HRONI OTION 070311 1
Hooke w hrterzton of Sme Hrmon oton (SH) Veoty n eerton n hrmon moton. Exeme. Horzont n vert rng Sme enuum Phy enuum Energy n hrmon moton Dme hrmon moton
Hooke w Srng ontnt The fore exerte by rng roorton to the tne to whh the rng trethe or omree from t rexe oton (for m x). 0 Rexe oton x kx x - 0 ement m Newton n w: m x kx m t t x k + x m 0 n orer Dfferent Eq.: t outon x( t) o( ω t + δ ) 3
hrterzton of hrmon moton. x k D.E. outon: + x 0 t m Phe x( t) o( ω t + δ ) ontnt,δ ngur frequeny ω mtue x + ω x t Int he k m (r/) 0 ω π T Pero () f 1 T reueny (Hz) 4
hekng for the outon of the DE of hrmon moton x k By ervng x( t) o( ω t + δ ) nubttutngn + x 0 t m rt ervtve: Seon ervtve: ω + k m 0 x t ω en( ωt + δ ) x ω o( ωt + δ ) ω x t ω + k m x 0 ω x + k m x 0 x k x( t) o( ω t + δ ) outon of + x 0 t m when ω k m 5
y o( ωt) ωt 0 π ω t ωt π 5π ω t ωt 4π t T ω t π 3π ω t ωt ωt 3π 7π ω t 6
y o( ωt + δ ) δ > 0 o(δ ) π ω t + δ t T ω t + δ π ωt + δ π 3π ω t + δ ωt 5π ω t + δ t T ωt + δ 3π ωt + δ 4π 7π ω t + δ 7
y o( ωt δ ) δ > 0 o( δ ) ωt δ 0 π ω t δ t T ω t δ π ωt δ π 3π ω t δ ωt 5π ω t δ t T ωt δ 3π ωt δ 4π 7π ω t δ 8
btrt y y y o( ωt) o( ω t + δ ) he o( ωt δ ) Behn The oton n re he of the bk one ωt The oton n bue behn the bk one 9
Veoty n eerton n hrmon moton x( t) o( ω t + δ ) x( t) Veoty x& ω en( ωt + δ ) t x( t) eerton && x ω o( ωt + δ ) ω x( t) t x Zero veoty x 0 xmum eerton Zero eerton xmum veoty 10
Exme. Srng yng on horzont ne. n e rng of et ontnt 0 N/m jon bok of m 31.5 g on ft horzont urfe wthout frton. The rng trethe 8 m, then t reee n the m ote freey roun t equbrum oton. n: ) Equton for the reutng hrmon moton n the ero of the oton. b) eerton of the bok when t rehe the utmot oton, n t eerton when t ng through the equbrum oton. ) Veoty n eerton of the bok when 1 h gone by. x( t) o( ω t + δ ) x ( 0) o(0 + δ ) 0.08 m ω m k 0 0.315 8 r/ x( t) o( ω t + δ ) o δ 1 x( t) oωt 0.08 o8t π π π T 0.785 ω 8 4 o δ ±nπ ( n 0,1,...) (I.S. unt) 11
x0 0 0 x Rexe rng oton, ytem t ret Strethe rng oton: the oton begn -k x& -k x && x && x ( ω x ) t x ω x 8 0.08 5.1 m/ Intermete ont: the otor goe to the orgn T > t > 0 4 x 1
x0 x& x t 0 ω enωt 0 x& & x( x 0) When x0 one qurter of ero h gone by. x x& -k x The otor e through the equbrum oton: mxmum veoty, zero eerton T t 4 π T ω en T 4 en π ω 0.08 8 0.64 m/ The otor overe the equbrum oton menwhe t veoty eree n t eerton nree 13
Wht hen fterwr, unt t T? Wht re veoty n eerton? Wht re veoty n eerton when 1 h gone by? 14
btrt x 0 x v 0 mx t 0 x 0 v vmx 0 t T / 4 x v 0 + mx t T / x 0 v + vmx 0 t 3T / 4 x v t T 0 mx x x 15
Exme. Srng n vert oton Hooke w kx.b.d. mg m 0 x x - 0 n Newton w m g+ m t x + k m x g x mg kx m t 16
Equton for the vert rng The outon : t x x + t k m x g x + t k m x mg x( t) + o( ω t + δ ) k k mg ω o( ωt + δ ) + + o( ωt + δ ) m k k m x g Dertve: ω x t ω en( ωt + δ ) x ω o( ωt + δ ) t k ω + t + + g m o( ω δ ) k m g the equton tfe! 17
Sme enuum O θ T mg mg enθ T Y θ X mg oθ O θ mg enθ Torque O ten to retore the equbrum oton mg O mg enθ 18
oment of the X-omonent of weght (bout O): unment equton for rotton O J mv m θ θ v t θ t O mg J t O J t m enθ θ t mg enθ m θ t mg enθ v θ g + enθ 0 t 19
omre θ g + enθ 0 t or m nge en θ θ Then θ g + enθ 0 t orm of the outon: Pero: x k wth + x 0 t m Sme enuum Srng θ g n be ubttute by + θ 0 t θ ( t) o( ωt + δ ) T π π ω g ω g 0
Wht re re m nge? R 1 r R R R θ R ength of r θ R nge (rn) Ru θ (º) θ (r) n θ f % 0 0.0000 0.0000 0.0 0.0349 0.0349 0.0 5 0.0873 0.087 0.1 8 0.1396 0.139 0.3 10 0.1745 0.1736 0.5 1 0.094 0.079 0.7 15 0.618 0.588 1.1 18 0.314 0.3090 1.6 0 0.3491 0.340.0 0.3840 0.3746.4 5 0.4363 0.46 3.1 8 0.4887 0.4695 3.9 30 0.536 0.5000 4.5 3 0.5585 0.599 5.1 35 0.6109 0.5736 6.1 <1% % <5% m nge <15º 1
Phy Penuum O τ θ O θ mg mg enθ mg mg oθ τ m g τ I α I α mg enθ retorng θ mg + enθ 0 t I
θ mg + enθ 0 t I SH Equton. θ mg Sm nge + θ 0 t I ω mg I T π I mg ω Exme. homogenou of ru R 0 m hng from n very ner t ege. The ote wth m mtue. Wht the ero of the oton (negetng ny frton fore)? R 1 I mr + m 1 mr + mr 3 mr mg T I π π mg 3 mr mgr π 3 R g 3 0.0 π 1.1 9.80 3
ETRES O HRONI OTION 1º There retorng fore (or retorng torque) rety roorton to ement. º The ytem exee the equbrum oton, n from tht moment the retorng fore ten to en t bk to tht equbrum oton. 3º Whenever m frton fore re reent, the movement ow own tte by tte (me oton) 4º Whenever the frton fore re bg enough, the movement n our wthout ny oton (overme hrmon moton). 5º It obe to nfuene over the movement by n extern fore (rven hrmon moton). 4
Energy n hrmon moton x x kx W x kx x W x x 1 kx x kx x o 0 Potent energy: W 1 kx 5
KINETI ENERGY POTENTI ENERGY ENERGY E 1 1 1 E & m ω en ( ωt + δ ) mv mx 1 1 1 kx k o ( ω t + δ ) m ω o ( ωt + δ ) + 1 m ω { en ( ωt + δ ) + o ( ωt + δ )} 1 m ω ω 1 k k m 6
Dme hrmon moton x t 0 + γ& x + ω x x 0 R m v x& γ ω 0 x t + m b m k m R kx R bv bx& x kx bx& m t x b k + x + x 0 & t m m 7
x + γ& x + ω0 x 0 t orm of outon γt x ex o( ω t + δ ) γ k b ω ω0 4 m 4m Quty ftor ω Q 0 γ y oω0 t ω 0 0.5 r/ δ 0 0 10 0 30 40 50 y ± ex( γt / ) γ - 0.05 Q nree, the energy of the otor te owy 1 Q y ex( γt / ) oωt 0.5 0.05 ω 10 γ ω 0 4 8 t ()
rt mng uno γt x ex o( ω t + δ ) γ ω 0 4 γ - 0.05 1 γ k b ω ω0 4 m 4m There no oton 0 10 0 30 40 50 60 70 80 90 100 9
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