lec11-203-11-SHMmod3

Lec11-203-11-SHMmod3 - spring SHM F-kx k a x 0 m a 2 x 0 x A sin t v A cos t a =-A 2 sin t Pendulum g L 11-0 A=xmax vmax = A amax = A 2 k m 2 =2f =

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-kx F 0 x m k a m k ω 2 π ω = 2 π f = T T 1 f δ) ωt ( sin A x ( cos ω A v 2 a = -A ω sin ( ωt +δ) 0 x ω a 2 A=x max max v = A ω 2 max a = A ω spring SHM 11-0 L g ω Pendulum
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Examples: Spring, Pendulum… Most mechanical systems- displace from equilibrium- 2 possibilities restoring force 1 st approx F=-kx oscillates about equilibrium with natural frequency repulsive force - runs away Small displacement Simple Harmonic Motion Dynamics potential energy 1 st approx U=1/2 kx 2 3 Points to note hit (tickle) system – it “rings” at a natural frequency (f o ) vibrate system with frequency (f)- it responds with f but close to (or at) f o get sharp, big “resonant” response friction makes natural ring die of exponentially with time 11-1
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x = 0 Cyclic Transfer Between Potential Energy and Kinetic Energy Kinetic Energy Potential Energy Harmonic Motion position velocity +v max -v max -x max -x max +x max T period, time required for one complete cycle [seconds/cycle = s] Frequency, f f = [ Hz = cycle/s] 1 T 11-2 Galileo -clock
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-kx ma 0 x m k a 0 x ω a 2 m k ω F (Newton & Hook were bitter rivals) Newton says simple harmonic motion δ) ωt ( sin A x ( cos ω A v ( ω A a 2 Spring: Simple Harmonic Motion Dynamics f π 2 ω T 1 f angular freq. (rad./sec.) period (sec.) frequency (Hz = sec. -1 ) 11-1 Equilibrium (no stretch/compression) -kx F Hooks Law x = 0 F F compressed stretched more stretched F always acts toward x=0 F 11-3
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x = A sin( ωt +δ) v = A ω cos( 2 a = -A ω sin( Spring: Simple Harmonic Motion 11-4
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A = t ) ωt ( sin A x = t v v x = v cos( ) x a a = v 2 /A A ω v a = 2 /A uniform circular motion- simple harmonic motion analogy 11-5 http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=148
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This note was uploaded on 04/03/2012 for the course PHY 1020 taught by Professor Kodera during the Spring '11 term at University of Florida.

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Lec11-203-11-SHMmod3 - spring SHM F-kx k a x 0 m a 2 x 0 x A sin t v A cos t a =-A 2 sin t Pendulum g L 11-0 A=xmax vmax = A amax = A 2 k m 2 =2f =

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