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# chapter13 - Chapter 13 Vibrations and Waves Hooke's Law...

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Chapter 13 Chapter 13 Vibrations and Waves

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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 Hooke’s Law Reviewed F = - κξ
Sinusoidal Oscillation Sinusoidal Oscillation Pen traces a sine wave

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Graphing x vs. t Graphing x vs. t A : amplitude (length, m) T : period (time, s) A T
Some Vocabulary Some Vocabulary f = Frequency ϖ = Angular Frequency T = Period A = Amplitude φ = phase x = Α χοσ( ϖτ - φ 29 = Α χοσ(2 π φτ - φ 29 = Α χοσ 2 πτ Τ - φ f = 1 Τ ϖ = 2 π φ = 2 π Τ

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Phases Phases Phase is related to starting time 90-degrees changes cosine to sine x = Α χοσ 2 πτ Τ - φ = Α χοσ 2 π ( τ - τ 0 29 Τ ιφ φ = 2 πτ 0 Τ cos ϖτ - π 2 = σιν ϖτ ( 29
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  Velocity and  Acceleration vs. time Acceleration vs. time T T T

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v v  and   and  a a  vs. t  vs. t Find v max with E conservation Find a max using F=ma x = Α χοσ ϖτ ω = - ω μ αξ σιν ϖτ α = - α μ αξ χοσ ϖτ 1 2 kA 2 = 1 2 μω μ αξ 2 ω μ αξ = Α κ μ - kx = ma - kA cos wt = - ma max cos wt a max = A k m
What is  What is  ϖ ϖ ? ? Requires calculus. Since d dt A cos ϖτ = - ϖΑ σιν ϖτ ω μ αξ = ϖΑ = Α κ μ ϖ = k m

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Formula Summary Formula Summary f = 1 Τ ϖ = 2 π φ = 2 π Τ x = Α χοσ( ϖτ - φ 29 ω = - ϖΑ σιν( ϖτ - φ 29 α = - ϖ 2 Α (χοσ ϖτ - φ 29 ϖ = k m
Example13.1 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 2

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Example 13.2 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.
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chapter13 - Chapter 13 Vibrations and Waves Hooke's Law...

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