EK Physics Ch5 Waves: Sound and Light

Question 1

simple harmonic motion

Answer 1

a specific type of oscillation

• in particular it happens as a result of a “restoring force” that is proportional to your displacement from equilbirum
• ex. spring, has an equilbrium length and when we stretch the spring out; that is we stretch it out some delta X force of spring pulls ups back the other way, and that force from teh spring is equal to K times delta x, it is proportional to distance from equilibrium, if compress will push me back hte other way will be K delta x how much have stretched or compressed the spring
• WHY DOES THIS MATTER= if pull object out and release it get oscillations!

Question 2

simple harmonic motion 2

Answer 2

MAIN THING TESTED= conservation of energy and energy transfers importantly not momentum!!! its the energy of object conserved not the momentum! what will happen? well at beginning I stretch it out, all energy in potential energy, I have potential energy of the string* what will happen? object gets pulled to teh left and starts speeding up, eventually it will pass through equilbirum position at equilbiurm position there is no longer any potential energy of the string, instead all that KE of the block no more elastic potential energy, now I do have a speed; block over shoots equilbrium position

delta x should be the same, same size displacement; amplitude of oscillation* oscillations keep going and keep going because total energy conserved, just converting from kinetic to potential to kinetic to potential back and forth back and forth*

Question 3

period of oscillations

Answer 3

• amount of time it takes for one complete cycle*
• notice not the amount of time it takes to return to equilbrium position, fully stretched out compressed b ack to eq position back to fully stretched out, have to go one full cycle
• equation= period for a spring= 2pie square root mass/k
• for spring

Question 4

period of oscillations for pendulum

Answer 4

Question 5

REMEMBER

Answer 5

The frequency of a sound wave does not depend on the medium. In other words, the frequency of the wave in propane is the same as its frequency in air or glass.*** find wavelength by v=F X wavelength

Question 6

mechanical waves

Answer 6

obey the laws of classical mechanics and require a medium or substance through which to travel

ex sound waves (longitudinal**) and waves on a string (traverse**) are mechanical!

Question 7

electromagnetic waves

Answer 7

do not require a medium through which to travel; they can propagate in cavuo, or in a vacuum. LIGHT is an ex of an electromagnetic wave*

remember= sound requires a medium to travel through*

Question 8

velocity

Answer 8

• describes the distance over which the wave travels per unit time and is determined by the medium through which the wave travels
• v=FX wavelength

Question 9

wavelength

Answer 9

• measure of distance from any point in teh wave to the point where the wave begins to repeat itself
• usually expressed in meters or nanometers

Question 10

frequency

Answer 10

• is the number of wavelengths that pass a fixed point in one second
• it is the inverse of time and expressed in s^-1 or hertz (Hz)
• describes how often a vibration occurs
• the frequency of a wave is determined by the frequency with which the wave source vibrates
• it does not change as a wave moves from one medium to another**

Question 11

REMEMBER=

Answer 11

• frequency is determined by the source of the wave
• velocity is determined by the medium*** through which it travels
• as a wave travels within a medum, it has a fixed velocity determined by the characteristics of the medium; when a wave travels form one medium to another its velocity changes accoridng to the features of the new medium
• electromagnetic waves such as light are able to travel in the absence of a medium
• when they do travel through a medium they are affected just like other types of waves. The velocity of light in different media wil be discussed in detail later in this lecture in context of refraction

Question 12

period

Answer 12

time it takes the wave to travel the distance of one wavelength adn is the receiprocal of frrequency*** T= 1/f

it is the number of seconds required for one wavelength to pass a fixed point. This period fo a wave describes the length of time the wave source requires to compelte on virbatory cycle. Just as frequency does not change when a wave moves from one medium to another, neither does the period***

Therefore= velocity = wavelength/T

Question 13

v= square root B/p

Answer 13

• velocity of a sound wave given by this equation where B is the bulk modulus of a medium is a measure of elasticity (resistance to change in shape) and pdensity of the medium is a measure of the inertia, resistance to change in motion*
• these are two characteritics of a given medium determine teh velocity of waves traveling through it!: elasticity and inertia
• equation does not mention frequency or wavelength, becuase velocity of a wave within a medium depends only on the characteristics of the medium, not the characteristics of the wave
• sound waves of diff frequencies, set by the wave source will still have the same velocity within a medium, posisbl becuase two waves have distincr wavelengths
• the bulk modulus of solids is far higher than for gases because tehre are more + stronger intermolecualr bonds btw their molecuels** a disturbance in on emoelcuel will qucikly be propagated and disturb other moelcules
• for a gas, which has far weaker and fewer intermolecular interactions a disturbance will not be propagated as easily*

Question 14

if temp inc…

Answer 14

• if temperature does change it can effect velocity of a sound wave traveling through the medium
• within a gas velocity inc with temperature
• influence of temp inducates that the random velocity fo the gas meolcules is a limiting fator for the velocity of a sound wave
• greater the temp, the greater the random velocity, and the greater the sound wave velocity
• the velocity of a sound wave through a gas is on teh order of magntide of (but slightly less than) the random velocity of its moelcules
• velocity of sound wves in a gas is limited by the average speed of the moelcules iwthin that gas**
• sound waves move more quickly through hot gases than through cold gases*

Question 15

heavier mediums…

Answer 15

tend to slow waves down, while stiffer mediums tend to speed waves up

since a wave must move the medium in order to pass through it, the inertia of the medium (its resistance to motion) tends to slow it down

Question 16

consider sound waves traveling from air to water….

Answer 16

Since water is denser than air one might htink it should slow sound waves. Water more than makes up for its higher density though with a much greater bulk modulus, and sound waves travel significantly faster in water***

In general:

velocity sound in solid> velocity sound in l> velocity sound in gas***

Question 17

summary of v=fwavelength

Answer 17

• frquency and period are determined by wave source adn do not change as a wave passes from one medium to another
• velocity is determined by medium through which the wave travels, adn velcoity cahgnes as a wave moves throguh a boundary btw media
• with this equation, for a given freqquency, an inc in velcoity inc wavelength or dec in velcoity dec wavelength

Question 18

constructive inteference

Answer 18

• occurs when the sum of hte displacements results in a greater displacement, gets bigger**

Question 19

destructive interference

Answer 19

• occurs when the sum of the displacmemnts results in a smaller displamcement, gets smaller!!!
• after passing through each other, waves that interfere wil revert to their original shape unaffected by the interference

Question 20

“in phase”

Answer 20

• whether interference is constructive or destructive depends on the PHASE of the wave, which realtes to its wavelength, frequency and place and time of origin
• think of it as a shift on graph
• two waves that have the same wavelength and beign at hte same point and time are said to be in phase
• two waves that havve the same wavelength but travel different distances ot arrive at the same point will be out of phase unless that distance is some multiple

Question 21

sound

Answer 21

• transfer of energy through oscillations btw high and low pressure
• as a sound source vibrates back adn forth, ti does work on its surroundings, creating regions of high and low density
• sound becomes audible when oscillations in pressure within a certain frequency range cause the sensory elements of the ear to vibrate, triggering the transmission of electrical impules to the brain

Question 22

pitch

Answer 22

measure of how high or low a note sounds, correlates with frequency; a high note has a high pitch and high frequency

Question 23

intensity level

Answer 23

• a measure of loudness, describes how intense a sound seems to be
• artifical scale for intensity level (B) has been created based upon a lagrithmic scale of intensities, untis fo this scale are decibels (dB) Relationship btw B and I is given in this equation= B= 10log (I/I0)
• where I0 is the threshold intensity of human bearing, the lowest intensity audible by the typical human
• although intensity is measured on a linear scale, humans do not perceive it that way
• ex sound waves created by the rustling of leaves are abotu 10x more intense than thsoe created by normal breathing, but we do not experience them as 10 times louder**
• decibel system created to rpvoidue

Question 24

decibel system

Answer 24

• created to provide an intuitive system that reflects how our human brain perceives the varying intensity of sound
• if intensity inc by a factor of 10, the decibels incease by teh addition of 10 decibels** in other words an inc in intensity from 30 W/m2 to 3000 W/m2 is equivalent to an inc of 20 decibels
• 2 zeros were added to intensity, so I add 20 decibels to the decibel levle
• if i add 3 zeros to the itnensity, I would have aded 30 decibels to the decibel level and so on***
• change in I, X 10= + 10
• X10² = +20
• X 10³ = +30
• X 10⁴= +40
• X 10⁵ = +50

Question 25

humans threshold

Answer 25

• represents the min sound intensity that can be heard, for individuals with normal hearing that is around 10^-12 W/m^2
• we cannot hear vs what we can hear depends on freuqnecy and intensity, frequency is one fo the factors that contributes to intensity adn its square is directly related to teh intensity fo sound**
• audible, infrasonic or ulatra sonic(can’t hear it beyond our sonic) sensory receptors respond to frequencies btw 20 to 20,000 hz
• waves with frequencies in that range are called audible waves, waves with frequencies above this range are called utlrasonic

Question 26

ultrasound machine

Answer 26

• probe of machine contains crystals that vibrate adn generte high frequency, ultrasonic sound waves when subject ot a high frequency alternating current
• waves reflected off boundaries within the body return to the ultrasound probe, returnign waves cause crytal to vibrate generating a current that can be processed by the ultraosund machine
• based on time ti takes reflected waves to return to the probe, an ultrsound machine determines the distance of hte boundary from the probe
• based on intensity of reflected waves, the machine infers the relative densities of the media on either side of hte boundary
• the greater the difference in density, the greater the intensity of reflected sound
• in this wave, information provided by reflected sound waves can be used to generate sonographic images*

Question 27

physical characteristics of sound….

Answer 27

Both ptich and intensity level are ways of relating physical characteristics of sound to the way that we perceive it. Frequency (a physical property) is peceived as pitch; intensity (a physical property) is perceved as intensity level or “loudness”***

Question 28

resonance

Answer 28

• Standing waves cause the string to resonate or vibrate at its natural frequency or resnonant frequencys
• Since velocity is constant for a given medium, the equaton v=frequency X wavelength, can be used to find the resonant frequency for any given wavelength. All mechanical structures have natural frequencies at which they resonante. If an outside driving force is applied
• Condition where the natural frequnecy and dirivign frequency are equal is also called resonance
• Resonatign string reveals the driving force to be the reflected wave

Question 29

resonance in pipes and strings

Answer 29

• resonance will most likely appear in this context
• in teh case of a string, one or both ends may be fixed to a point
• for a pipe, one or both ends may be open
• if a string is fixed at both ends as on a guitar the ends of the string will be nodes. Those points are held in place and cannot vibrate
• For a string secured at only one end, an anntidnode will be located at the unfixed end becuase, unlike a fixed point, this point wil be in motion
• if a standing wave has been estbalished no point on the string will have a larger displacement than the unfixed end so its displacement will equal to teh amplitude of the wave

Question 30

attentuation

Answer 30

• resonance is impacted by the fact that real waves undergo attentuation (damping)
• which is the decrease in intensity of a wave propagating through a medium. One cause of attenutation is reflection. Attenutation of sound waves due to reflection makes it difficult to create ultrasonic images of a bone’s interior. Because boen is so much denser than teh surrounding tissue, almost all of hte ultrasound waves are relfected at its boundary. As a result, v few sound waves continue into adn reflect off of hte interior of the bone.
• Another possible cause of attentuation is spreading; recall that as a sound wave spreads overa larger and larger area, its intensity is reduced. Yet, another cause of attenuation is absorption which is described later in this lecture.
• when a wave attenuates, its intensity decreases
• frquency is constant, so its the amplitude adn velocity that decrease as intensity decreases. the loss of energy of teh wave is most evident in its loss of amplitude***

Question 31

doppler effect

Answer 31

1. when source or observe is moving toward the other, observed frequency is higher than the source frequency b/c the subsequent wave front is emitted closer to the observe and observer perceives a higher frequency, distance btw the two less
2. when soruce of observer is moving away from the other, the observed frequency is lower than the source frequency; distance traveled is greater than ti wold be if both were stationary. Each wave rfont is emitted farther from teh observer, and hte observer perceives a lower frequency
3. source frequency**= frequency at which the waves are emitted= doesnt change!
4. only the observed frequency changes!!

Question 32

doppler effect 2

Answer 32

1. recall that speed of a wave is determined by the medium in which it travels. once emitted, a sound wave with a given speed will continue trveling at tha tspeed, regardless of hte source. becuase wave spee dis constnat, a change in distance traveled will change the time interval btw wave fronts!
2. the system is possible b/c velocoty is a vector, once you label direction positive for one vector, it must be so for all vectors@!
3. in doppler equation c doenst have ot be spee dof light (340), easy way to solve problem follow these 3 steps:
   
   1. assume the observer is not moving
   2. if hte source is moving toward the object, label that direction negative adn use the minus sign for Vs
   3. check the direction that the observer is moving. if the direction is the same, use the same sign for vo; if not, use hte opposite sign for vo

Question 33

doppler effect 3

Answer 33

1. tips for signs–> when a soruce or observer is moving towards the other, Fo> Fs, therefore the factor by which fs is multiplied must be greater than one. So in this instance, we add vo to c in teh numerator and/or substract vs from c in teh denominator
2. When teh numerator is alrger thant c and/or the denominator is smaller than c, the ratio in the equation (c+/- v0/ c+/- vs) is greate rthan one!
3. on the other hand, when a soruce or observer is movign away fromt eh otehr, fo< fs, and the factor by which fs is multiplied is less than one. In this case, we subtrct vo frm c in teh numerator and or add vs to c in teh denominator***
4. Another way to figure it out= only change in this equation i stha thte signsin the denimaotor are flipped* now for hte source adn the object, use the top siogn if they are movign “towards” and hte bottom sign if they are movign “away”
5. ex a bat that is gaining on a fleeing bug and emitting utlrasonci cries
6. if you wanted ot calcualte the frequency of teh cries the insect expereicnes consdier the bat the source adn the bug the obejct. teh bat is moving toward the bug so use the negative sign on top in the denominator. the bug the obejct is movign way from the bat, so use the negative sign on the bototm in the numerator**

Question 34

doppler effect 4

Answer 34

can be approximatd by change f/fs= v/c

and change in wavelength/wavelengths = v/c

**remember v is the net speed at which the source and object are approachign each other. For objects movign in teh same direction, substract their individual speeds; for objects movign in opposite directions, add their individual speeds

when an observer and source move toward each other, the freq the observer perceives increases b/c it encounters more waves in a given amount of itme. Add change in freq to freq S

when observer adn source increase the distance btw them, the observer encounters fewer waves for each moment in time. Subtract delta f frm fs

fo= fs +/- change in f

wavelength o= wavelength s +/- change in wavelength

Question 35

beats

Answer 35

1. occur when two waves with slgiuthly different freq are superimposed. at some points they will be narly in phase and experience constructive interference, at other points out of phase and exerpeince destructive interference
2. interfernece btw emitted sound wave with the freuqnecy set by teh soruce adn hte returning sound wave with an altered frequency causes beats
3. these points will aternate at hte beat frequency, or the frequency equal to teh difference btw hte frequencies of the original two waves
4. use his equaiton absolute value [fo-fs]
5. beat frequency provides a measure of change in frequency, speed of a vehicle can then by calculated by equation= change f/fs= 2v/c
6. or v=c(change in f)/2fs

Question 36

beats 2

Answer 36

this is the diff btw observed freuqnecy and frequency ommitted by teh source

it can also represent hte difference in the frequencies of two sounds, like tuning forks that are close in pitch**

Question 37

shock wave

Answer 37

is a conical wave front produced when the velocity of the sound soruce exceeds the velocity of the sound wave*

Ratio of velocity of the source to teh velocity of hte wave is vs/v, is known as the Mach number. this inc as the velocity of the sound soruce inc. kike any soudn wave, a shock wave consists of oscillations between high and low pressures. when teh source mvoes faster than the speed of sound, many wave fronts overlap, generating a region of very low air pressure. Air moves in response to the presure gradient, and the rapid movement creates a loud sonic boom.

Question 38

law of reflection

Answer 38

theta 1 = tehta 2, angle is the same

key to this is that angles are not measured from te surface* we alwas measure our angles from what is called the normal line; normal because perpendicular to surface, imaginary line perpendicular to surface at hte point of contact***

if give angles to ground, must convert to angles relative to normal**

Question 39

refraction

Answer 39

• light “bends” and changes its path as a result of entering the new medium*
• this happens because the light is changing speed, the new medium equals a change in speed, because of the change in speed this is what is causing the light range to bend*
• so upon entering the new medium light ray will have a new path

Question 40

index of refraction

Answer 40

property of material, each material has its own property of refraction, it is measuring basically some people refer to it as the optical density of a material, how hard it is for light to travel through a material**

higher index of refraction means light travels slower* letter= n

as n inc, velocity dec

n=c/v

Question 41

index of refraction equation

Answer 41

n= c/v

c= speed of light in vacum, 3 x 10⁸

v= speed in a specific medium, so n will always be greater than 1, speed in vacuum is the largest the speed of light can be**

n> or equal to 1***

so we have an equation that tells us about relationship of theta 1 and theta 2 and how it depends on idnex of refraction fo two mateirals, n1 and n2

Question 42

common n values:

Answer 42

for air= n=1

for water= n=1.3

for glass = n=1.5

diamond very high = n=2.4

Question 43

what is relationship btw n and thetas for index of refraction?

Answer 43

Snell’s law

n1sin θ1 = n2sin θ2

n inc θ dec

if n1> n2 that n goes down then the other part theta has to go up, so that means light gets bent away from normal**

theta 1 is directly proportionalto theta 2*** if sin theta 1 goes up on one side changes thetas on other side*** changes n is inversely to change theat, inc one theta inc the other theta*

Question 44

n1 V n2 angle

Answer 44

Question 45

total internal reflection

Answer 45

n1>n2

critical angle- at this angle and above light can only be relfected**

some light above this angle will only be reflected! this will nto happen in the other case where n2> n1, all light goes fromf irst medium into teh second; in this case portion of the light that cannot travel from one medium into the next medium* so if you want to solve for the critical angle we can do it, only occurs when n1 is greater than n2, when start out in denser medium* so now can solve for it, n1sin θc = n2sin 90

now sin θc= n2/n1, now know n2 has to be greater than 1 becuase sign of any angle is less than 1! so cannot hav sin2 greater than sin 1**

*fiber optics cabel works on this principle that I have my cabel with a very high n value, so I can shine a light over here at that end and it cannot escape cable bounces down enitre length of hte cable none of it can get out* advantage of this is you can send signals at basically the speed of light with very very little loss of energy*

Question 46

why diamonds sparkle*

Answer 46

only light that can escape has to have an angle ery veyr close to normal., when light exits gem goes in very straight direction, every other ligth bouncing around until coming straight at eye when look at gem face

Question 47

mirror

Answer 47

• light hits my body, scattered in all directions, why if can look at my foot, light goes into my eye allows me to perceive my foot
• can see myself in a mirror for the same basic region, except light coming off my foot gets scattered toward the mirror and then hitting the mirror* and then bouncing back to my eye* JUST reflection must obey* law of reflection** to say that I have a theta 1 adn theta 2 adn they have to be equal to each other*
• when look into mirror look like you are behind wall, when light comes to eye eye doesn’t know light has been reflectd, eye preceives light as having traveled in a straight line always* so what I see then is I see my foot way back behind the mirror, my image ends up back there
• for plane mirror angle of incidence and anlge of reflection are equal!
• do= di

Question 48

curved mirrors

Answer 48

curved mirrors and curved lens focus light! what they are useful for, focus light towards or away from a point

Question 49

real image

Answer 49

• light is really there! light is actually present!!!
• inverted always!
• need a screen to view them!!! need somethign to reflect light back to eye, otherwise travels through space and you cannot see it

Question 50

image summary

Answer 50

converging

1. object is very far away, so far that it is greater than 2x focal length away- in this case do > 2f real reduced image
2. if between 2f> do> f

between focal point but not super far away, case we did before= real and enlarged

3. f > do, said that we would have a virtual enlarged image*
4. diverging- always does the same thign! ALWAYS makes virtual and reduced**

Question 51

myopia

Answer 51

nearsightedness

CANNOT SEE FAR AWAY

can see near**

image is overconverged in front of retina

focal length of lens is too short, so need to correct it becuase light is being converged to much* so need diverging correction so that the light can come in and be spread out focused onto retina, notice diverging correction will have a negative focal length therefore will have a negative power becuase power is 1/focal length this is important becuase glasses perscription is -3, -4, -5; -13 that perscription is telling you the power of the lens being used!!! it is negative because it is diverging*

Question 52

hyperopia or hypermetropia

Answer 52

why ppl hold menu far away to read it, cannot see close!! opposite problem

farsightedness

image is behind retina

focal length +

power = + (1/f)