Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu1PHYS 1441 – Section 004Lecture #23Wednesday, Apr. 28, 2004Dr. Jaehoon Yu• Period and Sinusoidal Behavior of SHM• Pendulum• Damped Oscillation• Forced Vibrations, Resonance•WavesToday’s homework is #13 and is due 1pm, next Wednesday !Final Exam Monday, May. 10!Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu2Announcements• Final exam Monday, May 10– Time: 11:00am – 12:30pm in SH101– Chapter 8 – whatever we cover next Monday– Mixture of multiple choices and numeric problems– Will give you exercise test problems next Monday• Review next Wednesday, May 5.Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu3Sinusoidal Behavior of SHMWednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu4The Period and Sinusoidal Nature of SHMv0Consider an object moving on a circle with a constant angular speed ωIf you look at it from the side, it looks as though it is doing a SHMsinθ201xvvA⎛⎞=−⎜⎟⎝⎠0v02ATvπ=2012mv2mTkπ=112kfTmπ==Since it takes T to complete one full circular motionFrom an energy relationship in a spring SHM0kvAm=Thus, T isNatural Frequency0vv=22AxA−=21xA⎛⎞=−⎜⎟⎝⎠2ATπ=2Afπ=212kA=Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu5Example 11-5Car springs. When a family of four people with a total mass of 200kg step into their 1200kg car, the car’s springs compress 3.0cm. The spring constant of the spring is 6.5x104N/m. What is the frequency of the car after hitting the bump? Assume that the shock absorber is poor, so the car really oscillates up and down. fT2mkπ=4140020.926.5 10sπ==×1T=12kmπ=416.5101.092 1400Hzπ×==Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu6Example 11-6Spider Web. A small insect of mass 0.30 g is caught in a spider web of negligible mass. The web vibrates predominantly with a frequency of 15Hz. (a) Estimate the value of the spring constant k for the web.f(b) At what frequency would you expect the web to vibrate if an insect of mass 0.10g were trapped?fkSolve for k12kmπ=15Hz=224 mfπ=()2244 3 10 15 2.7 /Nmπ−=⋅×⋅ =12kmπ=412.7262110Hzπ−==×Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu7The SHM Equation of Motionv0The object is moving on a circle with a constant angular speed ωHow is x, its position at any given time expressed with the known quantities?cosxAθ=tθϖ=cosxA tϖ=sincecos2Aftπ=How about its velocity v at any given time? 2 fϖπ=andv0sinvθ=−()0sinvtϖ=−()0sin 2vftπ=−How about its acceleration a at any given time? 0kvAm=aFm=kxm−=()cos 2kAftmπ⎛⎞=−⎜⎟⎝⎠()0cos 2aftπ=−From Newton’s 2ndlaw0kAam=Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu8Sinusoidal Behavior of SHM()cos 2xAftπ=()0sin 2vv ftπ=−()0cos 2aa ftπ=−Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu9The Simple PendulumA simple pendulum also performs periodic motion.The net force exerted on the bob is ∑rFxLθ=Satisfies conditions for simple harmonic motion!It’s almost like Hooke’s law with.Since the arc length, x, is tFF==∑mgkL=The period for this motion isTThe period only depends on the length of the string and the gravitational accelerationAmgTθcos−=0=∑tFAmgθsin−=mgθ≈−mgxL−2mkπ=2mLmgπ= 2Lgπ=Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu10Example 11-8Grandfather clock. (a) Estimate the length of the pendulum in a grandfather clock that ticks once per second. Since the period of a simple pendulum motion isTThe length of the pendulum in terms of T is 224πgTL =Thus the length of the pendulum when T=1s is 224TgLπ=gLπ2=(b) What would be the period of the clock with a 1m long pendulum?gLπ2=T1.022.09.8sπ==219.80.254mπ×==Wednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu11Damped OscillationMore realistic oscillation where an oscillating object loses its mechanical energy in time by a retarding force such as friction or air resistance.How do you think the motion would look?Amplitude gets smaller as time goes on since its energy is spent.Types of dampingA: OverdampedB: Critically dampedC: UnderdampedWednesday, Apr. 28, 2004 PHYS 1441-004, Spring 2004Dr. Jaehoon Yu12Forced Oscillation; ResonanceWhen a vibrating system is set into motion, it oscillates with its natural frequency f0.However a system may have an external force applied to it that has its own particular frequency (f), causing forced vibration.For a forced vibration, the amplitude of vibration is found to be dependent on the different between f and f0. and is maximum when f=f0.A: light dampingB: Heavy dampingThe amplitude can be large when f=f0, as long as damping is small.This is called resonance. The natural frequency f0is also called resonant
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