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PHYSICS 231 Lecture 34: Oscillations & Waves

PHYSICS 231 Lecture 34: Oscillations & Waves . 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 . Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom. Harmonic oscillations vs circular motion. v 0. t=0. t=1.

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PHYSICS 231 Lecture 34: Oscillations & Waves

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  1. PHYSICS 231Lecture 34: Oscillations & Waves Period T 63 2 Frequency f1/61/3 ½ (m/k)6/(2)3/(2) 2/(2)  (2)/6 (2)/3(2)/2 Remco Zegers Question hours: Thursday 12:00-13:00 & 17:15-18:15 Helproom PHY 231

  2. Harmonic oscillations vs circular motion v0 t=0 t=1 t=2 v0=r=A =t =t t=3 t=4 v0  vx  A PHY 231

  3. velocity v kA/m -kA/m A xharmonic(t)=Acos(t) x time (s) -A =2f=2/T=(k/m) A(k/m) vharmonic(t)=-Asin(t) -A(k/m) aharmonic(t)=-2Acos(t) a 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 (<150) radians!!! F=-mg also =s/L so: F=-(mg/L)s PHY 231

  5. pendulum vs spring * 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: • The mass is hung higher • The mass is replaced by a heavier mass • The clock is brought to the moon • The clock is put in an upward accelerating • elevator? PHY 231

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

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

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

  12. A single pulse velocity v time to time t1 x0 x1 v=(x1-x0)/(t1-t0) PHY 231

  13. describing a traveling wave : wavelength distance between two maxima. While the wave has traveled one wavelength, each point on the rope has made one period of oscillation. v=x/t=/T= f 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: • amplitude • period • frequency 2m • amplitude: difference between maximum (or minimum) • and the equilibrium position in the vertical direction • (transversal!) A=2m/2=1m • v=1m/s, =2*2m=4m T=/v=4/1=4s • f=1/T=0.25 Hz 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 PHY 231

  16. Speed of waves on a string F tension in the string  mass of the string per unit length (meter) screw tension T example: violin L M v= /T= f=(F/) so f=(1/)(F/) for fixed wavelength the frequency will go up (higher tone) if the tension is increased. 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. • What is the mass per unit length? • What is v if M=2.0 kg? a) Tension F=mg=3*9.8=29.4 N v=(F/) so =F/v2=0.05 kg/m b) v=(F/)=(2*9.8/0.05)=19.8 m/s 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 Ffriction<Fmovement sn-maP=0 smg=maP so sg=aP ap=-2Acos(t) so maximally 2A=2fA sg=2fA A= sg/2f=0.62 m PHY 231

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