2012_1061_Lecture_Ch15

2012_1061_Lecture_Ch15 - Simple Harmonic Mo/on F = kx dp F...

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Simple’Harmonic’Mo/on’ F = ma = dp dt F = kx m d 2 x dt 2 = d 2 x dt 2 + k m x =0
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Simple’Harmonic’Mo/on’ d 2 x dt 2 + k m x =0 x = A cos ( ω t + φ ) solu/on’ ω = k m
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T is the Period φ is the phase angle A is the amplitude f = 1 T is the frequency
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ω = 2 π T =2 π f
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x = A cos ( ω t + φ ) v = dx dt = ω A sin ( ω t + φ ) a = d 2 x dt 2 = dv dt = ω 2 A cos ( ω t + φ ) v max = ω A = ° k m A a max = ω 2 A = k m A
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Energy’in’the’Simple’ Harmonic’Oscillator’ Since’total’energy’is’conserved’ F = kx U = ° Fdx = 1 2 2 Total’Mechanical’Energy’ E = 1 2 mv 2 + 1 2 2 E = 1 2 kA 2 = 1 2 mv 2 max
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since v max = A ° k/m v = ± v max ° 1 x 2 A 2 v = ± ° k m ( A 2 x 2 )
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The’Simple’Pendulum’ A’ simple’pendulum’ consists’of’a’ small’object’suspended’from’the’ end’of’a’light’(negligible)’weight’ cord’that’doesn±t’stretch.’ F = mg sin θ ≈− mg θ substitute x = l θ or θ = x/l F mg l x = m d 2 x dt 2 θ = θ max cos ( ω t + φ ) ω = ° g l f = ω 2 π = 1 2 π ° g l T = 1 f =2 π ° g l
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The’Physical’Pendulum’and’the’Torsion’Pendulum’ τ = mgh sin θ ° τ = I α = I d 2 θ dt 2 I d 2 θ dt 2 = sin θ I d 2 θ dt 2 + sin θ =0 For’small’angular’amplitudes’
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2012_1061_Lecture_Ch15 - Simple Harmonic Mo/on F = kx dp F...

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