Chapter 15 Oscillations and Waves

Chapter 15 Oscillations and Waves

Chapter 15 Oscillations and Waves Oscillations and Waves • Simple Harmonic Motion • Energy in SHM • Some Oscillating Systems • Damped Oscillations • Driven Oscillations • Resonance MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 2 Simple Harmonic Motion Simple harmonic motion (SHM) occurs when the restoring force (the force directed toward a stable equilibrium point ) is proportional to the displacement from equilibrium. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 3 Characteristics of SHM • Repetitive motion through a central equilibrium point. • Symmetry of maximum displacement. • Period of each cycle is constant. • Force causing the motion is directed toward the equilibrium point (minus sign). • F directly proportional to the displacement from equilibrium. Acceleration = - ω2 x Displacement MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 4 A Simple Harmonic Oscillator (SHO) Frictionless surface The restoring force is F = −kx. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 5 Two Springs with Different Amplitudes Frictionless surface MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 6 SHO Period is Independent of the Amplitude MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 7 The Period and the Angular Frequency 2π T = . The period of oscillation is ω where ω is the angular frequency of k the oscillations, k is the spring ω = m constant and m is the mass of the block. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 8 Simple Harmonic Motion At the equilibrium point x = 0 so, a = 0 also. When the stretch is a maximum, a will be a maximum too. The velocity at the end points will be zero, and it is a maximum at the equilibrium point. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 9 Representing Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 10 Representing Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 11 Representing Simple Harmonic Motion Position - x max = A Velocity - v max = ωA 2 Acceleration - a max = ω A MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 12 A simple harmonic oscillator can be described mathematically by: x() t = Acos ωt dx v() t = = -A ωsin ωt dt dv a() t = = -A ω2cos ωt dt where A is the amplitude of the motion, the maximum Or by: displacement from equilibrium, A ω = v , and x() t = Asin ωt max Aω2 = a dx max. vt=() =A ωcos ωt dt dv a() t = = -A ω2 sin ωt dt MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 13 Linear Motion - Circular Functions MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 14 Projection of Circular Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 15 Circular Motion is the superposition of two linear SHO that are 90 0 out of phase with each other y= Asin(ω t ) x= Acos(ω t ) MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 16 Shifting Trig Functions sin The minus sign means that the x = A ω t - φ cos phase is shifted to the right. sin t A plus sign indicated the phase x=A 2 π - ϕ {}cos T is shifted to the left π x = Asinω t - 2 π π x= A() sin ωt cos2 - sin 2 cos ωt x= A() sin ωt (0) - (1)cos ωt x = -Acos ωt MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 17 Shifting Trig Functions π sinω t- =0 Shifted Trig Functions 2 1.50 sin( ωt) π 1.00 ω δ ωt- =0 sin( t- ) 2 0.50 π ωt = 0.00 2 -3.00 -2.00 -1.00 0.00 1.00 2.00 3.00 -0.50 π 1 1 T t= ; = -1.00 2ω ω 2π π T T -1.50 t = = Time 2 2 π 4 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 18 Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 19 Equation of Motion & Energy Assuming the table is frictionless: F = - kx = ma ∑ x x k 2 Classic form for SHM a() t=- xt=-()() ω x t x m 12 1 2 Also, EKt=+=()() Ut mvt() + kxt() 2 2 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 20 Spring Potential Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 21 Spring Total Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 22 Approximating Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 23 Approximating Simple Harmonic Motion MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 24 Potential and Kinetic Energy MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 25 The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point? At equilibrium x = 0: 1 1 1 E = K +U = mv 2 + kx 2 = mv 2 2 2 2 Since E = constant, at equilibrium (x = 0) the KE must be a maximum. Here v = v max = A ω. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 26 Example continued: The amplitude A is given, but ω is not. 2π 2π ω = = =12 .6 rads/sec T 0.50 s and v = Aω = (5.0 cm )(12 .6 rads/sec ) = 62 .8 cm/sec MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 27 The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of 2.0 kHz by moving back and forth with an amplitude of 1.8 ×10 −4 m at that frequency. (a) What is the maximum force acting on the diaphragm? 2 2 2 2 ∑ F = Fmax = ma max = m()Aω = mA ()2πf = 4π mAf The value is F max =1400 N. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 28 Example continued: (b) What is the mechanical energy of the diaphragm? Since mechanical energy is conserved, E = K max = U max. 1 The value of k is unknown so use K . U = kA 2 max max 2 1 1 1 1 2 2 2 2 2 K = mv Kmax = mv max = m()Aω = mA ()2πf max 2 max 2 2 2 The value is K max = 0.13 J. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 29 Example: The displacement of an object in SHM is given by: y()()()t 8= . 00cm sin1[] . 57 rads/sec t What is the frequency of the oscillations? Comparing to y(t) = A sin ωt gives A = 8.00 cm and ω = 1.57 rads/sec. The frequency is: ω1 . 75 rads/sec f = = = .0 250 Hz 2π 2π MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 30 Example continued: Other quantities can also be determined: 2π 2π The period of the motion is T = = 4= . 00 sec ω1 . 75 rads/sec xmax = A = 8.00 cm vmax = Aω = (8 .00 mc )(1.57 rads/sec ) =12 .6 cm/sec 2 2 2 amax = Aω = (8 .00 cm )(1.57 rads/sec ) =19 .7 cm/sec MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 31 What About Gravity? When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system. The effect of gravity is canceled out. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 32 Why We Ignore Gravity with Vertical Springs MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 33 The Simple Pendulum A simple pendulum is constructed by attaching a mass to a thin rod or a light string. We will also assume that the amplitude of the oscillations is small. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 34 The Simple Pendulum The pendulum is best described using polar coordinates. The origin is at the pivot point. The coordinates are (r, φ). The r-coordinate points from the origin along the rod. The φ- coordinate is perpendicular to the rod and is positive in the counter clockwise direction. MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 35 F= − mgsin φ = ma φ Apply Newton’s 2 nd ∑ φ Law to the pendulum v2 bob. ∑ Fr = T − mgcos φ = m r If we assume that φ <<1 rad, then sin φ ≈ φ and cos φ ≈1, the angular frequency of oscillations is then: ∑ Fφ =− mgsin φ = maφ = mL α −mgsin φ = mL α g α= − (g / L )sin φ ω = L α= − (g / L ) φ 2π L The period of oscillations is T = = 2π ω g MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 36 Example: A clock has a pendulum that performs one full swing every 1.0 sec. The object at the end of the string weighs 10.0 N. What is the length of the pendulum? L T = 2π g gT 2 9.8 m/s 2 ()1.0 s 2 L = = () = 0.25 m Solving for L: 4π 2 4π 2 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 37 The gravitational potential energy of a pendulum is U = mgy. Taking y = 0 at the lowest point of the swing, show that y = L(1-cos θ). θ Lcos θ L L y = L 1( − cos θ ) y=0 MFMcGraw-PHY 2425 Chap 15Ha-Oscillations-Revised 10/13/2012 38 The Physical Pendulum A physical pendulum is any rigid object that is free to oscillate about some fixed axis.

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