angular frequency: ω = 2πf ω = (k/m) ½

20. Sect. 11-3: x(t) Find x(t) From diagram: x = A cos(θ) Mass rotating with angular velocity ω  θ = ωt, ω = 2πf  For SHO: x = Acos( ω t) = Acos(2 π ft) = Acos[(2 π t)/(T)] At t = 0, x = A. At t = T, x = A -A  x  A since -1  cos(θ)  1

21. x = Acos(ωt) = Acos(2πft) = Acos[(2πt)/(T)] Assumes initial conditions that x = A at t = 0. Can show, if x is something else at t = 0, x(t) is still a sine or cosine function: If, at t = 0, x = 0, find: x = Asin(ωt) = Asin(2πft) = Asin[(2πt)/(T)]

22. x(t), v(t), a(t) For x = A at t = 0: x(t) = Acos(2πft) = Acos[(2πt)/(T)] –x is clearly a function of time x = x (t)! –Velocity and acceleration are ALSO functions of time: v = v(t), a = a(t). Also sinusoidal functions. v(t): We had v = v 0 [1 - (x 2 /A 2 )] ½. Put x(t) in this to get v(t): [1 - (x 2 /A 2 )] = 1 - cos 2 [(2πt)/(T)] = sin 2 [(2πt)/(T)] (Trig identity!)  v(t) = v 0 sin[(2πt)/(T)] Only for x = A at t = 0!

23. For x = A at t = 0: x(t) = Acos(2πft) = Acos[(2πt)/(T)] a(t): We had (Hooke’s “Law”): F = -kx = -kx(t) We also have Newton’s 2 nd Law: F = ma Combining gives a = a(t) = -(k/m) x(t) Using above form for x(t): a(t) = -(k/m) Acos[(2πt)/(T)] NOTE! The acceleration is a function of time and is NOT constant! a  constant !

24. For x = A at t = 0: a(t) = -(k/m) Acos[(2πt)/(T)] a = a(t)  constant ! NOTE!! This means that the 1 dimensional kinematic equations for constant acceleration (from Chapter 2) DO NOT APPLY!!!!!! That is, THROW THESE AWAY FOR THIS CHAPTER!! THESE ARE WRONG AND WILL GIVE YOU WRONG ANSWERS!!

25. Summary We’ve shown that for x = A at t = 0: x(t) = Acos[(2πt)/(T)], v(t) = Asin[(2πt)/(T)] a(t) = -(k/m) Acos[(2πt)/(T)] From conservation of energy:(½)k(A) 2 = (½) m(v 0 ) 2

26. Here: x(t) = Asin[(2πt)/(T)] v(t) = Acos[(2πt)/(T)] a(t) = -(k/m) Asin[(2πt)/(T)] x(t), v(t), a(t) : Functional forms depend on initial conditions!