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Chapter_15

Course: PHYSICS 113, Spring 2011
School: Wake Forest
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15: Chapter Oscillatory motion Reading assignment: Chapter 15.1 to 15.5; 15.6 & 15.7 cursory Homework : QQ1, QQ2, QQ3, QQ4, QQ6, OQ1, OQ3, OQ, 4, OQ5, OQ7, OQ9, OQ10, AE1, AE6, 4, 5, 8, 9, 15, 18, 19, 27, 29, 37 Due date: Tuesday, April 26 Well deal mainly with simple harmonic oscillations where the position of the object is specified by a sinusoidal function. Mass on a spring Energy...

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15: Chapter Oscillatory motion Reading assignment: Chapter 15.1 to 15.5; 15.6 & 15.7 cursory Homework : QQ1, QQ2, QQ3, QQ4, QQ6, OQ1, OQ3, OQ, 4, OQ5, OQ7, OQ9, OQ10, AE1, AE6, 4, 5, 8, 9, 15, 18, 19, 27, 29, 37 Due date: Tuesday, April 26 Well deal mainly with simple harmonic oscillations where the position of the object is specified by a sinusoidal function. Mass on a spring Energy Pendulum Simple harmonic motion/oscillation - Block attached to a spring - Motion of a swing - Motion of a pendulum (mathematical, physical) - Vibrations of a stringed musical instrument - Motion of a cantilever - Oscillations of houses, bridges, - All clocks use simple harmonic motion Piezoelectric (quartz) tuning fork from a wrist watch (In a piezoelectric material, distortion creates voltage and vice versa) Brief Aside: The Importance of Time & The Longitude Problem 1) 2) Harrisons H4 1759 Harrisons H1 1735 http://www.rog.nmm.ac.uk/museum/harrison/longprob.html 3) Determine local time through sun. Compare with time (accurate clock!) at port of departure (London). Each hour difference corresponds to 15o longitude (360o/24 hours). Simple harmonic motion/oscillation Restoring force: F = - kx Acceleration: k a = x m Acceleration and restoring force: - proportional to x - directed toward the equilibrium position Simple harmonic motion/oscillation An object moves with simple harmonic motion whenever its acceleration is proportional to its displacement from some equilibrium position and is oppositely directed. a = 2 x d 2x 2 = x 2 dt Solution to this 2nd order differential equation is: x(t ) = A cos(t + ) or: x(t ) = A sin(t + ) Properties of simple harmonic motion Position of particle at time t: x(t ) = A cos(t + ) Aamplitude Angular frequency phase constant, phase angle Tperiod (t + ) phase x(t ) = A cos(t ) Properties of simple harmonic motion The period T of the motion is the time it takes for the particle to complete one full cycle Periodic motion The value of x at time t is equal to the value of x at time t +T Properties of simple harmonic motion Displacement: Period T: 2 T= x(t ) = A cos(t + ) Angular frequency: Frequency: 1 f= = T 2 2 = 2f = T Units: 1/s = 1 Hz Velocity: v(t ) = A sin(t + ) Acceleration: a (t ) = A cos(t + ) 2 i-clicker-1 & 2 Youre standing at the end of a springboard, bouncing gently up and down without leaving the boards surface (performing simple harmonic motion). If you bounce harder (larger amplitude), the time it takes for each bounce will A. become shorter B. become the longer C. remain same How about if your friend walks up and bounces with you? Black board example 15.1 An mass oscillates with an amplitude of 4.00 m, a frequency of 0.5 Hz and a phase angle of /4. (a) (b) (c) What is the period T? Write an equation for the displacement of the particle. Calculate the velocity and acceleration of the object at any time t. (d) Determine the position, velocity and acceleration of the object at time t = 1.00s. (e) Calculate the maximum velocity and acceleration of the object. The block-spring system 2 m T= = 2 k 1 1 f= = T 2 k m The frequency depends only on: - the mass of the block - the force constant of the spring. The frequency does not depend on the amplitude. Black board example 15.2 i-clicker 3 A spring stretches by 3.90 cm when a 10.0 g mass is hung from it. A 25.0 g mass attached to this spring oscillates in simple harmonic motion. (a) Calculate the period of the motion. (b) Calculate frequency and the angular velocity of the motion. A) 1.00 s B) 1.21 s C) 1.43 s D) 1.50 s E) 1.59 s Energy of harmonic oscillator Kinetic energy: 121 K = mv = m 2 A2 sin 2 (t + ) 2 2 Potential energy: 1212 U = kx = kA cos 2 (t + ) 2 2 Total energy: E = K +U = 121 kA = mvmax 2 = constant 2 2 Black board example 15.3 A 0.200 kg mass is attached to a spring and undergoes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. (a) What is the force constant of the spring? (b) What is the amplitude of the motion? (c) What is the velocity of the mass when the displacement is 8.00 cm? (d) What is the kinetic and potential energy of the system when the displacement is 8.00 cm? The (mathematical) pendulum (point mass) 2 L T= = 2 g For small motion (less than about 10). The physical pendulum 2 I T= = 2 mgd For small motion (less than about 10 I moment of interia m mass of object g acceleration due to gravity d distance from pivot point to center of mass Black board example 15.4 Find the period of a 14.7 inch (0.37 m) long stick that is pivoted about one end and is oscillating in a vertical plane. Simple harmonic motion and uniform circular motion x - component : x(t ) = R cos = R cos t y - component : y (t ) = R sin = R sin t Damped, simple harmonic motion x(t ) = [ A cos( ' t + )] e k b2 ' = m 4m 2 bt b is damping constant 2m Forced (Driven) Oscillations and Resonance b is damping constant A damped, harmonic oscillator (ang. frequency ) is driven by an outside, sinusoidal force with ang. frequency d Resonance when d = (can get very large amplitudes)
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