The Spring. Consider a spring, which we apply a force F A to either stretch it or compress it

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The Spring Consider spring, which we pply force F A to either stretch it or copress it F A - unstretched -F A 0 F A k k is the spring constnt, units of N/, different for different terils, nuber of coils

Fro Newton s 3 rd Lw, the spring eerts force tht is equl in gnitude, but opposite in direction F k s Hooke s Lw for the restoring force of n idel spring. (It is conservtive force.) Chpter 4, proble P9 Five identicl springs, ech with stiffness 390 N/, re ttched in prllel (tht is side-by-side) to hold up hevy weight. If these springs were replced by n equivlent single spring, wht should be the stiffness of this single spring?

Oscilltory Motion We continue our studies of echnics, but cobine the concepts of trnsltionl nd rottionl otion. In prticulr, we will re-eine the restoring force of the spring (lter its potentil energy). We will consider the otion of ss, ttched to the spring, bout its equilibriu position. This type of otion is pplicble to ny other kinds of situtions: pendulu, tos, plnets,...

Siple Hronic Motion If we dd ss to the end of the (ssless) spring, stretch it to displceent 0, nd relese it. The spring-ss syste will oscillte fro 0 to 0 nd bck. k - 0 0 0 Without friction nd ir resistnce, the oscilltion would continue indefinitely This is Siple Hronic Motion (SHM) SHM hs iu gnitude of 0 A, clled the Aplitude

One wy to understnd SHM is to consider the circulr otion of prticle nd rottionl kinetics (The Reference Circle) The prticle trvels on circle of rdius ra with the line fro the center to the prticle king n ngle θ with respect to the -is t soe instnt in tie Now, project this D otion onto 1D is

A cosθ -A 0 A but θ ω t Therefore, A cos( ω t) A θ is the displceent for SHM, which includes the otion of spring SHM is lso clled sinusoidl otion A 0 plitude of the otion (iu) ω is the ngulr frequency (speed) in rd/s. It reins constnt during the otion. A

ω nd the period T re relted Define the frequency # cycles f sec π ω π T f T ω π T As n eple, the lternting current (AC) of electricity in the US hs frequency of 60 Hz Now, lets consider the velocity for SHM, gin using the Reference circle 1 Units of 1/s Hertz (Hz) Reltes frequency nd ngulr frequency

v v v t -v sinθ t rω Aω -Aω sin( ω t) v t Aplitude of the velocity is v A θ v Aω Accelertion of SHM r v r r t cosθ r ω r r θ

r rω Aω Aω The plitude of the ccelertion is Sury of SHM Acos( ω t) v Aω sin( ω Aω t) cos( ω t) v cos( ω t) Aω A, ω t 0, π,π,... π 3π Aω, ω t,... Aω, ω t 0, π,π,... (A1) ωt (π rds)

v (Aω1) ωt (π rds) (Aω 1)

Eple Proble Given n plitude of 0.500 nd frequency of.00 Hz for n object undergoing siple hronic otion, deterine () the displceent, (b) the velocity, nd (c) the ccelertion t tie 0.0500 s. Solution: Given: A0.500, f.00 Hz, t0.0500 s. ω πf π (.00 Hz) 4.00π rd/s ωt (4.00π rd/s)(0.0500 s) 0.00π rd 0.68 rd () Acos( ωt) (0.500 )cos(0.68 rd)

v v 0.405 -Aω sin( ω t) (0.500 )(4π rd/s)sin(0.68 rd) v 3.69 /s Aω cos( ω t) (0.500 )(4π rd/s) cos(0.68 rd) 63.9 /s Frequency of Vibrtion Apply Newton s nd Lw to the spring-ss syste (neglect friction nd ir resistnce)

k F s 0 A F s F k k[ k ω Acos( ω t)] Consider -direction only Substitute nd for SHM [ Aω cos( ω t)] ω k Angulr frequency of vibrtion for spring with spring constnt k nd ttched ss. Spring is ssued to be ssless.

ω T π 1 f f π f k 1 π T This lst eqution cn be used to deterine k by esuring T nd T Note tht ω (f or T) does not depend on the plitude of the otion A k 4π k k 4π Cn lso rrive t these equtions by considering derivtives Frequency of vibrtion Period of vibrtion T

The Siple Pendulu An ppliction of Siple Hronic Motion A ss t the end of ssless rod of length L There is restoring force which cts to restore the ss to θ0 θ L F g sinθ Copre to the spring F s -k The pendulu does not disply SHM gsinθ T g

But for very sll θ (rd), we cn ke the pproition (θ<0.5 rd or bout 5 ) siple pendulu pproition sinθ θ since s F gθ rθ Lθ Arc length This is SHM F g s L g L s Looks like spring force ( F k) k s g L Like the spring constnt Now, consider the ngulr frequency of the spring

ω f k 1 π g L g / L g L ω Siple pendulu frequency Siple pendulu ngulr frequency With this ω, the se equtions epressing the displceent, v, nd for the spring cn be used for the siple pendulu, s long s θ is sll For θ lrge, the SHM equtions (in ters of sin nd cos) re no longer vlid ore coplicted functions re needed (which we will not consider) A pendulu does not hve to be point-prticle