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Physics 207

by: Mark Hedinger

Physics 207 PHSX 207

Mark Hedinger

GPA 3.97
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About this Document

In-class notes, not book notes. Includes in-class sample equations.
College Physics ll
Gregory Francis
Class Notes




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This 3 page Class Notes was uploaded by Mark Hedinger on Friday September 2, 2016. The Class Notes belongs to PHSX 207 at Montana State University - Bozeman taught by Gregory Francis in Fall 2016. Since its upload, it has received 74 views.


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Date Created: 09/02/16
29Aug2016-31Aug2016 Physics 207—Chapter 11: Simple Harmonic Motion: Read Chapter 11 Homework: Chapter 11, Problems 2, 5, 6 Introduction Exam Times are at 6:00 p.m. a. Tutorials are out of the same text-book as Physics 205 i. Due Tuesdays by 5:00 p.m. (The Tuesday following the lab). b. Homework i. Place on 5x8 index card. ii. One problem per card iii. Your name/ TA’s name and section number iv. Due by the beginning of the next class period Physics 207 “A 500 g mass hangs from a spring and is found to stretch the spring by 12 cm from its un-stretched length. What is the spring constant of the spring?” F=-k ∆ s (You can ignore the minus sign) ∆ S = 21cm-9cm=12cm mblock00 g = 0.5 kg Fs,block5 N |F|=k| ∆ s | = 5N=k(0.12m) = 41 N/m W E, blockg= 0.5kg (10N/kg) = 5 N SMH characterized by:  Timing o Period (T): How long it takes to go through one cycle o Frequency (f): Number of cycles in a second  Amplitude o Maximum distance from equilibrium Frequency = 1/T = 1 k 2π √ m Period (T) = 1/f =π k √ Frequency = # cycles / seconds Things that go in a circle have the same up and down motion as any other SMH A larger mass will increase the period. To increase the frequency (# of cycles) , use a smaller mass or make the spring more stiff. *** The period or frequency of an oscillator is related only to the stiffness (k) of the restoring force and the inertia (m) of the system. It does not tell you anything about how the system got started and is independent of the amplitude. 2 π ~10 31Aug2016—Homework for Friday Ch 11. Problems 9, 15, 21 “Simple” in SHM Doesn’t depend on amplitude Resonance makes it a problem Static approach to solve for “k” |Fs|=k| ∆ s| Dynamic approach to solve for “k” m T=2 π √k 2 2 T = 4π m = k = 4π m k T2 Amplitude The location the object is at the turnaround point At the turnaround pointΔ s= Δ k At Equilibrium F net 0 A completely compressed or stretched Spring F = kA s,object Acceleration is at zero when the spring is at equilibrium, and the acceleration is at its maximum when the spring is at its turnaround point “In oscillation problems, there is a force trying to get an object back to equilibrium but once it gets there it shoots past the equilibrium point” – Francis Newton’s second law—F net= ma (The net force causes a mass to accelerate) The oscillation period will be the exact same on the moon as it would on earth because mass is the same and so is the spring constant Remember that in a system where no friction or outside force acts on the system, the total energy always remains the same 2 2 E total1/2mv + 1/2k Δ s At turn around point v =0 and the Δ s = A (aka. E total 0 + 1/2kA ) At equilibrium s = 0 and v is max (aka E total1/2mv +0) Resonance


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