Table Of ContentS1CS
Part II
Sections VI-X
Section VI
Sound and
Doppler Effect
Section VII
Fluids and Solids
Section VIII
Electrostatics and
Electromagnetism
Section IX
Electricity and
Electric Circuits
Section X
Light and Optics
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Table of Contents
6. Sound and Doppler Effect
Production and Detection of Sound page 6
Speed of Sound page 7
Pitch and Intensity page 9
Doppler Effect page 12
Resonance in Strings and Pipes page 16
Sound and Doppler Effect Review Questions page 23
Detailed Answer Explanations page 28
Sound and Doppler Effect Practice Exam page 33
Detailed Answer Explanations page 44
7. Fluids and Solids
Fluid Properties page 55
Fluids in Motion page 66
Solids page 78
Fluids and Solids Review Questions page 83
Detailed Answer Explanations page 89
Fluids and Solids Practice Exam page 93
Detailed Answer Explanations page 104
8. Electrostatics and Electromagnetism
Electrostatics page 115
Electric Fields page 120
Electromagnetism page 131
Electrostatics and Electromagnetism Review Questions page 141
Detailed Answer Explanations page 147
Electrostatics and Electromagnetism Practice Exam page 151
Detailed Answer Explanations page 170
9. Electricity and Electric Circuits
Currents page 173
Voltage and Resistance page 176
Capacitors page 184
Electric Circuits page 188
Alternating Current page 199
Electricity and Electric Circuits Review Questions page 201
Detailed Answer Explanations page206
Electricity and Electric Circuits Practice Exam page211
Detailed Answer Explanations page 222
10. Light and Optics
Electromagnetic Radiation page 233
Reflection and Refraction page 238
Optics page 246
Interference Phenomena page 259
Light andOptics Review Questions page 267
Detailed Answer Explanations page 272
Light and Optics Practice Exam page 277
Detailed Answer Explanations page 288
Sound and
Doppler Effect
Physics Chapter 6
'source A
A)
Compressed Wave: Elongated Wave:
Shorter X Longer X
Higher/ Lower/
Elongated Wave://;^^--^N^ComPrtesed Wave:
LongerK / / /O-—0\\\ Shorter^
Lower/ / / / /0-=^\\\\ HiSher/
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by
the Berkeley Review
Sound and Doppler Effect
Selected equations, facts, concepts, and shortcuts from this section
O Important Equations
6 = 10 log (Jsound/j ) /shifted = /unshiftedVwave±Vreceiver (Doppler shift)
° vwave ± vsender
/beat =l/l - /2I (Heard as a"destructive beat") Fixed Strings: Xn =2L/n .\ /n =nv/2L
Open Pipes: Xn =2L/n •'• /n =nv/2L closed P*!** ^n =4L/n •'• /n =nv/4L> where nis 1. 3» 5» etc-
© Important Concept
Echolocation (Using the time to detect echoing sound waves to establish position of an object)
Echo Detector
If material is not uniform, refraction
of the waves can create blind spots. IP
tj (time for reflection to return from 1st boundary):
used to determine distance to first interface
Waves normal to surface minimize §§te m
the impact of refraction. Non-parallel IHSf g^|Me|ir reflection to return from 2nd boundary):
surfaces are more difficult to analyze :M*s£ ^;v:u|e3^Metermine distance to second interface
than parallel ones. FS^ •.-. .v-;;
m
© Doppler EffectCalculation Approach
1) First determine whether the objects are moving towards or away from one another
If the objects are moving towards one another, then the frequency of the wave will increase. This
will require either adding to the numerator in the Doppler equation (if the receiver is in motion) or
subtractingfrom the denominator in the Dopplerequation (if the sender is in motion). If the objects
are moving away from one another, then the frequency of the wave will decrease. This will
require either subtracting from the numerator in the Doppler equation (if the receiver is in motion)
or adding to the denominator in the Doppler equation (if the sender is in motion).
2) Second determine if the scenario involves one direction transmission or an echo
If it's a one-direction wave only, then you need only plug into the Doppler equation once. If it's
an echo, then you'll need to plug into the Doppler equation a second time, using the first solved
frequency as the input frequency and reversing the roles of the sender and receiver from the first
wave. In other words, the sender of the outbound wave becomes the receiver of the inbound wave.
Physics Sound Production and Detection of Sound
Sound
In Chapter 5, we studied standing waves on a string. In this chapter, we shall
expand our coverage of waves to include sound waves, one example of a
longitudinal wave.
Production and Detection of Sound
What is sound and how is it produced? Sound is simply a longitudinal wave that
can be produced from the disturbance of a solid, a liquid, or a gas. A sound wave
arises from the vibrations and collisions of molecules within a particular
substance. Even though these molecules maintain their same average position
within that substance, the collision of one molecule with another molecule results
in the transmission of energy in the form of a wave. Sound waves cannot travel
through a vacuum.
Sinusoidal waves are the simplest type of sound waves. They have a defined
wavelength, amplitude, and frequency. In order to detect a sound wave, the
mechanical vibrations of that wave must be analyzed in terms of intensity and
frequency.
The human ear is an excellent sound detector and can distinguish sounds that
differ in frequency from one another by as little as 0.3%. A simplified version of
the human ear is shown in Figure 6-1.The pinna (outer ear) collects sound waves
and funnels them into the auditory canal. As the sound waves press on the
tympanic membrane (eardrum), they cause the membrane to vibrate. This
vibrational energy is transmitted through the ossicles, three small bones
(hammer, anvil, and stirrup) connecting the tympanic membrane to the oval
window of the inner ear.
Pinna Tytripinric
Codi Itii (iiiicni led)
(outer car) I .
(eardrum)
Oval ^_ Basilar
Auditory window/ membrane
canal
Low
High
frequencies
frequencies
Figure 6-1
As the ossicles vibrate, they amplify the vibrational energy set in motion at the
tympanic membrane. This vibrational energy is passed from the oval window to
a fluid in the cochlea of the inner ear. The fluctuations in pressure in this fluid
cause hair cells in the basilar membrane of the cochlear duct to move. Movement
of the hair cells generates a nerve impulse that is sent to the brain and is
interpreted as sound. High frequency sounds are generated by nerves closest to
the oval window, while low frequency sounds are generated by nerves farthest
away from the oval window.
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Physics Sound Speed of Sound
Speed of Sound
When a sound wave moves forward, it compresses the medium in front of it.
This results in compressional waves moving outward in all directions from the
source of the sound. Consider the compressional wave in Figure 6-2. Note that as
the wave moves outward, there are regions of high density (or high pressure)
and low density (or low pressure) within the medium. Alsonote that the volume
of space associated with the regions of high density decreases relative to the
volume of space associated with the regions of low density. The region of
reduced pressure is also called a rarefaction.
Propogation of wave
•=>
•/:
high low
density density
Figure 6-2
Solids, Liquids, and Gases
The speed with which particles can return to their original position following a
disturbance in the medium dictates the speed of sound in that medium. The
strength of the forces between the molecules in a given medium and the density
of the particles in that medium determine how fast sound will travel in that
medium. The forces between the molecules in a solid are greater than the forces
between the molecules in a liquid or a gas. The stronger the force between any
two molecules, the greater the restoring forcebetween those molecules. The more
rapidly molecules are restored to their original condition, the more rapidly they
can participate in another compression wave, and the faster the propagation of
sound in that particular medium. Asa result, we see the general trend for speed
orsound as vsoun(j jn solid > vsouncj mliquid > vsoun£i jn gas.
The speed of longitudinal sound waves in various ideal gaseous media can be
calculated from the formula in equation (6.1):
(6.1)
In this equation, pis the density (of agas), Pis the pressure (of a gas), yis Cp/Cv
(where Cp is the molar heat capacity at constant pressure, and Cv is the molar
heat capacity at constant volume for a gas), R is the ideal gas constant, T is
temperature in kelvins, and Mis the molecular mass of the gas.For a monatomic
gas y = 1.67, for a diatomic gas y = 1.40, and for a polyatomic gas y = 1.33. This
means that for a given pressure and gas density, a monatomic gas will permit
sound to travel through it at a greater speed than a diatomic or polyatomic gas
will. For example, if the temperature of the air is 0 °C, the pressure is 1 arm (or
1.01 x105 N/m2), and the density is1.29 kg/m3, then the speed ofsound inair is
331 m/s. In this case, we are making the assumption that the air is composed of
gases in the diatomicstate,becauseair is predominantly N2 and O2.
(1.40)(1.01 xl05N/m2
v = =331 m/s
1.29 kg/m3
Copyright © by The Berkeley Review The Berkeley Review
Physics Sound Speed of Sound
In Table 6.1, we see the speed of longitudinal sound waves in various media.
These values may vary from textbook to textbook, depending on whether the
author assumed isothermal or adiabatic conditions. For the MCAT, knowing that
thespeed ofsoundin air at roomtemperature isabout340 m/s is goodenough.
Table 6.1
Material Temperature / Pressure Speed (ms)
Air (0°C, 1 arm) 331
Air (20°Q 1 arm) 340
Air (100°C, 1 arm) 386
Hydrogen (0°C, 1 arm) 1286
Helium (0°C, 1 arm) 965
Water (lake) 1497
Water (sea) 1531
Aluminum 5100
Iron 5130
Glass 5500
Granite 6000
In Table 6-1, note that the speed of sound in metals and liquids is much greater
than the speed of sound in air. This is due to the fact that metals and liquids are
not as compressible as gases. Therefore, the restoring forces for a metal or a
liquidare much greater than they are for air.
Example 6.1a
Which of the following changes will increase the speed of sound in a fixed
volume of diatomic hydrogen gas that is enclosed in a fixed-volume container?
I. Increasethe temperature of the gas.
n. Replace the hydrogen molecules with diatomic oxygen molecules, while
holding the temperature constant.
III. Increase the intermolecular attraction between the molecules.
A. I and II only
B. II and III only
C. I and III only
D. I, II, and HI
Solution
A general rule regarding the speed of sound is the following:
v oc Restoring Force or Molecular Kinetic Energy
v Molecular Inertia
Increasing the temperature of the gas in this enclosed volume will increase the
molecular kinetic energy without changing the molecular inertia (i.e., mass
density). Thus, Statement I is true, and choice B is wrong. Replacing the
hydrogen with oxygen will noticeably increase the massbut, presumably, change
the molecular interaction forces little. This would reduce the speed of sound,
invalidating Statement II. We are left to conclude that choice C is the correct
choice. Statement III must be valid, if the above equation is to hold true.
The best answer is choice C.
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PhysiCS Sound Speed of Sound
Example 6.1b
An acoustic scientist measures the properties of three unknown molecules,
labeled A, B, and C. He records the following relationships for the molecular
weights and cohesive forces of the molecules:
MWa > MWb = MWc and Fa < Fr < Fc
These relationships hold true, regardless of the state of the material. Noting these
results, which of the following relationships could he predict regarding the speed
of sound in these materials? (Assume all solids have the same lattice structure
and all gases have the same molarity.)
I. Speed of sound is greater in Gas A than Gas B, when both are at the same
temperature and pressure.
II. Speed of sound is greater in Solid C than Solid B, when both are at the
same temperature.
III. Speedofsound in Liquid C is greater than that in LiquidA,when Liquid A
is hotter than Liquid C.
A. II only
B. I and III only
C II and III only
D. I, II, and III
Solution
The speed ofsound within a medium depends on the ability ofthe particles that
make up the medium to quickly return to their original position following a
disturbance. The restoring speed of the particles depends on the force holding
particles together (cohesive force) and the mass of the particles. Lighter particles
move faster than heavier particles, so speed of sound is faster in mediums that
are made of lighter particles, assuming all other factors are equal. The question
also mentions forces. The stronger the forces between particles (stronger
intermolecular forces), the faster the particles can return to their original position
following a disturbance, which makes the speed of sound increase. Because
Particle Cis the lightest and has the strongest intermolecular forces, the speed of
sound will be fastest in Medium C when the temperature is uniform. This makes
Statement II a true statement, which eliminates choice B.
Statement III is invalid, because without knowing the temperature (or knowing
thatifs equal for all mediums), we donothave enough information todetermine
the relative speeds of sound. This eliminates choices Cand D, leaving only choice
A. Statement I is false, because Particle A has the weakest intermolecular forces
and is the heaviest particle, resulting inMedium A having the smallest speed of
sound. Thisfurthersupports theelimination of choices Band D.
This makes choice A the best answer.
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