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lecture notes-Chapter10-posted

# lecture notes-Chapter10-posted - Chapter 10 Simple Harmonic...

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1 Chapter 10 Simple Harmonic Motion SHM Examples: Mass-Spring System, Simple Pendulum Energy Conservation Applied to SHM

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2 § 10.5 Simple Harmonic Motion Condition for oscillations around a point: restoring force
3 § 10.5 Simple Harmonic Motion Simple harmonic motion (SHM) occurs when the restoring force (the force directed toward a stable equilibrium point ) is proportional to the displacement from equilibrium.

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4 The motion of a mass on a spring is an example of SHM. The restoring force is F=-kx. x Equilibrium position x y
5 Assuming the table is frictionless: ( 29 ( 29 t x m k t a ma kx F x x x - = = - = Also, ( 29 ( 29 ( 29 ( 29 ( 29 2 2 2 1 2 1 t kx t mv t U t K t E + = + = x Equilibr ium positio n x y

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6 § 10.6-7 Representing Simple Harmonic Motion When a mass-spring system is oriented vertically, it will exhibit SHM with the same period and frequency as a horizontally placed system.
7 SHM demo http://www.ngsir.netfirms.com/englishhtm/Sprin http://www.phy.ntnu.edu.tw/ntnujava/index.php

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8 At the equilibrium point x =0 so a =0 too. When the stretch is a maximum, a will be a maximum too. The velocity at the end points will be zero, and it is a maximum at the equilibrium point.
9 A simple harmonic oscillator can be described mathematically by: ( 29 ( 29 ( 29 t A t v t a t A t x t v t A t x ϖ ϖ ϖ ϖ ϖ cos sin cos 2 - = = - = = = Or by: ( 29 ( 29 ( 29 t A t v t a t A t x t v t A t x ϖ ϖ ϖ ϖ ϖ sin cos sin 2 - = = = = = where A is the amplitude of the motion, the maximum displacement from equilibrium, A ϖ =v max , and A ϖ 2 =a max.

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10 SHM graphically
11 The period of oscillation is . 2 ϖ π = T where ϖ is the angular frequency of the oscillations, k is the spring constant and m is the mass of the block. m k = ϖ

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12 Example (text problem 10.28): The period of oscillation of an object in an ideal mass-spring system is 0.50 sec and the amplitude is 5.0 cm. What is the speed at the equilibrium point? At equilibrium x=0: 2 2 2 2 1 2 1 2 1 mv kx mv U K E = + = + = Since E=constant, at equilibrium (x = 0) the KE must be a maximum. Here v = v max = A ϖ .
13 ( 29 ( 29 cm/sec 8 . 62 rads/sec 6 . 12 cm 5.0 and rads/sec 6 . 12 s 50 . 0 2 2 = = = = = = v T π π ϖ The amplitude A is given, but ϖ is not. Example continued:

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14 Example (text problem 10.41): The diaphragm of a speaker has a mass of 50.0 g and responds to a signal of 2.0 kHz by moving back and forth with an amplitude of 1.8 × 10 -4 m at that frequency.
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