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Physics 111Lecture 26 (Walker: 13.1-3)

Oscillations IApril 6, 2009

Quiz Wednesday - Chaps. 10 & 11 (but not 11.8-9)

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Periodic Motion (Oscillations)

Oscillations have some common characteristics:1. They take place around an equilibrium position;2. The motion is periodic and repeats with each cycle.

If an object vibrates or oscillates back and forth over the same path, each cycle taking the same amount of time, the motion is called periodic.

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Periodic Motion CharacteristicsPeriod T: time required for one cycle of periodic motion (unit: s)Frequency f: number of cycles per unit time

The frequency unit is called a hertz (Hz):Amplitude A: maximum distance object moves from equilibrium (unit: m)

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Frequency and Period1/ and 1/f T T f= =

is the frequency (units: Hz oscillations per second)f

≡

is the period (units: s)T

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Example: Frequency and PeriodWhat is the frequency of a pendulum swing that takes 2 seconds to complete a cycle?

Note that 1 Hz = 1/s

f = 1/T = 1/(2s) = 0.5 s-1 = 0.5 Hz

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Simple Harmonic Motion (SHM)

A spring exerts a restoring force that is proportional to the displacement from equilibrium:

SHM is a special kind of periodic motion that results when a mass is acted on by a force that is proportional to the mass’s displacement from equilibrium.

Look at mass attached to spring of spring constant k:

The minus sign on the force indicates that it is a restoring force – it tries to restore the mass to its equilibrium position.

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We assume that the surface is frictionless. There is a point where the spring is neither stretched nor compressed; this is the equilibrium position. We measure displacement from that point (x = 0).The force exerted by the spring depends on the displacement:

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SHM-Mass & Spring

A-A 0

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Simple Harmonic MotionA mass on a spring has a displacement as a

function of time that is a sine or cosine curve:

Here, A is called the amplitude of the motion.

A-A

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Position vs. time in SHMIf we call the period of the motion T (this is

the time to complete one full cycle) we can write the position as a function of time as:

For this position function, the position at time t + T is the same as the position at time t (one period earlier), as we would expect.

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Connections between Uniform Circular Motion and Simple Harmonic Motion

An object in simple harmonic motion has same motion as one component of an object in uniform circular motion:

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Uniform circular motion projected into one dimension is simple harmonic motion (SHM).

Connections between Uniform Circular Motion and Simple Harmonic Motion

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Consider particle rotating ccw, with the angle φincreasing linearly with time:

cosx A φ=

, so if 0 at 0.t ttφω φ ω φ∆

= = = =∆

( ) cosx t A tω=

Uniform Circular Motion vs. SHM

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Connections between Uniform Circular Motion and Simple Harmonic Motion

Here, the object in circular motion has an angular speed of

where T is the period of motion of the object in simple harmonic motion.

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Summary: Position Function for SHMThe position as a function of time for object in SHM started at x=A at time t=0:x = A cos (ωt) = A cos (2πft) = A cos (2πt/T)

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Example: Finding the TimeA mass, oscillating in simple harmonic motion, starts

at x = A and has period T.At what time, as a fraction of T, does the mass first

pass through x = ½A?

11 162

cos2 2 3T Tt Tππ π

⎛ ⎞−⎜ ⎟⎝ ⎠

= = =

12

2cos tx A ATπ

= =

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Velocity in SHMThe velocity as a function of time for object in SHM started at x=A at time t=0:

ωA

vmax = ωA

ωA

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Example: A System in SHMAn air-track glider is attached to a

spring, pulled 20 cm to the right, and released at t=0. It makes 15 completeoscillations in 10 s.

a. What is the frequency of oscillation?b. What is the object’s maximum speed?c. What is its velocity at t=0.50 s?

f = 15 oscillations / 10s = 1.5 Hzvmax = ωA = 2πfA = 2(3.14)(1.5/s)(0.2m)

= 1.89 m/sv =-ωAcos[ω(0.8s)] =-(1.89m/s)cos[2π(1.5/s)(0.8s)] = -.59 m/s

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Acceleration in SHMThe acceleration as a function of time for object in SHM started at x=A at time t=0: :

ω2A

-ω2A

amax = ω2A

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Example: Acceleration in SHM• A can in a paint can shaker is in SHM with an

amplitude of 0.5m and ω=3 rad/s. What is the maximum acceleration experienced by the can?

• amax = ω2A = (3 rad/s)2(0.5m) = 4.5 m/s2

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The Period of a Mass on a SpringSince the force on a mass m on a spring is proportional to the displacement x from equilibrium, Newton’s 2nd law gives

.

Substituting the time dependencies of a and xgives:

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The Period of a Mass on a SpringTherefore, the period is:

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End of Lecture 26For Wednesday, read Walker, 13.4-5.

Homework Assignment 13a is due at 11:00 PM on Wednesday, April 8.

Quiz Wednesday - Chaps. 10 & 11 (but not 11.8-9)