AP Physics B(Princeton 15 &

Giancoli 11 & 12)Waves and Sound

Assignments

• Reading: 11.1-4,7-9,11-13 and 12.1,2,4-7 and Princeton 15

• Problems: Waves: 11.42,43,55,56• Sound: 12.4,5,10,11 • Strings/AirColumns12.29,30,35• Interference 12.42,43• Doppler 12.51,52• SHM: 11.4,5,20, 25• Pendulum: 11.31,32

Preview

• What are the two categories of waves with regard to mode of travel?– Mechanical– Electromagnetic

• Which type of wave requires a medium?– Mechanical

• An example of a mechanical wave?– Sound

Velocity of a Wave

• The speed of a wave is the distance traveled by a given point on the wave (such as a crest) in a given internal of time.

• v = d/td: distance (m)t: time (s)

• v = f v: speed (m/s): wavelength (m)f : frequency (s-1, Hz)

Period of a Wave

• T = 1/f

• T : Period = (s)

• F : frequency (s-1, Hz)

Problem: Sound travels at approximately 340 m/s, and light travels at 3.0 x 108 m/s. How far away is a lightning strike if the sound of the thunder arrives at a location 5.0 seconds after the lightning is seen?

Light travels almost instantaneously from strike location to the observer.

The sound travels much more slowly:

d = vs t = (340 m/s)(5.0 s) = 170m

Problem: The frequency of a C key on the piano is 262 Hz. What is the period of this note? What is the wavelength? Assume speed of sound in air to be 340 m/s at 20 oC.

T = 1/f = 1/262 s-1 = 0.00382 s

V = f = v/f

= 340 m/s / 262 /s = 1.30 m

Problem

• A sound wave traveling through water has a frequency of 500 Hz and a wavelength of 3 m. How fast does sound travel through water?

• v = f = 3m (500 Hz) = 1500 m/s

Wave on a Wire

v = FT

m / L

v, velocity, m/s

FT, tension on a wire, N

m/L mass/unit length, kg/m

m/L may be shown as

Problem Ex. 11-11

A wave whose wavelength is 0.30 m is traveling down a 300 m long wire whose total mass is 15 kg. If the tension of the wire is 1000N, what are the speed and frequency of the wave?

Using equation on prior slide:v = √[( 1000N) / (15kg)(300m)] = 140m/s

f = v / = 140 m/s / 0.30 m = 470 Hz

Types of Waves

• A transverse wave is a wave in which particles of the medium move in a direction perpendicular to the direction which the wave moves.– Example: Waves on a guitar string

• A longitudinal wave is a wave in which particles of the medium move in a direction parallel to the direction which the wave moves. These are also called compression waves.– Example: Sound– http://einstein.byu.edu/~masong/HTMstuff/

WaveTrans.html

What are two types of wave shapes?

• Transverse

• Longitudinal

• http://www.school-for-champions.com/science/sound.htm

Transverse Wave Type

Longitudinal Wave Type

Longitudinal vs Transverse

Other Waves Types Occurring in Nature

• Light: electromagnetic

• Ocean waves: surface

• Earthquakes: combination

• Wave demos:

• http://www.kettering.edu/~drussell/Demos/waves/wavemotion.html

• http://www.kettering.edu/~drussell/Demos/doppler/mach1.html

Properties of Waves

• Reflection occurs when a wave strikes a medium boundary and “bounces back” into the original medium.

• Those waves completely reflected have the same energy and speed as the original wave.

Types of Reflection

Fixed-end Reflection-

The wave reflects with

inverted phase.

Open-end Reflection-

The wave reflects with

The same phase.

www.iop.org/activity/education/Teaching_Resources

Refraction of Waves Wave is transmitted

from one medium to

another. Refracted waves may

change speed and

Wavelength Almost always is accompanied

by some reflection. Refracted waves do not

change frequency.

Sound - a longitudinal wave

• Sound travels through air about 340 m/s.• Sound travels through other media as well,

often much faster than 340 m/s.• Sound waves are started by vibration of

some other material, which starts the air vibrating.

• www.silcom.com/~aludwig/musicand.htm

Hearing Sounds• We hear a sound as “high” or “low” pitch depending on

the frequency or wavelength. High-pitched sounds have short wavelengths and high frequencies. Low-pitched sounds have long wavelengths and low frequencies. Humans hear from about 20 Hz to about 20,000 Hz.

• The amplitude of a sound’s vibration is interpreted as its loudness. We measure loudness (also known as sound intensity) on the decibel scale, which is logarithmic.

http://www.allegropianoworks.com/assets/rare_compress.jpg

Doppler Effect• The Doppler Effect is the apparent change in pitch of a

sound as a result of the relative motion of an observer and the source of a sound. Coming toward you a car horn appears higher pitched because the wavelength has been effectively decreased by the motion of the car relative to you. The opposite occurs when you are behind the car.

http://people.finearts.uvic.ca/~aschloss/course_mat/MU207/images/Image2.gif

Pure Sound

• Sounds are longitudinal waves, but they can be shown to look like transverse waves.

• When air motion is graphed in a pure sound tone versus position, we get what looks like a sine or cosine function.

• A tuning fork produces a relatively pure tone as does a human whistle.

Graphing a Sound Wave

Complex Sounds

• Because of superposition and interference, real world waveforms may not appear to be pure sine or cosine functions.

• This is because most real world sounds are composed of multiple frequencies.

• The human voice and most musical instruments are examples.

The Oscilloscope• With an Oscilloscope we can view waveforms. Pure tones will

resemble sine or cosine functions, and complex tones will show other repeating patterns that are formed from multiple sine and cosine functions added together. (Amplitude vs time.)

The Fourier Transform• The Fourier transform has long been used for

characterizing linear systems and for identifying the frequency components making up a continuous waveform. This mathematical technique separates a complex waveform into its component frequencies.

• The Fourier Transform´s ability to represent time-domain data in the frequency domain and vice-versa has many applications. One of the most frequent applications is analysing the spectral (frequency) energy contained in data that has been sampled at evenly-spaced time intervals. Other applications include fast computation of convolution (linear systems responses, digital filtering, correlation (time-delay estimation, similarity measurements) and time-frequency analysis.

Fourier Transform - showing “time domain” and “frequency domain”.

Superposition Principle

• When two or more waves pass a particular point in a medium simultaneously, the resulting displacement of the medium at that point is the sum of the displacements due to each individual wave.

• The waves are said to interfere with each other.

Superposition of Waves

• When two or more waves meet, the displacement at any point of the medium is equal to the algebraic sum of the displacements due to the individual waves.

Types of Interference

• If the waves are in phase, when crests and troughs are aligned, the amplitude in increased and this is called constructive interference.

• If the waves are “out of phase”, when crests and troughs are completely misaligned, the amplitude is decreased and can even be zero. This is called destructive interference.

Constructive Interference

Crests are

Aligned the waves are

“in phase”

Destructive Interference

Crests are aligned with troughs

Waves are “out of phase”

Constructive & Destructive Interference

Interference Problem: Draw the waveform from the two components shown below.

Standing Waves

• A standing wave is one which is reflected back and forth between fixes ends of a string or pipe.

• Reflection may be fixed or open-ended.

• Superposition of the wave upon itself results in a pattern of constructive and destructive interference and an enhanced wave. Let’s see a simulation

• http://www.5min.com/Video/The-Rubens-Tube-Frequency-of-Fire-1858291

Fixed-end standing waves - guitar or violin string

• Fundamental

• 1st harmonic• = 2L

• First overtone

• 2nd harmonic• = L

• Second Overtone

• 3rd harmonic• = 2L/3

http://id.mind.net/~zona/mstm/physics/waves/standingWaves/standingWaves1/StandingWaves1.html

Problem

• A string of length 12 m that’s fixed at both ends supports a standing wave with a total of 5 nodes. What are the harmonic number and wavelength of this standing wave?

• L = 4(1/2 ) = 2L/4 4th harmonic since it matches n = 2L/n for n = 4

• wavelength4 = 2(12m) / 4 = 6 m

Open-ended standing waves - flute & clarinet

= 4L

= (4/3)L

= (4/5L

= 2L

= L

= (2/3)L

physics.indiana.edu/~p105_f02/standing_waves_...

http://upload.wikimedia.org/wikibooks/en/3/32/

Fhsst_waves40.png• open ends one end both ends

– closed closed

Sample Problem• 12-30. a) Determine the length of an organ pipe that

emits middle C (262Hz). The air temp. is 21oC.• A) v = 331m/s + 0.6 m/soC(21oC) = 344m/s• A) = 2L v = ff L = v/2f = 344m/s/{2(262/s)]• L = 0.656m• B) What are the wavelength and frequency of the 1st

harmonic? • Frequency is 262 Hz• Wavelength is twice the length of the pipe, 1.31 m.• C) What is the wavelength and frequency in the traveling

sound wave produced in the outside air?• They are the same because it is air that is resonating in

the organ pipe: 262Hz and 1.31 m

Superposition of 2 sound waves http://www.ece.utexas.edu/~nodog/me379m/superposition.html

Resonance and Beats

• Resonance occurs when a vibration from one oscillator occurs at a natural frequency for another oscillator.

• The first oscillator will cause the second to vibrate.

• See next slide.

Resonance

• http://www.isd-dc.org/ISD-Wash/GIFS%20Pictures%20&%20Whatnots/tuningforkresonance.jpg

Beats

• The word physicists use to describe the characteristic loud/soft pattern that characterizes two nearly matched frequencies.

• Musicians call this “being out of tune”.

Beats

• When two sound waves whose frequencies are close but not exactly the same, the resulting sound modulates in amplitude changing from loud to soft to loud. This is called beat frequency and is shown by:

• fbeat = f 1 - f 2

Diffraction

• Bending of a wave around a barrier

• Diffraction of waves combined with interference of the diffracte waves causes “diffraction patterns”.

• Here is an example using a “ripple tank”.

• http://www.falstad.com/ripple/

Double-slit or multi-slit diffraction

• micro.magnet.fsu.edu/.../doubleslit/• Remove frame 

Single Slit Diffraction

• n = s sin n -- dark band number -- wavelength (m) s -- slit width (m) -- angle defined by central band, slit, and

dark bank

Sample Problem• Light of wavelength 360 nm is passed

through a diffraction grating that has 10,000 slits per cm. If the screen is 2.0 m from the grating, how far from the central bright band is the first order bright band?

Sample Problem• Light of wavelength 560 nm is passed through two

slits. It is found that, on a screen 1.0 m from the slits, a bright spot is formed at x = 0, and another is formed at x = 0.03m. What is the spacing between the slits?

Sample Problem• Light is passed through a single slit of width 2.1 x 10-6

m. How far from the central bright band do the 1st and 2nd order dark bands appear if the screen is 3.0 m away from the slit?

Mathematical Description of a Traveling Wave

• Y = A sin ( t + x )

• Y dependent of x and t; y(x,t) or “y of x & t”

• If the - sign is used, wave is traveling in +x direction

• A is amplitude of the wave(omega) is angular frequency ( = 2f) angular wave # ( = 2 k, k = 1/)

Other forms

• Important features of the wave: amplitude, frequency f (through ), period T (which is 1/f = 2/wavelength ( = 2/) and wave speed v (which is f = /k)

• y = Asin2[ft + (1/x] or

• y = Asin(2/)(vt + x)

Sample Problem:

The vertical position y of any point on a rope that supports a transverse wave traveling horizontally is given by the equation

y = 0.1 sin (6 t + 8 x) Find:

amplitude: 0.1

angular frequency: = 6 s -1

frequency: f = / (2 ) = (6 s -1 )/ (2 ) = 3 Hz

angular wave number: k = / (2 ) = 8 m -1) / ((2 ) = 4 m-1

wavelength: = 1/ k = 1/ (4 m-1) = 0.25 m

period: T = 1/f = 1 / 3 Hz = 0.33s

wave speed: v = f = 0.25m (3 Hz) = 0.75 m/s

Assignment

• P15/MC1-4

Sound Level

• Intensity: Rate at which sound waves transmit energy is measured in energy per unit area: watt/m2 or watt/cm 2

• Intensity level or loudness level, • B = 10log I/Io where Io = 1x10-12 w/m2

• or 1x10-16 w/cm2

More math…

= 10 log I

Io 10-16

w/cm 2

Intensity level = 10 log Intensity / threshhold of hearing

We all don’t hear the same, so this is a comparative measurement in decibels

Flow chart for problem

• If I = 4.7 x 10^-10 w/cm^2:

• 10xlog(4.7 2nd EE -10 / 1 2nd EE -16) = 66.7 dB

• If I = 2.9 x 10^-3 w/cm^2:

• 10xlog(2.9 2nd EE -3 / 1 2nd EE -16) = 135 dB

Problem

• Now we are going backwards from intensity level (dB) to intensity (w/cm2)

• If the intensity level is 83 dB, convert that to intensity in w/cm2.

• B = 10 log I / Io get to a working eqtn:

• B /10 = log I / Io

• Log-1(B/10) = Log-1(log I/Io)

• Log-1(B/10) = I/Io

• Let’s say that the intensity level of a sound is 25.3 dB. What is the intensity of the sound in w/cm2?

• B = 10 log I / Io

• 25.3 dB = 10 log (I/10-16 w/cm2)

• 2.53 = log I – log 10-16

• 2.50 + log 10-16 = log I

• 2.50 – 16 = log I

• 25.3 dB = 10 log (I/10-16 w/cm2)• 2.53 = log I – log 10-16

• 2.50 + log 10-16 = log I what power do you raise 10 to, to get 10-16?

• 2.50 – 16 = log I• Adding on the left -13.5 = log I• Raise 10 to the -13.5 power by this sequence:• 2nd 10x (-13.5) 3.16 x 10-14 w/cm2 = I

Doppler Effect

• Doppler Effect is the apparent change in frequency as a result of relative motion between the source of a sound and an observer.

•

• f’ frequency heard by observer• f frequency of source• v velocity of sound in air• vd velocity of detector• vs velocity of source

Sample Problem

• A source of 4 kHz sound waves travel at 1/9 the speed of sound toward a detector that’s moving at 1/9 the speed of sound, toward the source.– a. what is the frequency of the waves as

they’re received by the detector?– b. how does the wavelength of the detected

waves compare to the wavelength of the emitted waves?

= v + 1/9 v x f = 5/4 f = 5/4 (4 kHz) = 5 kHz

v - 1/9 v

+ sign on top as detector moves toward source

-sign on bottom as source moves toward det.

-Frequency is shifted up by a factor of 5/4, the will shift down by the same factor.

d =s / 5/4 = 4/5 s

Sample

• A person yells, emitting a constant frequency of 200 Hz, as he runs at 5m/s toward a stationary brick wall. When the reflected waves reach the person, how many beats per second will he hear? (Use 343 m/s for the speed of sound.)

Determine what f will be reflected and heard by the runner. person is source and wall is detector

fwall = v f

v - vrunner

Reflected sound wave (no change in f): wall is the source and runner is the detector.

frunner = v + v runner f

v

Combine these two formulae: frunner = v + vrunner f

v - vrunner

f = (343+5)m/s (200Hz) = 206 Hz

(343-5)m/s

Beat frequency is fbeat = 206Hz - 200 Hz = 6 Hz.

p. 343 Princeton

Doppler Effect for Light

• wps.prenhall.com/.../ch25_SWA/images/8.gif

• c = 3 x 10 8 m/s

• u = speed of the source and the detector.

www.sv.vt.edu/.../class95/physics/doppler.gif

Periodic Motion …• Motion that repeats itself over a fixed and

reproducible period of time

• An example is a planet moving about its sun. This is called the period (T) or year of the planet

• Mechanical devices on earth that have periodic motion are useful timers and are called oscillators.

Simple Harmonic Motion• Attach a weight to a spring, stretch the spring

past its equilibrium point and release. The weight bobs up and down with a reproducible period, T

• Plot position vs time to get a graph that resembles a sine or cosine function. The graph is “sinusoidal”, so the motion is referred to as simple harmonic motion.

• Springs and pendulums undergo simple harmonic motion and are referred to as simple harmonic oscillators.

Graph Analysis

Graph Analysis

Oscillator Definitions

• Amplitude– Maximum displacement from equilibrium– Related to energy

• Period– Length of time required for one oscillation

• Frequency– How fast the oscillator is oscillating– f = 1/T in Hz or s-1

Problem 11-5• An elastic cord vibrates with a frequency

of 3.0 Hz when a mass of 0.60 kg is hung from it. What is its frequency if only 0.38 kg hangs from it?

Springs

• Springs are a common type of simple harmonic oscillator.

• Our springs are “ideal springs”, which means: They are massless

They are both compressible and extensible.

• They will follow Hooke’s Law.

F = -kx

Hooke’s Law Review

The force constant of a spring can be determined by attaching a weight and seeing how far it stretches.

Period of a Spring

T = period (s)

m = mass (kg)

k = force constant (N/m)

Sample Problem

• Calculate the period of a 350 g mass attached to an ideal spr5ing with a force constant of 35 N/m.

Sample Problem• A 300 g mass attached to a spring

undergoes simple harmonic motion with a frequency of 25 Hz. What is the force constant of the spring?

Sample Problem• An 80 g mass attached to a spring hung

vertically causes it to stretch 30 cm from its unstretched position. If the mass is set into oscillation on the end of the spring, what is its period?

Combinations of Springs

• In parallel, springs work together

• In series, springs work independently

What do you think?

Does this combination of springs act as parallel or series?

Answer: parallel !

Problem

• If you want to double the force constant of a spring, you

• A. double its length by connecting it to another one just like it.

• B. cut it in half.

• C. add twice as much mass.

• D. take half of the mass off.

• Answer: B

Conservation of Energy

• Springs and pendulums obey conservation of energy

• The equilibrium position has high kinetic energy and low potential energy.

• The positions of maximum displacement have high potential energy and low kinetic energy.

• Total energy of the oscillating system is constant.

Problem 11-20A block of mass m is supported

by 2 identical parallel vertical

springs, each with spring stiffness

constant k. What will be the

frequency of vibration?

AF

mgmg

BF

AF

BF

x

A B A B0 yF F F mg F F mg

A B A B A B2 2y netF F F F mg F kx F kx mg kx F F mg kx

General form of a restoring force producing SHM with spring constant of 2k

Sample Problem

Spring Problem

Another spring problem

Pendulums

• Pendulums can be thought of as simple harmonic oscillators.

• The displacement needs to be small for it to work properly.

Conservation of Energy

• Pendulums also obey conservation of energy.

• The equilibrium position has high kinetic energy and low potential energy.

• The positions of maximum displacement have high potential energy and low kinetic energy.

• Total energy of the oscillating system is constant.

Pendulum Forces

2005 Pearson Prentice Hall Fig 11-12

Period of a Pendulum

T -- period (s)

l -- length of string (m)

g -- gravitational acceleration (m/s2)

Problem:• Predict the period of a pendulum consisting

of a 500 g mass attached to a 2.5 m long string.

Problem:

• Suppose you notice that a 5 kg weight fixed to a string swings back and forth 5 times in 20 seconds. How long is the string?

Last problem!

• The period of a pendulum is observed to be T. Suppose you want to make the period 2T. What do you do to the pendulum?