Physics 9 — Monday, January 27, 2014

I Handed out HW2 Friday, due this Friday in class. Problemsfrom Ch16 (waves) and Ch17 (2D/3D waves/sound).

I Skim Chapter “G14” (heat) Wednesday. It’s short!

I I reserved DRL 3W2 from 6:30pm to 9:30pm on Wednesdays(usually Zoey) and Thursdays (usually Bill) for HW help.

I Today: finish sound waves. Rest of week: fluids.

Beats

I On Friday, we saw two sound waves interferingconstructively/destructively as a function of position, as weadjusted the relative path length with a trombone-like slide.

I We can see something similar happen as a function of time.

When you add together two tones of comparable amplitude andslightly different frequency, instead of hearing two separate tones,your ear hears the average of the two frequencies, modulated bythe difference of the two frequencies:

sin(ω1t) + sin(ω2t) = 2 cos

(ω1 − ω2

2t

)sin

(ω1 + ω2

2t

)

For frequencies f1 ≈ f2, the “beat frequency” that you hear is atthe difference frequency, |f1 − f2|.

(Two-tuning-forks demo. Graphic on next page.)

11 Hz sine, 10 Hz sine, and sum of the two sines

Doppler effect (illustrate)

http://www.physics.purdue.edu/class/applets/phe/dopplereff.htm

http://www.phy.ntnu.edu.tw/ntnujava/index.php?topic=21

f ′ = f

(c ± vobserver

c ∓ vsource

)Where f is the emitted frequency and f ′ is the observed frequency.Upper sign (f ′ > f ) when moving toward one another; lower sign(f ′ < f ) when moving away from one another.

At the end of the term, we’ll learn about radio waves and visiblelight, which are other examples of waves in three dimensions. Thefrequency (i.e. color!) of light is shifted for distant stars: inferrelative velocity. Police radar uses Doppler shift of radio wavesbounced off of your car to infer your speed.

Reminder: Guitar string frequencies

We combined three ingredients:

I An integer number of half-wavelengths must fit in length L ofwire, since ends are clamped: nλ/2 = L

I Speed of transverse waves propagating on string isc =

√T/µ where µ is mass per unit length of string

I λ = c/f relates wavelength, wave speed, and frequency

We got

fn =n

2L

√T

m/L=

n

2L

√T

µ

More massive wire → lower f . Higher tension → higher f .Make string shorter with fingertip → higher f .

Question: if you change L with your fingertip, is it m that staysthe same, or is it µ that stays the same?

Sound: longitudinal waves propagating in a medium.

csound =√

Bρ =

√bulk modulusmass/volume = 343 m/s in air at 20◦C.

(331 m/s in air at 0◦C.)

Rule of thumb: stiffer → higher fresonant & higher wave speed;more massive → lower fresonant & lower wave speed.

Question: Helium has a much lower density (lower mass/volume)than air, but a similar bulk modulus. If you exhale helium insteadof air, do you expect your airway to resonate at a higher or lowerfrequency than usual? Hint: the wavelength is fixed by the size ofmy airway, which doesn’t depend on whether I exhale air or helium.

Since sound waves are longitudinal waves, I have one moreanimation (this time I wrote it in Processing), to illustratelongitudinal waves. Sketch name is “wave longitudinal 1”

Intensity Level

The human ear is sensitive to a huge range of sound intensity.

Commonly encountered soundintensities differ by many powers often! So it makes sense to define alogarithmic measure of sound intensity.

“Sound level” a.k.a. “intensity level”β is measured in decibels.

β = 10 log10

(I

10−12W/m2

)

It’s good to understand the use of decibels to quantify soundintensity, because you will encounter decibels in architecturalacoustics, soundproofing, etc.

“Threshold of human hearing” = I0 ≡ 10−12 W/m2

For intensity I (unit = W/m2), the sound level (unit = dB) is

β = 10 log

(I

Io

)dB

And remember

log(1000) = +3

log(100) = +2

log(10) = +1

log(1) = 0

log(0.1) = −1

log(0.01) = −2

log(0.001) = −3

etc.

Question

If the intensity is I = 10−5 W/m2, what is the sound level β?

Question

If the intensity is I = 10−5 W/m2, what is the sound level β?

β = 10 log

(10−5 W/m2

Io

)dB

β = 10 log

(10−5 W/m2

10−12 W/m2

)dB

β = 10 log(107)

dB

β = 70 dB

One thing you’ll see in soundproofing discussions is that a ratioI2/I1 of two intensities (like inside vs. outside the closed window)can be expressed as a difference in decibels:

β1 = 10 log

(I1Io

)dB

β2 = 10 log

(I2Io

)dB

β2 − β1 = 10

(log

(I2Io

)− log

(I1Io

))dB = 10 log

(I2I1

)dB

(Remember that log(A/B) = log(A)− log(B).)

So you can use ∆β = 10 log(I2/I1) dB for the ratio after/beforepassing through a wall or window. −∆β for a wall, window, etc., iscalled “transmission loss” by the acoustics cognoscenti.

Question

If a wall permits only 110000 of the sound intensity to reach the

other side, what is the change ∆β in the sound level from one sideto the other?

Notice that a 30 dB window (pretty typical for a window, atfrequencies around 500 Hz) will reduce 70 dB busy street trafficdown to 40 dB level of quiet radio.

What fraction of sound intensity (power/area) is transmitted by a30 dB window?

A pretty good wall will reduce sound intensity (for frequenciesaround 500 Hz) by about 50 dB. That reduces the 65 dB soundlevel of someone talking on the other side of the wall to about15 dB, which is quieter than a whisper.

What fraction of sound intensity (power/area) is transmitted by a50 dB wall?

Soundproofing!

When a sound wave hits a wall, most of it is reflected back. Someof it is transmitted through to the other side of the wall.

Have you ever lived in a poorly soundproofed room?

Putting up a wall as a sound barrier is not so different fromputting a mass onto the middle of the wave machine.

Soundproofing ↔ wave machine

Boundary conditions at the point where I add the mass:

I displacement is same on both sides of mass

I tension × (change in slope) = mass × acceleration

Working out the math∗, I get fraction of intensity transmitted is

Itransmitted

Iincident=

1

1 +(πµfcρ

)2≈(

cρ

πµf

)2

where ρ is density of air, c is wave speed, µ is mass/area of wall.

(On the wave machine, ρ becomes mass/length and µ becomes theadded mass in the middle.)

* positron.hep.upenn.edu/wja/phys009_2014/masslaw.pdf

The “mass law” of preventing sound transmission

For a single wall or window, the transmitted intensity scales like

Tintensity ∝1

µ2f 2

So higher frequencies are easier to stop than lower frequencies, andmore mass-per-unit-area stops more sound.

Doubling the mass/area blocks 4× the intensity: reduction of 6 dBper doubling of mass. 10 log10(4) ≈ 6

Much better trick: use two independent layers of materialseparated by several inches (e.g. notice that a studio sound boothwindow has two widely separated layers of glass).

On previous slide’s graph, notice:

I Two separate panes (same total amount of glass) stop hugelymore sound than one pane.

I Low-frequency cutoff is quite sensitive to separation distance.

I Double-wall trick suffers from “resonances” at frequencies forwhich an integer number of half-wavelengths just fits betweenthe two walls. At these resonant frequencies (sometimes called“coincidence dips” in acoustics books), sound gets through.

In an open field, if I go 3 times as far away from a sound source,by what factor is the intensity reduced? What if I go 10 times asfar away from a sound source?

So the transmitted intensity is reduced by a factor of 4 (a 6 dBreduction in sound level) if

I you double your distance from the sound source;

I you double the mass-per-unit-area of the wall or window;

I the frequency of interest is an octave higher (when partiallyblocked by a typical wall or window).

I Going farther away decreases intensity like 1/r2.

I More massive partitions typically reduce intensity like 1/(ρL)2.

I Higher frequencies are easier to stop than lower frequencies:typically intensity transmitted through wall or window scaleslike 1/f 2.

Double-wall or double-window trick can get much better reductionof noise transmitted through wall or window.

For comparison, typical earplugs have transmission loss ≈ 30 dB

Physics 9 — Monday, January 27, 2014

I Handed out HW2 Friday, due this Friday in class. Problemsfrom Ch16 (waves) and Ch17 (2D/3D waves/sound).

I Skim Chapter “G14” (heat) Wednesday. It’s short!

I I reserved DRL 3W2 from 6:30pm to 9:30pm on Wednesdays(usually Zoey) and Thursdays (usually Bill) for HW help.

I Today: finish sound waves. Rest of week: fluids.