Physics 140 Sound

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Physics 140

Sound

Chapter 12 Sound waves

Sound is composed of longitudinal pressure waves.

wave propagaon

Compression è when Compression Compression parcles come together

Rarefacon Rarefacon Rarefacon è when

parcles are farthest apart 2 Physics 140, Prof. M. Nikolic Sound waves

Described by the gauge pressure – diﬀerence between the pressure at a given point and average pressure

3 Physics 140, Prof. M. Nikolic The speed of sound waves

T Let’s remember Chapter 11 and the speed of string waves v = µ The speed of sound B waves v = B is the bulk modulus of the medium ρ and ρ its density.

The speed of sound Y v = waves in thin rods Y is the Young’s modulus of the ρ medium and ρ its density.

All other equaons from Chapter 11 sll apply. 2π v = f λ k = 4 Physics 140, Prof. M. Nikolic λ Conceptual queson – Speed of sound Q1

Given the informaon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density Speed (GPa) (kg/m3) (m/s) Aluminum 70 2700 Brass 61 8400 Copper 14 8900 Mercury 27 13600

A. aluminum B B. brass v = C. copper ρ D. mercury

5 Physics 140, Prof. M. Nikolic Conceptual queson – Speed of sound Q1

Given the informaon in the table below, which substance would have the highest speed of sound? Substance Bulk Modulus Density Speed (GPa) (kg/m3) (m/s) Aluminum 70 2700 5.1 x 103 Brass 61 8400 2.7 x 103 Copper 14 8900 1.25 x 103 Mercury 27 13600 1.4 x 103

A. aluminum B B. brass v = C. copper ρ D. mercury

6 Physics 140, Prof. M. Nikolic B Exercise: Speed of sound in gold v = ρ What’s the speed of sound in a gold ingot? Density of gold is ρ = 19,360 kg/m3 and its bulk modulus is 220 GPa.

What’s given: B ρ = 19,360 kg/m3 v = B = 220 GPa = 220x109 Pa ρ

9 220×10 Pa 3 v = = 3.38 ×10 m/s 19300 kg/m 3

7 Physics 140, Prof. M. Nikolic Characteriscs of sound

Sound can travel through any kind of maer, but not through a vacuum.

The speed of sound is diﬀerent in diﬀerent materials; in general, it is slowest in gases, faster in liquids, and fastest in solids.

The speed depends somewhat on temperature, especially for gases.

T v = v0 T V0 is the absolute speed at 0 absolute temperature T0. e.g. v (air) = 343 m/s at T = 200 C. 0 0 8 Physics 140, Prof. M. Nikolic The speed of sound waves T Speed of sound in ideal gases: v = v0 T0 SI units of temperature: [K]

To convert temperature from 0C to Kelvins (SI unit of temperature) you just need to add 273 K.

T(in K) = TC (in C)+ 273

For the speed of sound in air (and only in air) we can use an approximate formula:

v = (331+0.606TC ) m/s

0 where Tc is the air temperature in C.

9 Physics 140, Prof. M. Nikolic Exercise: A lightning ﬂash A lightning ﬂash is seen in the sky and 8.2 seconds later the boom of thunder is heard. The temperature of the air is 12.0 0C. (a) What is the speed of sound in air at that temperature? What is given: Since we are looking for the speed of sound in air, we can use the t = 8.2 s approximate formula: 0 TC = 12 C v = (331+ 0.606TC ) m/s v = (331+0.606⋅12 0C) m/s = 338 m/s (b) How far away is the lightning strike? d = vt d = (338 m/s)(8.2 s)= 2800 m = 2.8 km

d 2800 m −6 t = t = 8 = 9.3×10 s v 3.0×10 km/s 10 Physics 140, Prof. M. Nikolic The amplitude and intensity

The intensity of a wave is the energy transported per unit me across a unit area.

It can be given as a funcon Or by looking at the amplitude of a pressure amplitude (P0) of the displacement of parcles P2 I = 0 P = ωvρs 2ρv 0 max

ω is the angular frequency of sound v is the speed of sound ρ is the density of the medium

smax is the maximum displacement Units of intensity: W/m2 11 Physics 140, Prof. M. Nikolic Loudness and decibels

Loudness of a sound is measured by the logarithm of the intensity.

Sound level (intensity level) is measured in decibels (dB) and is deﬁned:

I β = (10 dB)log I0

I0 is taken to be the threshold of hearing: −12 2 I0 = 1.0 × 10 W/m

Sound level is not the same as the intensity of the sound wave! 12 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: I r = 25 m β = (10 dB)log β = 71 dB -12 2 I0 I0 = 10 W/m

We can solve for the intensity of the sound wave: I I β I β β = (10dB)log log = =1010dB I I 10dB 0 0 I0

71 dB I =10−12 W/m 2 ⋅1010 dB =1.3×10−5 W/m 2

13 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. a) What is the rate at which sound energy is being produced by the loudspeaker, assuming it to be an isotropic source? What is given: r = 25 m β = 71 dB And we calculated: I = 1.3x10-5 W/m2

We need to ﬁnd the rate at which sound energy is being produced – that’s power, but we know intensity of sound and distance.

OK, let’s go back to Chapter 11 and use the equaon for intensity of the isotropic source:

P 2 I = 2 P I4 r 4πr = π P = (1.3×10−5 W/m2 )4π (25 m)2 = 0.10 W

14 Physics 140, Prof. M. Nikolic Exercise: Decibels from a speaker The sound level 25 m from a loudspeaker is 71 dB. b) What is the pressure amplitude of a sound wave? Density of air is 1.2 kg/m3 and the speed of sound is about 343 m/s. What is given: r = 25 m P2 β = 71 dB I = 0 3 P I 2 v ρ = 1.2 kg/m 2ρv 0 = ⋅ ρ v = 340 m/s And we calculated: -5 2 −5 2 3 I = 1.3x10 W/m P0 = 1.3×10 W/m ⋅2⋅1.2 kg/m ⋅343 m/s = 0.1 Pa

15 Physics 140, Prof. M. Nikolic Standing waves of sound

All of these musical instruments are eﬀecvely sealed oﬀ on one end and open on the other.

The air molecules right next to the closed end are essenally not moving è displacement nodes

The molecules at the open end are exposed to the atmosphere. This gives them the greatest freedom of movement è displacement annodes

16 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube closed at one end - L

First harmonic - fundamental Second harmonic Third harmonic

λ1 5 4 L 4L 3λ1 4 λ1 = ⇒ λ1 = L = ⇒ λ = L L = ⇒ λ1 = L 4 4 1 3 4 5

17 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube closed at one end - The general result for standing waves in a tube open at one end and closed at the other is 4L λn = n where n = 1, 3, 5,… (odd values only!!)

v nv fn = = = nf1 λn 4L f1 is the fundamental frequency.

Or 4L λn' = 2n'−1 where n’ = 1, 2, 3,…

Frequencies have only odd mulples of the fundamental one 18 Physics 140, Prof. M. Nikolic è Even harmonics are not present Exercise: A bugle What are the ﬁrst 3 harmonic frequencies emied by a bugle that has a tube 50 cm long aer it was le out in the cold, cold night (assume the temperature is -12oC)? What is given: (2n −1)v L = 50 cm = 0.5 m fn = = (2n −1) f1 0 TC = -12 C 4L

Since we are looking for the speed of sound in air, we can use the approximate formula:

v 331 0.606T m/s 0 = ( + C ) v = (331+0.606⋅(−12 C)) m/s = 323.7 m/s

(2n −1)v v n = 1 - fundamental frequency: f 323.7 m/s 1 = = f1 = =161.8 Hz 4L 4L 4⋅0.5 m

f = (2 ⋅ 2 −1) f = 3 f n = 2 2 1 1 f2 = 485.6 Hz

n = 3 f3 = (2⋅3−1) f1 = 5 f1 f3 = 809.3 Hz

19 Physics 140, Prof. M. Nikolic Pressure variaons - Tube closed at one end - Displacement node at the closed end corresponds to the pressure annode è molecules at that locaon are the ones geng maximally compressed

Displacement annode at the open end corresponds to the pressure node è molecules at that locaon are the ones geng maximally stretched out

20 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube open at both ends -

First harmonic - fundamental Second harmonic Third harmonic

λ1 3 2 L 2L 2λ1 λ1 = ⇒ λ1 = L = ⇒ λ = L L = ⇒ λ1 = L 2 2 1 2 3

21 Physics 140, Prof. M. Nikolic Standing waves of sound - Tube opened at both ends -

The general result for standing waves in a tube open at one end and closed at the other is

2L v nv λ = fn = = = nf1 n 2L n λn f1 is the fundamental frequency.

Unlike the pipe open at only one end, all harmonics are present.

Since both ends are open, air molecules at each end have maximum possible displacements è displacement annodes

22 Physics 140, Prof. M. Nikolic Pressure variaons - Tube opened at both ends -

Displacement annodes at the open ends correspond to the pressure nodes è molecules at that locaon are the ones geng maximally stretched out

23 Physics 140, Prof. M. Nikolic Exercise: An organ pipe An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0 0C. a) What is the fundamental frequency for this pipe at 200 C?

What is given: v nv v f1,0 = 382 Hz f = = = nf f = 0 n 1 1 T0 = 0 C λn 2L 2L 0 T20 = 20 C 0 The speed of sound in air at: T0 = 0 C is v0 = 331 m/s 0 T20 = 20 C is v20 = 343 m/s Note that in case you need to ﬁnd the speed of sound at any other temperature not provided in the table, you would need to use the eqn. for the speed of sound (Slides 8 and 9).

v We can ﬁnd the rao: 0 0 The fundamental frequency at 0 C: f1,0 = 2L v20 f v 1,20 2L v20 0 20 = = The fundamental frequency at 20 C: f1,20 = v 2L f1,0 0 v0 343 m/s 2L f1,20 = 382 Hz = 396 Hz 24 Physics 140, Prof. M. Nikolic 331 m/s Exercise: An organ pipe

An organ pipe that is open at both ends has a fundamental frequency of 382 Hz at 0 0C. b) How long is the pipe?

What is given: v0 v f1,0 = 382 Hz f = 0 0 1,0 2L = T0 = 0 C 2L f 0 1,0 T20 = 20 C 331 Hz L = = 0.43 m 2⋅382 m/s

You could also choose the other equaon, for f1,20 and v20 and sll get the same answer.

25 Physics 140, Prof. M. Nikolic Timbre

Standing wave is a superposion of many standing wave paerns with diﬀerent frequencies

• Fundamental frequency (1st harmonic) • Overtones (higher harmonics)

The same note sounds differently on different instruments è fundamental frequency might be the same but the overtones appear in different intensies

Timbre (tone quality)

27 Physics 140, Prof. M. Nikolic Beats Let’s consider two diﬀerent waves, each with the same wave speed but with slightly diﬀerent frequencies (and, hence, wavelengths).

The superposion of the waves will produce a pulsaon called beats.

28 Physics 140, Prof. M. Nikolic Beats If the beat frequency exceeds about 15 Hz, Beat frequency is Δf = f1 − f2 the ear will perceive two diﬀerent tones instead of beats.

Example: A violin is tuned by adjusng the tension in the strings. Brian’s A string is tuned to a slightly lower frequency than Jennifer’s, which is correctly tuned to 440 Hz. What is the frequency of Brian’s string if beats of 2 Hz are heard when they bow the strings together?

Δf = f1 − f2 f2 = f1 − Δf = 438Hz

What would happen if the temperature where Brian seats drops from 250 C to 200 C?

T The frequency of Brian’s violin would change: v = v T 0 f = f0 T0 T0 (25 + 273) f = 438 = 442 Hz (20 + 273) 29 Physics 140, Prof. M. Nikolic The Doppler eﬀect The Doppler eﬀect occurs when a source of sound is moving with respect to an observer.

• When source is moving toward an observer the sound has a higher frequency and shorter wavelength • When a source is moving away from an observer the sound has a lower frequency

and longer wavelength 30 The Doppler eﬀect 1 v = f λ = λ sound T

λ = vsoundT

In the same me T, the source will move distance d toward the Crest observer:

d = vsourceT The observer will hear a sound wave with a diﬀerent wavelength:

λ0 = λ − d = λ − vsourceT λ " v % λ0 = λ − vsource source λ0 = λ$1− ' vsound # vsound &

31 Physics 140, Prof. M. Nikolic The Doppler eﬀect

In general, when both the observer and the source are moving: ! $ vsound ± vobserver f0 = # & f " vsound ± vsource %

When the observer is moving When the source is moving towards towards the source use the observer use

vsound + vobserver vsound – vsource

When the observer is moving away When the source is moving away from the source use from the observer use

vsound - vobserver vsound + vsource

Note, that if the observer or the source are staonary, just set their speeds to zero. 32 Physics 140, Prof. M. Nikolic Exercise: Rocked sleds A maniac on a rocket sled is moving along a train track. She hears a train whistle at a frequency of 1200 Hz. The maniac, being a big fan of trains, knows that a train travels at an average speed of 30 m/s and the whistle is normally at a frequency of 720 Hz. What is the speed of the maniac if she starts sledding towards the train? Assume that the speed of sound is 340 m/s.

! $ vsound ± vobserver 1200 Hz f0 = # & f 720 Hz " vsound ± vsource %

340 m/s 30 m/s

Since the observer (maniac) is moving towards the source (train): vsound + vobserver

Since the source (train) is moving towards the observer (maniac): vsound − vsource " v + v % f = $ sound observer ' f 0 v − v # sound source & 33 Physics 140, Prof. M. Nikolic Exercise: Rocked sleds A maniac on a rocket sled is moving along a train track. She hears a train whistle at a frequency of 1200 Hz. The maniac, being a big fan of trains, knows that a train travels at an average speed of 30 m/s and the whistle is normally at a frequency of 720 Hz. What is the speed of the maniac if she starts sledding towards the train?

" % vsound + vobserver vsound + vobserver f0 f0 = $ ' f = # vsound − vsource & vsound − vsource f

f0 vsound + vobserver = (vsound − vsource ) f

1200 Hz 340 m/s+vobserver = (340 m/s−30 m/s)= 516.67 m/s 720 Hz

vobserver =176.67 m/s

34 Physics 140, Prof. M. Nikolic Echolocaon

Sound waves can be sent out from a transmier of some sort; they will reﬂect oﬀ any objects they encounter and can be received back at their source. The me interval between emission and recepon can be used to build up a picture of the scene.

If the sound takes me t to go from the source (bat, dolphin,…) to the object and back then it travels the distance: t d = v ⋅ sound 2

And all equaons for calculang the speed of sound sll apply! 35 Physics 140, Prof. M. Nikolic Exercise: A sonar A boat is using sonar to detect the boom of a freshwater lake. If the echo from a sonar signal is heard 0.540 s aer it is emied, how deep is the lake? Assume the lake’s temperature is uniform and at 25 0C. What is given: t = 0.54 s t d = vsound ⋅ T = 25 0C 2

From table 12.1 in our textbook → the speed of sound in freshwater is 1493 m/s. d =1493 m/s⋅0.27 s = 403.1 m

36 Physics 140, Prof. M. Nikolic How do I solve for the logarithm:

I I β I β β = (10dB)log log = =1010dB I0 I0 10dB I0

General rule:

C logB A = C A = B

37 Physics 140, Prof. M. Nikolic Summary

The speed of sound B waves v = B is the bulk modulus of the medium ρ and ρ its density.

The speed of sound Y waves in thin rods v = Y is the Young modulus of the medium ρ and ρ its density.

2π v f k = = λ λ

T The speed depends V is the absolute speed at v = v0 0 temperature - gases T0 absolute temperature T0. 38 Physics 140, Prof. M. Nikolic Summary

Intensity if a sound wave as a 2 P0 funcon of a pressure amplitude I = 2ρv

P0 = ωvρsmax Sound level (intensity level) I I = 1.0 × 10−12 W/m2 β = (10dB)log 0 I0 The Doppler eﬀect T(in K) = T (in C)+ 273 C ! v ± v $ f = # sound observer & f Beat frequency is Δf = f − f 0 1 2 " vsound ± vsource % 39 Physics 140, Prof. M. Nikolic