PHYS 1443 – Section 003Lecture #23AnnouncementsSimple Harmonic and Uniform Circular MotionsExample for Uniform Circular MotionDamped OscillationMore on Damped OscillationWavesSpeed of Transverse Waves on StringsSpeed of Waves on Strings cont’dExample for Traveling WavePHYS 1443 – Section 003Lecture #23Monday, Dec. 1, 2003Dr. Jaehoon Yu1. Simple Harmonic Motion and Uniform Circular Motion2. Damped Oscillation3. Waves4. Speed of Waves5. Sinusoidal Waves6. Rate of Wave Energy Transfer7. Superposition and Interference8. Reflection and TransmissionMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1Announcements• Homework # 12– Due at 5pm, Friday, Dec. 5• The final exam– On Monday, Dec. 8, 11am – 12:30pm in SH103.– Covers: Chap. 10 not covered in Term #2 – Ch15.• Need to talk to me? I will be around this week.Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu2Simple Harmonic and Uniform Circular MotionsUniform circular motion can be understood as a superposition of two simple harmonic motions in x and y axis.When the particle rotates at a uniform angular speed ω, x and y coordinate position becomext=0Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu3Since the linear velocity in a uniform circular motion is Aω, the velocity components arexyOPφAxyOPθAQωxyt=tθ=ωt+φxyOPθAQvvxxvxyOPθAQaaxθcosA=()φω+=tA cosyθsinA=()φω+=tAsinθsinv−=()φωω+−=tA sinyvθcosv+=()φωω+=tA cosxaθcosa−=()φωω+−= tA cos2yaθsina−=()φωω+−= tA sin2Since the radial acceleration in a uniform circular motion is v2/A=ω2Α, the components areExample for Uniform Circular MotionA particle rotates counterclockwise in a circle of radius 3.00m with a constant angular speed of 8.00 rad/s. At t=0, the particle has an x coordinate of 2.00m and is moving to the right. A) Determine the x coordinate as a function of time.Since the radius is 3.00m, the amplitude of oscillation in x direction is 3.00m. And the angular frequency is 8.00rad/s. Therefore the equation of motion in x direction isxMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Since x=2.00, when t=0However, since the particle was moving to the right φ=-48.2o, ()2.00 3.00 cos ;mφ=()()D2.4800.8cos00.3 −= tmxFind the x components of the particle’s velocity and acceleration at any time t.θcosA=()()φ+=tm 00.8cos00.3D2.4800.300.2cos1=⎟⎠⎞⎜⎝⎛=−φdtdx=()()()()D2.4800.8sin/0.242.4800.8sin00.800.3 −−=−⋅−= tsmtUsing the displcementxvLikewise, from velocitydtdv=()()()()D2.4800.8cos/1922.4800.8cos00.80.242−−=−⋅−= tsmtxaDamped OscillationMore realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.∑xFMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5The angular frequency ω for this motion isThe solution for the above 2ndorder differential equation is =xLet’s consider a system whose retarding force is air resistance R=-bv (b is called damping coefficient) and restoration force is -kxωbvkx−−=xma=dtdxbkx −−22dtxdm=()φω+tAcostmbe2−22⎟⎠⎞⎜⎝⎛−=mbmkDamping TermThis equation of motion tells us that when the retarding force is much smaller than restoration force, the system oscillates but the amplitude decreases, and ultimately, the oscillation stops.We express the angular frequency asωmk=0ω22⎟⎠⎞⎜⎝⎛−=20mbωWhere as the natural frequency ω0More on Damped OscillationThe motion is called Underdamped when the magnitude of the maximum retarding force Rmax= bvmax<kAHow do you think the damping motion would change as retarding force changes?As the retarding force becomes larger, the amplitude reduces more rapidly, eventually stopping at its equilibrium positionkAbv −→−maxMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu6The system is Critically damped0=ωmb2=0ωmkmb 22 ==0ωUnder what condition this system does not oscillate?If the retarding force is larger than restoration forcekAbvR >=maxmaxOnce released from non-equilibrium position, the object would return to its equilibrium position and stops.What do you think happen?The system is OverdampedOnce released from non-equilibrium position, the object would return to its equilibrium position and stops, but a lot slower than beforeWaves• Waves do not move medium rather carry energy from one place to another• Two forms of waves–Pulse– Continuous or periodic wave• Wave can be characterized by– Amplitude– Wave length–Period• Two types of waves– Transverse Wave– Longitudinal wave• Sound waveMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Speed of Transverse Waves on StringsHow do we determine the speed of a transverse pulse traveling on a string?If a string under tension is pulled sideways and released, the tension is responsible for accelerating a particular segment of the string back to the equilibrium position.So what happens when the tension increases? Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu8The speed of the wave increases.The acceleration of the particular segment increases Which means? Now what happens when the mass per unit length of the string increases? For the given tension, acceleration decreases, so the wave speed decreases.Which law does this hypothesis based on?Newton’s second law of motionBased on the hypothesis we have laid out above, we can construct a hypothetical formula for the speed of wave µTv =T: Tension on the stringµ: Unit mass per lengthIs the above expression dimensionally sound?T=[MLT-2], µ=[ML-1](T/µ)1/2=[L2T-2]1/2=[LT-1]Speed of Waves on Strings cont’dLet’s consider a pulse moving right and look at it in the frame that moves along with the the pulse.Since in the reference frame moves with the pulse, the segment is moving to the left with the speed v, and the centripetal acceleration of the segment is Now what do the force components look in this motion when θ is small? TTFrOθθθθ∆svRRvar2=∑tFθθcoscos TT−=0=∑rFθsin2T=θΤ2≈What is the mass of the segment when the line density of the string is µ?ms∆=µθµ2R=θµR2=Using the radial force componentma=Rvm2=RvR22θµ=θT2=∑rFµTv =Therefore the speed of the pulse isMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9Example for Traveling WaveA uniform cord has a mass of 0.300kg and a length of 6.00m. The cord passes over a pulley and supports a 2.00kg object. Find the speed of a
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