chapter13.key - Hooke's Law Reviewed Chapter 13 F = !kx...

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Chapter 13 Vibrations and Waves 1 When x is positive , F is negative ; When at equilibrium (x=0), F = 0 ; When x is negative , F is positive ; Hooke’s Law Reviewed F = ! kx 2 Sinusoidal Oscillation Pen traces a sine wave 3 Graphing x vs. t A : amplitude (length, m) T : period (time, s) A T 4 Some Vocabulary f = Frequency ! = Angular Frequency T = Period A = Amplitude " = phase x = A cos( ! t " # ) = A cos(2 $ ft " ) = A cos 2 t T " % ( ) * f = 1 T = 2 " f = 2 T 5 Phases Phase is related to starting time 90-degrees changes cosine to sine x = A cos 2 t T " $ % ( ) = A cos 2 ( t " t 0 ) T $ % ( ) if = 2 t 0 T cos t " 2 $ % ( ) = sin t ( ) 6
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x v Velocity is 90 ° “out of phase” with x: When x is at max, v is at min . ... Acceleration is 180° “out of phase” with x a = F/m = - (k/m) x Velocity and Acceleration vs. time T T T 7 v and a vs. t Find v max with E conservation Find a max using F=ma x = A cos ! t v = " v max sin t a = " a max cos t 1 2 kA 2 = 1 2 mv max 2 v max = A k m ! kx = ma ! kA cos " t = ! ma max cos t a max = A k m 8 What is ! ? Requires calculus. Since d dt A cos t = " A sin t v max = A = A k m = k m 9 Formula Summary f = 1 T = 2 f = 2 T x = A cos( t " # ) v = " A sin( t " ) a = " 2 A (cos t " ) = k the block has a mass of 0.50 kg, determine (a) the mechanical energy of the system (b) the maximum speed of the block (c) the maximum acceleration. a) 0.153 J b) 0.783 m/s c) 17.5 m/s 2 11 Example 13.2 A 36-kg block is attached to a spring of constant k=600 N/m. The block is pulled 3.5 cm away from its equilibrium positions and released from rest at t=0. At t=0.75 seconds, a) what is the position of the block? b) what is the velocity of the block? a) -3.489 cm b) -1.138 cm/s 12
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chapter13.key - Hooke's Law Reviewed Chapter 13 F = !kx...

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