PHYS 1443 – Section 003 Lecture #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 WaveMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu1PHYS 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 Yu2Announcements•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 Yu3Simple 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 , x and y coordinate position becomeSince the linear velocity in a uniform circular motion is A, the velocity components arext=0xyOPAxyOPAQxyt=t=t+xyOPAQvvxxvxyOPAQaaxSince the radial acceleration in a uniform circular motion is v2/A=2, the components arexacosA tAcosysinA tAsinsinv tA sinyvcosv tA coscosa tA cos2yasina tA sin2Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu4Example 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 isSince x=2.00, when t=0However, since the particle was moving to the right =-48.2o, Using the displcementx( )2.00 3.00 cos ;m f= 2.4800.8cos00.3 tmxFind the x components of the particle’s velocity and acceleration at any time t.xvLikewise, from velocityxacosA tm 00.8cos00.32.4800.300.2cos1dtdx 2.4800.8sin/0.242.4800.8sin00.800.3 tsmtdtdv 2.4 800.8cos/1922.4800.8cos00.80.242 tsmtMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu5Damped OscillationMore realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.xFThe angular frequency for this motion isThe solution for the above 2nd order differential equation is xWe express the angular frequency asThis 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.Let’s consider a system whose retarding force is air resistance R=-bv (b is called damping coefficient) and restoration force is -kxmk0Where as the natural frequency 0bvkx xmadtdxbkx 22dtxdm tAcostmbe222mbmk2220mbDamping TermMonday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu6More on Damped OscillationkAbv maxAs the retarding force becomes larger, the amplitude reduces more rapidly, eventually stopping at its equilibrium positionThe motion is called Underdamped when the magnitude of the maximum retarding force Rmax = bvmax <kAThe system is Critically dampedHow do you think the damping motion would change as retarding force changes?0Under what condition this system does not oscillate?If the retarding force is larger than restoration forcekAbvR maxmaxThe system is OverdampedWhat do you think happen?Once released from non-equilibrium position, the object would return to its equilibrium position and stops.Once released from non-equilibrium position, the object would return to its equilibrium position and stops, but a lot slower than beforemb20mkmb 22 0Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu7Waves•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 Yu8Speed 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.The speed of the wave increases.So what happens when the tension increases? Which law does this hypothesis based on?Based on the hypothesis we have laid out above, we can construct a hypothetical formula for the speed of wave For the given tension, acceleration decreases, so the wave speed decreases.Newton’s second law of motionThe acceleration of the particular segment increases Tv Which means? Now what happens when the mass per unit length of the string increases? T: Tension on the string: Unit mass per lengthIs the above expression dimensionally sound?T=[MLT-2], =[ML-1](T/)1/2=[L2T-2]1/2=[LT-1]Monday, Dec. 1, 2003 PHYS 1443-003, Fall 2003Dr. Jaehoon Yu9Speed 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 What is the mass of the segment when the line density of the string is ?Using the radial force componentNow what do the force components look in this motion when is small? TTFrO svRRvar2tFmrFTherefore the speed of the
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