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Periodic Motion - 1. Oscillatory/Periodic Motion Repetitive Motion Simple Harmonic Motion SHM special case of oscillatory. Mass on spring Pendulum with low amplitude. Vocabulary. Period T : time required for one cycle of periodic motion (sec).

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## Periodic Motion - 1

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**Oscillatory/Periodic MotionRepetitive MotionSimple Harmonic**Motion SHMspecial case of oscillatory. • Mass on spring • Pendulum with low amplitude.**Vocabulary**• Period T: time required for one cycle of periodic motion (sec). • Frequency: number of oscillations per unit time. unit is Hertz:**Equilibrium (0): the spot the mass would come to rest when**not disturbed – Fnet = 0. • Displacement: (x) distance from equilibrium. • Amplitude (xo) – max displacement from eq. • Angular frequency w, multiply f by 2p. Reported as cy/sec or rad/sec. 2p = 1 full cycle.**Angular frequency w is a measure that comes from circular**motion. • It is measured in radians per second. • f, Hz is cycles per second.**An object makes 1 revolution per second. How many radians**does it complete in 1 second? • 2p rad s-1 = 1 revolution per second.**2p rad s-1 = also describes 1 non-circular cycle or**oscillation.**To find w, from frequency f, or period T, use:w = 2pf**or w = 2p T**Ex 1. A pendulum completes 4 swings in 0.5 seconds. What is**its angular frequency? • T = 0.125 s w = 2p/T = 50 rad/sec.**Free Body Diagram Mass on Spring**• Complete sheet.**Simple Harmonic Motion**A spring exerts a restoring force proportional to the displacement fr. eq: Remember Hooke’s Law Negative sign – x opposite Fnet.**Simple Harmonic Motion2 Conditions.**• 1. Acceleration/Fnet proportional to displacement. • 2. Acceleration/Fnet directed toward equilibrium.**Simple Harmonic Motion On a bobbing mass:**• No friction • Weight stays constant. • Tension increases with stretch. • Fnet / accl toward equilibrium position. • Fnet increases with is proportional to displacement.**Graphical Treatment**Equations of SHM**Displacement, x, against time**x = xo cos wt start point at max ampl. ** Set Calculator in Radians.**Displacement against time**x = xosinwt**Velocity against time v = vocoswt**Midpoint = max velocity. Starting where?**Equations of Graphs**• x = xocoswt x = xo sin wt • v = -vo sin wt v = vocoswt • a = -aocoswt -aosin wt • Released from Released equilibrium. top.**Ex 2. A mass on a spring is oscillating with f = 0.2 Hz and**xo = 3 cm. What is the displacement of the mass 10.66 s after its release from the top? • x = xo cos wt xo = 3 cm • w = 2pf. = 0.4 p Hz =1.26 rad/s. • t = 10.66 s • x = 0.03 cos (1.26 x 10.66) = 0.019 m • You must use radians on calculator.**Mechanical Oscillations**• A mechanical oscillation is a periodic conversion of E from P.E. to K.E to P.E. etc. • If a body oscillates it must be acted on by a force directed toward the equilibrium position. This force is called the restoring force. • Hooke’s law expresses the relationship btw F & x. • SHM is the simplest type of oscillatory motion.**SHM: body’s a, & restoring force are directly**proportional to its displacement from equilibrium & is directed toward that point.**Simple Harmonic Motion**A spring exerts a restoring force proportional to the displacement equilibrium:**SHM is related to Circular Motion**• Use the relationship to derive equations.**If an object moving with constant speed in a circular path**is observed from a distant point (in the plane of the motion), it will appear to be oscillating with SHM. The shadow of a pendulum bob moves with s.h.m. when the pendulum itself is either oscillating (through a small angle) or moving in a circle with constant speed, as shown in the diagram.**For any s.h.m. we can find a corresponding circular motion.**When a circular motion "corresponds to" a given s.h.m., i) the radius of the circle is equal to the amplitude of the s.h.m. ii) the time period of the circular motion is equal to the time period of the s.h.m.**Derive Relationship between accl & w for SHM.**From circular motion ac = v2/r Oscillating systems have acceleration too. • But w = 2p/T • vc = 2pr/T • vc = wr • But ac = v2/r • So ac = (wr)2/r**a = -w²x**The negative sign shows Fnet & accl direction opposite displacement. Know the Derivation of accl in Hamper pg 76. Or use 2nd derivative of displacement.**Ex 3. A pendulum swings with f = 0.5 Hz. What is the size &**direction of the acceleration when the bob has displacement of 2 cm right? • a = -w²x • w = 2pf = p • a = -(p)2 (0.02 m) = -0.197 m/s2. left.**Ex: A mass is bobbing on a spring with a period of 0.20**seconds. What is its angular acceleration at a point where its displacement is 1.5 cm? • w = 2p/T • .w = 31 rad/s • a = -w²x • a = (31rad/s)(1.5 cm) = 1480 cm/s2. • 15 m/s2.**For a mass undergoing SHM on a spring, what is the relation**between angular frequency w, and k the spring constant? • Use Hooke’s law and make substitutions to derive a relation in terms of angular frequency, k, and mass.**F= - kx.**ma = - k x. So a = -k x m a = -w² x So w2 = -k/m**To find the velocity of an oscillating mass or pendulum at**any displacement: When the mass is at equilibrium, x = 0, and velocity is maximum: vo = ± wxo. Derivation on H pg 77.**Ex 4. A pendulum swings with f = 1 Hz and amplitude 3 cm. At**what position will be its maximum velocity &what is the velocity? At max velocity vo = wxo. w = 2pf = 2p(1) = 2p rad/s vo = (2prad/s)(0.03) vo = 0.188 m/s vo = 0.2 m/s**Hwk Hamper pg 75- 77 Show equations and work, hand in**virtual solar system lab. Mechanical Universe w/questions http://www.learner.org/resources/series42.html?pop=yes&pid=565**Units of Chapter 4**• The Pendulum • Damped Oscillations • Driven Oscillations and Resonance**The Period of a Mass on a Spring**Since the force on a mass on a spring is proportional to the displacement, and also to the acceleration, we find that . Make substitutions to find the relationship between T and k.**The Period of a Mass on a Spring**Therefore, the period is How does T change as mass increases? Sketch it!**Energy Conservation in Oscillatory Motion**In an ideal system the total mechanical energy is conserved. A mass on a spring: Horizontal mass no PEg.**Determining the max KE & PE:**• PE = ½ k x2 for a stretched spring. • So PEmax = PE = ½ k xo • KE = ½ mv2 • vmax = wxo, • KEmax = ½ m(w2xo 2),**At any point:**• KE = ½ mw2 (xo2 - x2 ) • How could you determine PE from Etot? • Subtract KE from Etot.**Since the total E will always equal the max KE (or PE), we**can calculate the number of Joules of total E from the KE equation: ET = ½ mw2xo2**Ex 5: A 200-g pendulum bob is oscillating with Amplitude = 3**cm, and f = 0.5 Hz. How much KE will it have as it passes through the origin? • KEmax = ½ w2xo 2, • xo = 0.03 m • w = p. • KE = 8.9 x 10-4 J.**Energy Conservation in Oscillatory Motion**The total energy is constant; as the kinetic energy increases, the potential energy decreases, and vice versa.**Energy Conservation in Oscillatory Motion**The E transforms from potential to kinetic & back, the total energy remains the same.**The Pendulum**A simple pendulum consists of a mass m (of negligible size) suspended by a string or rod of length L (and negligible mass). The angle it makes with the vertical varies with time as a sine or cosine.**The Pendulum**The restoring force is actually proportional to sinθ,whereas the restoring force for a spring is proportional to the displacement x.

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