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chapter13.key

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

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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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a 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 m 10 Example13.1 An block-spring system oscillates with an amplitude of 3.5 cm. If the spring constant is 250 N/m and 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?
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