1. PHY 231 1 PHYSICS 231 Lecture 34: Oscillations & Waves Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom Period T 6 3 2 Frequency f 1/6 1/3 ½  (m/k) 6/(2  ) 3/(2  ) 2/(2  )  (2  )/6 (2  )/3 (2  )/2

2. PHY 231 2 Harmonic oscillations vs circular motion t=0 t=1t=2 t=3 t=4 v 0 =  r=  A v0v0 =t=t =t=t A  v0v0 vxvx 

3. PHY 231 3 time (s) A -A -kA/m kA/m velocity v a x A  (k/m) -A  (k/m) x harmonic (t)=Acos(  t) v harmonic (t)=-  Asin(  t) a harmonic (t)=-  2 Acos(  t)  =2  f=2  /T=  (k/m)

4. PHY 231 4 Another simple harmonic oscillation: the pendulum Restoring force: F=-mgsin  The force pushes the mass m back to the central position. sin  if  is small (<15 0 ) radians!!! F=-mg  also  =s/L so: F=-(mg/L)s

5. PHY 231 5 pendulum vs spring parameterspringpendulum restoring force F F=-kxF=-(mg/L)s period TT=2  (m/k) T=2  (L/g) * frequency ff=  (k/m)/(2  )f=  (g/L)/(2  ) angular frequency  =  (k/m)  =  (g/L) *

6. PHY 231 6 example: a pendulum clock The machinery in a pendulum clock is kept in motion by the swinging pendulum. Does the clock run faster, at the same speed, or slower if: a)The mass is hung higher b)The mass is replaced by a heavier mass c)The clock is brought to the moon d)The clock is put in an upward accelerating elevator? LL mm moonelevator faster same slower

7. PHY 231 7 example: the height of the lecture room demo

8. PHY 231 8 damped oscillations In real life, almost all oscillations eventually stop due to frictional forces. The oscillation is damped. We can also damp the oscillation on purpose.

9. PHY 231 9 Types of damping No damping sine curve Under damping sine curve with decreasing amplitude Critical damping Only one oscillations Over damping Never goes through zero

10. PHY 231 10 Waves The wave carries the disturbance, but not the water Each point makes a simple harmonic vertical oscillation position x position y

11. PHY 231 11 Types of waves Transversal: movement is perpendicular to the wave motion wave oscillation Longitudinal: movement is in the direction of the wave motion oscillation

12. PHY 231 12 A single pulse velocity v time t o time t 1 x0x0 x1x1 v=(x 1 -x 0 )/(t 1 -t 0 )

13. PHY 231 13 describing a traveling wave While the wave has traveled one wavelength, each point on the rope has made one period of oscillation. v=  x/  t= /T= f : wavelength distance between two maxima.

14. PHY 231 14 example 2m A traveling wave is seen to have a horizontal distance of 2m between a maximum and the nearest minimum and vertical height of 2m. If it moves with 1m/s, what is its: a)amplitude b)period c)frequency 2m a)amplitude: difference between maximum (or minimum) and the equilibrium position in the vertical direction (transversal!) A=2m/2=1m b)v=1m/s, = 2*2m=4m T= /v=4/1=4s c)f=1/T=0.25 Hz

15. PHY 231 15 sea waves An anchored fishing boat is going up and down with the waves. It reaches a maximum height every 5 seconds and a person on the boat sees that while reaching a maximum, the previous waves has moves about 40 m away from the boat. What is the speed of the traveling waves? Period: 5 seconds (time between reaching two maxima) Wavelength: 40 m v= /T=40/5=8 m/s

16. PHY 231 16 Speed of waves on a string F tension in the string  mass of the string per unit length (meter) example: violin LM screw tension T v= /T= f=  (F/  ) so f=(1/ )  (F/  ) for fixed wavelength the frequency will go up (higher tone) if the tension is increased.

17. PHY 231 17 example A wave is traveling through the wire with v=24 m/s when the suspended mass M is 3.0 kg. a)What is the mass per unit length? b)What is v if M=2.0 kg? a) Tension F=mg=3*9.8=29.4 N v=  (F/  ) so  =F/v 2 =0.05 kg/m b) v=  (F/  )=  (2*9.8/0.05)=19.8 m/s

18. PHY 231 18 bonus ;-) The block P carries out a simple harmonic motion with f=1.5Hz Block B rests on it and the surface has a coefficient of static friction  s =0.60. For what amplitude of the motion does block B slip? The block starts to slip if F friction <F movement  s n-ma P =0  s mg=ma P so  s g=a P a p = -  2 Acos(  t) so maximally  2 A=2  fA  s g=2  fA A=  s g/2  f=0.62 m