University Physics: Waves and Electricity

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University Physics: Waves and Electricity Ch15. Simple Harmonic Motion Lecture 2 Dr.-Ing. Erwin Sitompul http://zitompul.wordpress.com 2013

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University Physics: Waves and Electricity. Ch15. Simple Harmonic Motion. Lecture 2. Dr.-Ing. Erwin Sitompul. http://zitompul.wordpress.com. 2013. Homework 1: Plotting the Motions. Plot the following simple harmonic motions in three different plots: (a) x a ( t ) = x m  cos ωt - PowerPoint PPT Presentation

Transcript of University Physics: Waves and Electricity

Page 1: University Physics: Waves and Electricity

University Physics: Waves and Electricity

Ch15. Simple Harmonic MotionLecture 2

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2013

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2/2Erwin Sitompul University Physics: Waves and Electricity

Plot the following simple harmonic motions in three different plots:(a) xa(t) = xmcosωt (b) xb(t) = xmcos(ωt–π/2) (c) xc(t) = xm/2cos(ωt+π/2)(d) xd(t) = 2xmcos(2ωt+π)

Homework 1: Plotting the Motions

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2/3Erwin Sitompul University Physics: Waves and Electricity

Plot the following simple harmonic motions:(a) xa(t) = xmcosωt (b) xb(t) = xmcos(ωt–π/2) (c) xc(t) = xm/2cos(ωt+π/2)(d) xd(t) = 2xmcos(2ωt+π)

xa(t)T0.5T

xm

–xm

0

xb(t)

xa(t)T0.5T

xm

–xm

0xc(t)

xm/2

–xm/2

Solution of Homework 1: Plotting the Motions

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2/4Erwin Sitompul University Physics: Waves and Electricity

Plot the following simple harmonic motions:(a) xa(t) = xmcosωt (b) xb(t) = xmcos(ωt–π/2) (c) xc(t) = xm/2cos(ωt+π/2)(d) xd(t) = 2xmcos(2ωt+π)

xa(t)T0.5T

xm

–xm

0

xd(t)

2xm

–2xm

Solution of Homework 1: Plotting the Motions

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Object on a Vertical Spring When an object hangs from a vertical spring, there is a force

mg downward in addition to the force of the spring. If we choose the upward

direction to be positive, the spring’s force on the object is Fs = ky, with y as the difference in the stretched and unstretched position of the spring.

The Newton’s 2nd Law gives:

2

2

d ym ky mgdt

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Object on a Vertical Spring After some manipulations, the solution can be found as

( ) cos( )y t A t

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Energy in Simple Harmonic Motion The energy of a linear oscillator transfers back and forth

between kinetic energy and potential energy, while the sum of the two –the mechanical energy E of the oscillator– remains constant.

The potential energy is associated entirely with the spring. Its value depends on how much the spring is stretched or compressed.

2 2 21 12 2( ) cos ( )mU t kx kx t

The kinetic energy of the system is associated entirely with the block. Its value depends on how fast the block is moving.

2 2 2 21 12 2( ) sin ( )mK t mv m x t

2 212( ) sin ( )mK t kx t

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Energy in Simple Harmonic Motion The mechanical energy is then given by

( ) ( ) ( )E t U t K t 2 2 2 21 1

2 2( ) cos ( ) sin ( )m mE t kx t kx t 21

2( ) mE t kx

The mechanical energy of a linear oscillator is indeed constant and independent of time.

The potential energy and kinetic energy of a linear oscillator are shown next as the function of time and as the function of displacement.

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Energy in Simple Harmonic Motion

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Simple Harmonic Motion and Uniform Circular Motion

Simple harmonic motion is actually the projection of uniform circular motion on a diameter of the circle in which the circular motion occurs.

• Measurement of the angle between Jupiter and its moon Callisto as seen from Earth.

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x(t)

v(t)

Simple Harmonic Motion and Uniform Circular Motion

Next we can see, whether we look at the displacement, the velocity, or the acceleration, the projection of uniform circular motion is indeed simple harmonic motion.

( ) cos( )mx t x t ( ) sin( )mv t x t 2( ) cos( )ma t x t

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Damped Simple Harmonic Motion When the motion of an oscillator is

reduced by an external force, the oscillator and its motion are said to be damped.

An idealized example of a damped oscillator is shown on the next figure.

As the vane moves up and down, the liquid exerts an inhibiting drag force on it and thus on the entire oscillating system.

With time, the mechanical energy of the block-spring system decreases, as energy is transferred to thermal energy of the liquid and vane.

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Damped Simple Harmonic Motion Let us assume the liquid exerts a damping

force Fd that is proportional to the velocity v of the vane and block. Then, for the components along the x axis

→ →

dF bvwhere b is a damping constant, that depends on both the vane and the liquid, with the SI unit of kilogram/s.

The force on the block from the spring is Fs = –kx. Let us assume that the gravitational force on the block is taken out of calculation by setting the position where the mass rests as the equilibrium position of the system.

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Damped Simple Harmonic Motion Then we can write Newton’s 2nd Law for

the components along the x axis asbv kx ma

2

2 0d x dxm b kxdt dt

The solution of this equation is2( ) cos( )bt m

mx t x e t

where xm is the amplitude and ω’ is the angular frequency of the damped oscillation. This angular frequency is given by

2

24k bm m

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Damped Simple Harmonic Motion The solution can be seen as a cosine function whose

amplitude, which is xme–bt/2m, gradually decreases with time, as the next figure suggests.

2( ) cos( )bt mmx t x e t

If the damping is small, we can find the mechanical energy by replacing xm as found in the formula previously with xme–bt/2m, so that

212( ) bt m

mE t kx e

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Here are three sets of values for the spring constant, damping constant, and mass for the damped oscillator. Rank the sets according to the time required for the mechanical energy to decrease to one-fourth of its initial value, greatest first.

Set 1, Set 2, then Set 3

Checkpoint

0 02211 02( ) (2 ) b t m

mE t k x e

• The greater the magnitude of the exponential, the faster the decrease.

0 01( )2 2

0

b t m

mk x e

0 0(6 ) 2(4 )212 02( ) b t m

mE t k x e 0 06( )21 8

02

b t m

mk x e

0 0(3 ) 2213 02( ) (3 ) b t m

mE t k x e 0 03( )23 2

02

b t m

mk x e

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For the damped oscillator, m = 250 g, k = 85 N/m, and b = 70 g/s.

(a) What is the period of the motion?

Example 1

(b) How long does it take for the amplitude of the damped oscillation to drop to half its initial value?

2

24k bm m

2

2

85 (0.07)0.25 4(0.25)

18.439rad s

2T

2

18.439

0.34 s

2 12

bt me 2 1ln

2mtb

2(0.25) 1ln(0.070) 2

4.95 s

• Or after around 15 periods of oscillation

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For the damped oscillator, m = 250 kg, k = 85 N/m, and b = 70 g/s.

(c) How long does it take for the mechanical energy to drop to one-half its initial value?

Example 1

12

bt me 1ln2

mtb

(0.25) 1ln(0.070) 2

2.475 s

• Auto Shock- Absorber

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University Physics: Waves and Electricity

Ch16. Transverse WavesLecture 2

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

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2/20Erwin Sitompul University Physics: Waves and Electricity

Waves are of three main types:1.Mechanical waves. Common examples include water waves,

sound waves, and seismic waves. These waves are governed by Newton's laws and they can exist only within a material medium such as water, air, and rock.

2.Electromagnetic waves. Common examples include visible and ultraviolet light, radio and television waves, microwaves, x rays, and radar waves. These waves require no material medium to exist; for example light waves.

3.Matter waves. These waves are associated with electrons, protons, and other fundamental particles, and even atoms and molecules.

Much of what we discuss here applies to all kinds of waves. But for specific examples we shall refer to mechanical waves.

Types of Waves

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A wave sent along a stretched, taut string is the simplest mechanical wave.

A single up-and-down motion generates a single pulse. The pulse moves along the string at some velocity v.

Assumptions:• No friction-like forces within the string, where the energy of the

wave may dissipate → No energy loss• Strings are very long → No need to consider reflected waves from

the far end

• A typical string element (marked with a dot) moves up once and then down as the pulse passes.

• The element’s motion is perpendicular to the wave’s direction of travel → the pulse is a transverse wave

A Simple Mechanical Wave

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If a sinusoidal wave is sent along the string, we would find that the displacement of every string element is perpendicular to the direction of travel of the wave. (Figure on the left)

This motion is said to be transverse, and the wave is said to be a transverse wave.

If you push and pull on a piston in simple harmonic motion, a sinusoidal sound wave travels along the pipe. (Figure on the right)

Because the motion of the elements of air is parallel to the direction of the wave’s travel, the motion is said to be longitudinal, and the wave is said to be a longitudinal wave.

Transverse and Longitudinal Waves

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Both a transverse wave and a longitudinal wave are said to be traveling wave because they both travel from one point to another.

Note that it is the wave that moves from one end to another end, not the material (string or air) through which the wave moves.

Transverse and Longitudinal Waves

• Longitudinal wave

• Transverse wave

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To completely describe a wave on a string (and the motion of any element along its length), we need a function that gives the shape of the wave.

This means, we need a relation in the form y = f(x,t),

in which y is the transverse displacement of any string element as a function f of the time t and the position x of the element along the string.

If the wave is sinusoidal, then at time t, the displacement y of the element located at position x is given by:

( , ) sin( )my x t y kx t

Wavelength and Frequency

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Traveling wave:• Five snapshots of a sinusoidal wave traveling in

the positive right direction of an x axis is shown in the figure.

• The movement of the wave is indicated by the short arrow pointing to a peak of the wave.

• But, at the meantime, the string moves only parallel to the x axis. Examine the motion of the red-colored string element at x = 0.

Wavelength and Frequency

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The wavelength λ of a wave is the distance (parallel to the direction of the wave’s travel) between repetitions of the wave shape.

The angular wave number k is related to the wavelength by:2k

The SI unit of k is radian per meter or m–1. Differentiate this k with the one that represents a spring

constant as previously, where the unit is N/m.

Wavelength and Angular Wave Number

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The period of oscillation T of a wave is the time any string element takes to move through one full oscillation.

The angular frequency ω is related to the period by:2T

The SI unit of ω is radian per second or s–1. The frequency f of a wave is defined as 1/T and is related to

the angular frequency ω by:1

2fT

Period, Angular Frequency, and Frequency

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The figure is a composite of three snapshots, each of a wave traveling along a particular string.The phases for the waves are given by:(a) 2x–4t(b) 4x–8t(c) 8x–16tWhich phase corresponds to which wave in the figure?

(a) ↔ 2 (b) ↔ 3 (c) ↔ 1

Checkpoint

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Two sinusoidal traveling waves with identical amplitude, wavelength, and period, can also be shifted one from another.

To accommodate this, the wave function can be generalized by inserting a phase constant Φ:

( , ) sin( )my x t y kx t

• Now, estimate the phase constant of this wave.

5

Phase Constant

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The figure shows two snapshots of the sinusoidal wave, taken a small time interval Δt apart.

The wave is traveling in the positive direction of x (to the right), and the entire wave pattern moves a distance Δx during the time interval Δt.

After some derivation, we can write the wave speed v as:

vk

T

f

The Speed of a Traveling Wave

?dx dy dtvdt dy dx

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The Speed of a Traveling Wave

( , ) sin( )my x t y kx t

The equation that describes a wave moving in the positive direction is:

( , ) sin( )my x t y kx t

The equation that describes a wave moving in the negative direction is:

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Here are the equations of three waves:(1) y(x,t) = 2sin(4x–2t)(2) y(x,t) = sin(3x–4t)(3) y(x,t) = 2sin(3x–3t)Rank the waves according to their (a) wave speed and (b) maximum speed perpendicular to the wave’s direction of travel (the transverse speed), greatest first.

(a) Wave speed: 2, 3, 1

Wave speed: vk

(b) Transverse speed: 3, then 1 and 2 tie

Transverse speed: m mv x

The Speed of a Traveling Wave

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A wave traveling along a string is described byy(x,t) = 0.00327sin(72.1x–2.72t),

in which the numerical constants are in SI units.(a) What is the amplitude of this wave?

(b) What are the wavelength, period, and frequency of this wave?

0.00327 m 3.27 mmmy

2k

2k

272.1

0.0871 m 8.71 cm

2T

2T

2

2.72

2.310 s

1fT

1

2.310 0.433 Hz

Example 1

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(c) What is the velocity of this wave?

(d) What is the displacement y at x = 22.5 cm and t = 18.9 s?

A wave traveling along a string is described byy(x,t) = 0.00327sin(72.1x–2.72t),

in which the numerical constants are in SI units.

vk

2.7272.1

0.0377 m s 3.77 cm s

( , ) 0.00327sin(72.1 2.72 )y x t x t (0.225 m,18.9 s) 0.00327sin(72.1 0.225 2.72 18.9)y

0.00327sin( 35.1855 rad) (0.00327 m)(0.588)0.00192 m

1.92 mm

ˆ3.77i cm sv

• How come?

Example 1

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Email QuizA hungry scorpion can detect the presence of a nearby beetle by the waves the motion sends along the sand surface while the beetle moves.

The waves are of two types: transverse wave with the speed of vt = 50 m/s and longitudinal waves traveling at vl = 140 m/s.

If a careless motion of the beetle sends out such waves, the scorpion can determine the distance of the beetle from the difference Δt in the arrival times of the waves at its leg nearest to the beetle. If Δt is given as 4.0 ms, determine the beetle’s distance d (in cm).

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A sinusoidal wave of frequency 500 Hz has a speed of 350 m/s. (a) How far apart are two points that differ in phase by π/3 rad?(b) What is the phase difference between two displacements

at a certain point at times 1 ms apart?

Homework 2: Phase Differences

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2/37Erwin Sitompul University Physics: Waves and Electricity

Homework 2A: Phase Differences1. A sinusoidal wave of wavelength 280 cm has a speed of 60 m/s. (a) How many seconds apart are two displacements at a certain

point that differ in phase by 2π/3 rad?(b) What is the phase difference between two displacements at a

certain time at distances 14 cm apart?

2. In the figure below, two springs with the same value of spring constant 7580 N/m are connected to a box with the mass 0.245 kg. Determine the oscillation frequency on the frictionless floor?