Review Sheet for exam 3
Note that this is just a general “off the top of my head” review sheet. Anything we talked about
in class is fair game. Do not ask me “Do I have to know this?” If I talked about it in class, you
have to know it.
Beginning of Quantum Mechanics and Atomic Structure
Thomson’s “Plum Pudding Model”: Atom is a sphere with the positive charge “smeared out” all through
the sphere, with electrons (negative charges) like “chunks” embedded in the positively charged “smear”.
Rutherford - nuclear model of atom (replaces Thomson's "plum pudding model"). States that atom
is positively charged nucleus surrounded by electrons, mostly empty space (true). Used
scattering experiments . . .
Planck - Solved blackbody radiation problem - energy comes in "chunks" (quanta). Required
new constant of nature - Planck's constant h (= 6.636  10-34 J-s).
Einstein - Solved photoelectric effect - photons E = hf , work function , (see handout).
Old question, reopened: is light a particle, or a wave, or what??
Hydrogen spectrum - N. Bohr - quantization of angular momentum, radial orbits, energy
stationary states (these states didn't radiate electromagnetically)
Book stuff: 40% book, pgs 230 – 235
Back Questions: I don’t really have any good back questions on this stuff. Sorry.
Wave Motion
For specific definitions of various wave stuff, see the PPs. In particular, v = f λ
Waves on a string: v = sq rt ( T ÷ µ )
µ ≡ mass per unit length, T ≡ Tension in the string.
Standing waves on a string: fn = ( n ÷ 2L ) × [ sq rt ( T ÷ µ ) ]
; n = 1, 2, 3, 4, . . .
Standing waves in a closed pipe:
fn = ( n ÷ 4L ) × vsound
; n = 1, 3, 5, 7, . . .
Standing waves in an open pipe:
fn = ( n ÷ 2L ) × vsound
; n = 1, 2, 3, 4, . . .
For all standing waves . . .n = 1 is the fundamental, or the 1st harmonic.
The frequency of the nth harmonic is given by fn = n f1
Book stuff: Chapter on Vibrations and Waves; Musical Sounds (appropriate sections only). These book
sections also are relevant to the topic of “Sound” (next page).
Back questions on general wave motion (you already saw some of these in prep for exam 1)
1.
A human with good hearing can hear sound waves with frequencies ranging from 20.0 Hz up to
20,000 Hz. If the speed of sound in air is 344 m/s, what is the range of wavelengths of sound
waves that a human can hear? ( λ’s between 0.0172 meters and 17.2 meters )
2.
The human ear canal is approximately 6.00 cm long. What frequency of sound
(vsound = 344 m/s) would generate a wave that has a wavelength equal to the length of the ear’s
canal? ( f = 5733 Hz)
3.
We did an experiment called “standing waves on a string”. In this experiment, we generated
waves on a string. (Duh.) Let’s say that the string is 2.00 meters long, and on that string we
can fit four full waves. We use an oscillator to produce the waves; the frequency of that
oscillator is 120 Hz. What is the speed of waves on the string?
( v = 60 m/s)
4.
If the mass of the string in problem 3 was 8.00 grams, what was the Tension in the string that
produced the pattern described in problem 4? If we produced this tension by hanging a mass
from the end of the string (as we actually did in lab), what was the necessary mass?
( T = 14.4 N, mhanging = 1.47 kg )
I don’t have any great MC questions here . . . sorry.
Sound
Sound waves are longitudinal pressure waves, with wavelengths and frequencies just like any
other wave - its velocity depends on air temperature - is roughly 340 m/s. (1100 ft/sec)
Amplitude ( 10−11 to 10−5 meters) - determines intensity (10−12 to 1 watt/m2) determines loudness (0 - 120 dB (decibels)).
Loudness = 10 log ( I ÷ I0 ) where I0 = 10−12 watt/m2
Pitch of sound related to frequency of the sound wave; f = 440 Hz is an A note; humans detect
between 20 and 20,000 Hz, many animals can hear higher frequencies (ultrasonic).
Timbre (“quality”) of sound is related to the harmonics that are in a sound. It is why an “A” note
from a clarinet sounds different from an “A” note from a flute.
f < 20 Hz is infrasonic ; f > 20,000 Hz is ultrasonic.
Relative velocity between source and listener - Doppler shift
1.
If the distance between the source and the observer is decreasing, the frequency heard by the
observer is larger than the frequency emitted by the source.
2.
If the distance between the source and the observer is increasing, the frequency heard by the
observer is smaller than the frequency emitted by the source.
fL = [ ( vsound ± vL ) ÷ ( vsound ± vS ) ] × fs
If speed of source is bigger than the speed of sound (supersonic) - get a shock wave (or bow wave)
Back questions on Sound (MC)
1.
The loudness of a sound is determined mainly by its
a) amplitude.
b) speed.
2.
c) frequency.
d) pitch.
An ambulance siren at rest puts out a sound with a frequency of 440 Hz. If the ambulance is
moving towards me at 40 m/s, I will hear a frequency of
a) 440 Hz.
b) 394 Hz.
c) 220 Hz.
d) 498 Hz.
3.
The amplitude of a 440 Hz sound wave at the threshold of hearing will be approximately
a) equal to its wavelength.
b) the size of a nucleus.
4.
The word ultrasonic refers to a sound that has
a) a speed faster that 1100 ft/sec.
b) an amplitude greater than 1 mm.
5.
b) equal to
b) equal to
d) unrelated to
c) the size of a protein molecule.
d) the size of the tympanic membrane.
c) Newton.
d) Hertz (cycle/second)
A sound wave can be best described as a
a) transverse pressure wave.
b) transverse bow wave.
10.
c) lower than
The unit best equipped to measure the human response to a sound's loudness is the
a) Watt per square meter.
b) decibel.
9.
d) unrelated to
At the threshold of hearing, the vibrations of the air molecules striking the eardrum are
a) 2000 .
b) smaller than the size of an atom.
8.
c) lower than
If the source of a sound and a listener are moving towards each other, the pitch detected by the
listener is always _______ the pitch of the source.
a) higher than
7.
c) a frequency greater than 20,000 Hz.
d) a loudness greater than 120 dB.
If the source of a sound and a listener are moving away from each other, the pitch detected by
the listener is ____________ the pitch of the source.
a) higher than
6.
c) twice the size of an atom.
d) one-tenth the size of an atom.
c) longitudinal pressure wave.
d) longitudinal bow wave.
Unlike light, sound
a) can propagate through air.
b) can propagate through water.
c) can propagate through a solid.
d) cannot propagate through a vacuum.
11.
While driving away from you at 40 m/s, I play a C note on my keyboard, which is loud enough
for all to hear. If a standard C note has a frequency of 512 Hz, at what frequency do I hear the
note?
a) 458.5 Hz
12.
b) 512.0 Hz
b) 512.0 Hz
c) 579.6 Hz
c) 100 to 1000 Hz.
d) 200 to 20,000 Hz.
Assume I set off a nuclear bomb on the moon. I would quickly see the explosion from a
spaceship in lunar orbit, but not hear it, because
a) light travels faster than sound.
b) there is no air on the moon.
15.
d) 624.3 Hz
The hearing frequency range of an average human non-rock star is from
a) 1 to 100 Hz.
b) 20 to 20,000 Hz.
14.
d) 624.3 Hz
In the situation described in question 11, at what frequency do you hear the note?
a) 485.5 Hz
13.
c) 579.6 Hz
c) the light is refracted around the moon.
d) sound waves don't travel in straight lines.
The horn on a standard car has frequency (at rest) of 440 Hz (an A note). You see a car, but
it is so far away from you that you can't tell by looking whether the car is moving towards you,
away from you, or standing still. The car blows its (rather loud) horn, and you hear a C note
(512 Hz). What is the car doing?
a) Moving away from you.
b) Moving towards you.
c) Not moving at all.
d) There is not enough information.
Answer Key:
1. A
8. B
2. D
9. C
3. D
10. D
4. C
11. B
12. A
5. C
13. B
6. A
14. B
7. B
15. B
More numerical problems on waves, standing waves, sound, and the Doppler Effect
1. A string has a length of 80.0 cm and a mass of 4.00 grams. It is held at a tension of 2,000 N.
a)
b)
c)
d)
What is the frequency of the fundamental standing wave on this string? ( f1 = 395 Hz )
How about the 3rd harmonic? ( f3 = 1186 Hz )
And the 5th overtone? ( 5th o.t. ≡ 6th harmonic . . . f6 = 2372 Hz )
Which (if any) of these sounds would be able to be heard by a human with normal hearing?
( All of them )
2. A closed pipe has a length of 0.450 meters. The fundamental frequency of this pipe is 220 Hz.
What is the speed of sound in the region of the pipe? ( vsound = 396 m/s . . . a bit high  )
3. If the pipe in problem 2 was open instead of closed, what would be the fundamental frequency of
the pipe? In this problem, use the speed of sound that you calculated in problem 2. ( f1 = 440 Hz )
4. If we can treat a dog whistle as an open pipe, how long should the pipe be for a dog to hear the
whistle but not you? There are many possible answers depending on the assumptions you make. Be
clear about your assumptions; if the physics is correct, so is your answer. ( f = 35 kHz, L = 4.85 mm )
5. A person is standing on a street corner holding a tuning fork that is vibrating at 512 Hz. (A “C”
note.) I run past the person at 6.00 m/s. If the speed of sound is 340 m/s
a)
b)
c)
d)
What frequency do I hear from the tuning fork as I approach the person on the corner?
What frequency do I hear from the tuning fork as I move away from the person on the corner?
What is the change in frequency that I hear as I run past the person?
What frequency does the person holding the tuning fork hear?
Ans: (a) f = 521 Hz; (b) f = 503 Hz; (c) ∆ f = − 18 Hz; (d) f = 512 Hz
6. This time I am standing on a street corner and an ambulance races by me while sounding its siren.
The ambulance driver hears the frequency of the siren as 600 Hz, and notes that his speed is 35.0 m/s.
Take the speed of sound as 340 m/s.
a) What frequency do I hear from the siren as the siren approaches me?
b) What frequency do I hear from the siren as it moves away from me?
c) What is the change in frequency that I hear as the siren moves past me?
Ans: (a) f = 669 Hz; (b) f = 544 Hz; (c) ∆ f = − 125 Hz
7. In part (c) of questions 5 and 6, the change in frequency is negative; that is, the frequency
decreases as I run past a stationary source, or, while I am stationary, the source moves past me. Will
this always be the case? Prove your answer without numbers!! This can be done using the (written)
rules for the Doppler Shift from this review sheet . ( Think about “distance increasing”, etc. )
These problems are from Principles of Physics, 4th Edition, by Raymond J. Serway
13−35 Standing at a crosswalk, you hear a frequency of 560 Hz from the siren of an approaching
ambulance. After the ambulance passes, you hear a frequency of 480 Hz. Determine the speed of the
ambulance. Take the speed of sound to be 340 m/s. ( vamb = 26.2 m/s )
13−57 (revised) A person is driving a very fast car towards a wall (don’t ask why). The speed of the
car is 60.0 m/s. The person is also holding a source of sound that emits a frequency of 1800 Hz when
it is at rest. With vsound = 340 m/s, what is the frequency of the sound heard by the driver? ( 2571 Hz )
a) directly from the source? ( f = 1800 Hz )
b) reflected off of the wall back to the driver? ( f = 2571 Hz )
14−22 The top string of a guitar has a fundamental frequency of 330 Hz when it is played without
pressing down anywhere on the string (length = 66.0 cm). If the guitarist presses her finger on the fret
board so that only two−thirds of the string is vibrating when she plays the string, what is the new
fundamental frequency that she hears? ( f1,new = 495 Hz )
14−24 A violin string has a length of 35.0 cm and is tuned to “concert G”, fG = 392 Hz. Where must
the violinist place her finger in order to play “concert A”, which is fA = 440 Hz? ( Lnew = 31.2 cm )
14−30 The fundamental frequency of an open organ pipe corresponds to middle C (f = 261.6 Hz). The
2nd overtone of a closed organ pipe has the same frequency. Take vsound = 340 m/s.
a) What is the length of the open pipe? ( Lopen = 0.65 meters = 65 cm )
b) What harmonic is being sounded in the closed pipe? ( 2nd o.t. of closed pipe is 5th harmonic )
c) What is the length of the closed pipe? ( Lclosed = 3.25 meters )
14−52 A string fixed at both ends has a mass of 4.80 grams, a length of 2.00 meters, and is held under
a tension of 48.0 Newtons. It is vibrating in its 2nd harmonic. This vibrating string causes vibrations of
the air molecules in contact with the string, creating a sound wave that has the same frequency as the
frequency of vibration of the string. If vsound = 340 m/s, what is the wavelength of these sound waves?
( λ = 4.81 meters )
14−54 A string is 40.0 cm long and has a mass per unit length of 9.00 grams/meter (careful!!). What
must be the tension in the string if its second harmonic has the same frequency as the first overtone of
an open pipe whose length is 1.75 meters? Take vsound = 340 m/s. ( T = 54 Newtons )
Schaum’s problems: Chapter 22, probs 25, 26, 27, 30, 35, 36, 37 and
Chapter 23, probs 34, 35, 36, 37, 43, 44 (SA only)
Detailed solutions to
these on sols sheets.
And some relativity problems . . .
Remember!! This relativity stuff is optional. But if you want more problems on it, let me know!
1. A muon is formed high in the atmosphere, 4600 meters above the earth’s surface (as measured by
someone on earth), and is moving towards the earth at v = 0.99c, again as measured by someone on the
earth. After moving that 4600 meters (according to Earth), the muon decays.
a) How long did the muon live before decaying, as measured by observers on Earth?
b) How long did the muon live before decaying, as measured by the muon?
c) How far did the muon think the Earth moved between the time it was formed and time it decayed?
2. A spaceship is moving from Earth to Melmac, which is 5.00 light years away and at rest with
respect to the earth. The pilot of the ship measures his ship to be 80.0 meters long. People on Earth
think that the ship is moving at 0.8 c.
a)
b)
c)
d)
How fast is the ship moving as measured by people on Melmac?
How long does the trip take, as measured by people on Earth?
How long does the trip take, as measured by the pilot?
What is the length of the ship, as measured by people on Melmac?
3. The Sun puts out 3.85 × 1026 Joules of energy every second. The sun has a mass of 1.99 × 1030 kg.
If the sun “dies” after all of its mass has been converted to energy, how long will the sun live? Express
your answer in billions of years, knowing that one year is 3.15 × 107 seconds. Incidentally, the sun
won’t actually use up all of its mass before “dying”.
Answers:
1.
With v = 0.99 c, we get γ = 7.089
(a) ∆tmuon, measured by earth = ∆tNP = 4600 meters ÷ ( 0.99 × 3 × 108 m/s ) = 15.5 µs.
(b) ∆tmuon, measured by muon = ∆tP = ∆tNP ÷ γ = 15.5 ÷ 7.089 = 2.18 µs
(c) Lmeassured by earth = LP = 4600 meters. Lmeassured by muon = LNP = LP ÷ γ = 4600 ÷ 7.089 = 649 meters
2.
(a)
(b)
(c)
(d)
With v = 0.8 c, we get γ = 1.67
Melmac and Earth are at rest wrt each other, so vmelmac = vearth = 0.8 c
∆tearth = DistEarth ÷ speedEarth = 5 c years ÷ 0.8 c = 6.25 years = ∆tNP
∆tpilot = ∆tP = ∆tNP ÷ γ = 6.25 ÷ 1.67 = 3.74 years
Pilot measures LP for ship => LP = 80 meters. Earth folks (and Melmac folks) measure LNP
ship. LNP = LP ÷ γ = 80 ÷ 1.67 = 48 meters.
Answer to #3 on next page . . .
3.
Sun puts out 3.85 × 1026 Joules per second. Assume it gets that energy by changing mass to
energy through E = mc2. Then the Sun must convert m = 3.85 × 1026 ÷ ( 3 × 108 )2 = 4.28 × 109 kg of
its mass to energy each second. The Sun’s mass is 1.99 × 1030 kg, so at a rate of 4.28 × 109 kg/sec, the
Sun will run out of mass in t = ( 1.99 × 1030 kg ) ÷ ( 4.28 × 109 kg/sec ) = 4.65 × 1020 seconds.
At 3.15 × 107 seconds/year, and with One Billion ≡ 1 × 109 years, the sun should live for
[ ( 4.65 × 1020 sec ) ÷ ( 3.15 × 107 sec/year ) ] ÷ ( 1 × 109 ) = 14,768 billion years.
The actual expected life of the sun is roughly 10 billion years, so this is clearly an incorrect calculation.
But it does give us some idea of how long we have!