Quiz Samples for Chapter 17 General Physics I May 25, 2020 Waves...

General Physics IQuiz Samples for Chapter 17

Waves - IIMay 25, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

17-1 Speed of Sound

1. (�) Sound waves are longitudinal mechanicalwaves that can travel through solids, liquids, orgases. The speed v of a sound wave in a mediumhaving bulk modulus B and density ρ is

v =

√B

ρ.

In air at 20◦C, the speed of sound is 343 m/s.

2. Figure illustrates several ideas that we shall use inour discussions.

Ray

Ray

Wavefronts

Point S represents a tiny sound source, called apoint source, that emits sound waves in alldirections. The wavefronts and rays indicate thedirection of travel and the spread of the soundwaves.

(a) (�) Wavefronts are surfaces over which theoscillations due to the sound wave have thesame value; such surfaces are represented bywhole or partial circles in a two-dimensionaldrawing for a point source.

(b) (�) The wavefronts are concentric spheresand spread out in three dimensions.

(c) (�) As the wavefronts move outward andtheir radii become larger, their curvaturedecreases. Far from the source, weapproximate the wavefronts as planes (or lineson two-dimensional drawings), and the wavesare said to be planar.

(d) (�) Rays are directed lines perpendicular tothe wavefronts that indicate the direction oftravel of the wavefronts.

3. (�) The speed of a sound wave is determined bythe transmitting medium.

4. (�) The difference between transverse andlongitudinal waves depends on the direction ofoscillation of the medium relative to the directionof propagation of the wave.

5. (�) A set of points on a wave that all have thesame displacement is a wavefront.

6. The speed of any mechanical wave, transverse orlongitudinal, depends on both an inertial propertyof the medium (to store kinetic energy) and anelastic property of the medium (to store potentialenergy).

(a) (�) The speed of a transverse wave along astretched string is

v =

√τ

µ,

where τ and µ are the tension and linear massdensity of the string.

©2020 KPOPEEE All rights reserved. Korea University of 11



(b) (�) We can generalize the speed formula for avibrating string into

v =

√τ

µ→

√elastic property

inertial property.

If the medium is air and the wave islongitudinal, we can guess that the inertialproperty, corresponding to µ, is the volumedensity ρ of air.

µ =mass

string length→ ρ =

mass

volume.

(c) (�) As a sound wave passes through air,potential energy is associated with periodiccompressions and expansions of small volumeelements of the air. The property thatdetermines the extent to which an element ofa medium changes in volume when thepressure (force per unit area) on it changes isthe bulk modulus B, defined as:

B = − ∆p∆VV

,

where ∆VV is the fractional change in volume

produced by a change in pressure ∆p. Thesigns of ∆p and ∆V are always opposite and,therefore, B is positive definite.

(d) (�) Both B and ∆p have a common physicalunit, newton per square meter, which is givena special name, the Pascal(Pa).

7. Derivation of

v =

√B

ρ.

Let the pressure of the undisturbed air be p andthe pressure inside the pulse be p+ ∆p, where ∆pis positive due to the compression. Consider anelement of air of thickness ∆x and face area A,moving toward the pulse at speed v.

Pulse

(a) (�) As this element enters the pulse, theleading face of the element encounters a regionof higher pressure, which slows the element tospeed v + ∆v, in which ∆v is negative.

(b) (�) This slowing is complete when the rearface of the element reaches the pulse, whichrequires a time interval

∆t =∆x

v.

(c) (�) The average net force on the elementwithin the region [x, x+ ∆x] during ∆t is

F = pA− (p+ ∆p)A = −∆pA,

where pA is the force from the left and−(p+ ∆p)A is the force from the right.Because ∆p > 0, the net force is to the left.

(d) (�) The volume of the element is ∆V = A∆x.The mass is

∆m = ρ∆V = ρA∆x = ρAv∆t.

(e) (�) The average acceleration of the elementduring ∆t is a = ∆v

∆t . Thus Newton’s secondlaw reads

F = −∆pA = (ρAv∆t)∆v

∆t= ∆ma,

which can be written as:

ρv2 = − ∆p∆vv

.




(f) (�) The ratio ∆vv can be expressed as

∆v

v=

A∆v∆t

Av∆t

=A∆(∆x)

A∆x

=∆V

V.

(g) (�) Thus

ρv2 = − ∆p∆VV

= B,

which leads to

v =

√B

ρ.

8. (�) The physical dimensions of the bulk modulusB is the same as that of the pressure.

17-2 Traveling Sound Waves

1. (�) As a property of a sound wave, frequencydetermines its pitch.

2. (�) A sound wave causes a longitudinaldisplacement s of a mass element in a medium asgiven by

s = sm cos(kx− ωt).(a) (�) Here, sm is the displacement amplitude

(maximum displacement) from equilibriumposition,

k =2π

λ,

ω = 2πf,

λ and f being the wavelength and frequency,respectively, of the sound wave.

(b) (�) The sound wave also causes a pressurechange of the medium from the equilibriumpressure (the excess pressure):

δp = δpm sin(kx− ωt),

where the pressure amplitude is

δpm = (vρω)sm.

3. Consider an oscillating element of air ofcross-sectional area A and thickness ∆x, with itscenter displaced from its equilibrium position bydistance s, the displacement:

Equilibriumposition

Oscillating fluid element

Oscillating

s(x, t) = sm cos(kx− ωt),

where sm is the displacement amplitude.




(a) (�) The pressure variation in the displacedelement is

δp = −B∆V

V,

where V = A∆x is the volume of the airelement of cross-sectional area A andthickness ∆x.

(b) (�) We define

∆s = ∆(∆x),

the variation of the air thickness ∆x thatresults in the compression

∆V = A∆s.

(c) (�) The pressure difference ∆p can beexpressed as

δp = −B∆V

V= −BA∆s

A∆x→ −B ∂s

∂x.

(d) (�) The displacements(x, t) = sm cos(kx− ωt) can be substituted toobtain the variation of the air pressure:

δp = −B ∂s∂x

= Bksm sin(kx− ωt).

Thus the pressure varies as a sinusoidalfunction of time, and the amplitude of thevariation is equal to the terms in front of thesine function.

(e) (�) The amplitude of the pressure variation is

δpm = Bksm = v2ρksm,

where we have made use of the formula

v =

√B

ρ.

(f) (�) Because k = ωv , we find that

δpm = v2ρ(ω/v)sm = vρωsm.

4. (�) During a time interval of exactly one period ofvibration of a tuning fork, the emitted soundtravels a distance of one wavelength in air.

5. (�) At points in a sound wave where the gas ismaximally compressed, the pressure is maximum.

17-3 Interference

1. The interference of two sound waves with identicalwavelengths passing through a common pointdepends on their phase difference φ there. We usecosine functions to describe two interfering waves:

s1(x, t) = sm cos(kx− ωt),s2(x, t) = sm cos(kx− ωt+ φ).

(a) (�) The resultant wave is

s′ = s1(x, t) + s2(x, t)

= [2sm cos 12φ] cos(kx− ωt+ 1

2φ).

(b) (�) The amplitude of the resultant wave is

s′m = |2sm cos 12φ|.

(c) (�) If the sound waves were emitted in phaseand are traveling in approximately the samedirection,

φ is given by

φ =∆L

λ2π,

where ∆L is the path length difference and λis the wavelength.

(d) (�) Fully constructive interference occurswhen φ is an integer multiple of 2π,

φ = 2πm, m = 0, 1, 2, · · · .

∆L is related to the wavelength λ by

∆L

λ= m = 0, 1, 2, · · · .




(e) (�) Fully destructive interference occurs whenφ is an odd multiple of π,

φ = (2n+ 1)π, n = 0, 1, 2, · · · .

∆L is related to the wavelength λ by

∆L

λ=

2n+ 1

2=

1

2,

3

2,

5

2, · · · .

2. Two small identical speakers are connected (inphase) to the same source. The speakers are apartby a distance D and at ear level. An observerstands at X, in front of one speaker with thedistance L(� D) as shown. The amplitudes arenot changed.

(a) (�) The path length difference is

∆L =√L2 +D2 − L

= L

(√1 +

D2

L2− 1

)≈ D2

2L,

where we have made use of the binomialexpansion

√1 + x = 1 +

1

2x− 1

8x2 + · · · .

(b) (�) The sound he hears will be least intense ifthe wavelength is

λ =D2

(2n+ 1)L,

where n = 0, 1, 2, · · · .(c) (�) The sound he hears will be most intense if

the wavelength is

λ =D2

2mL,

where m = 1, 2, 3, · · · .

17-4 Intensity and Sound Level

1. (�) The intensity I of a sound wave at a surface isthe average rate per unit area at which energy istransferred by the wave through or onto thesurface:

I =P

A,

where P is the time rate of energy transfer (thepower) of the sound wave and A is the area of thesurface intercepting the sound.

2. Suppose that we can ignore echoes and assumethat the sound source is a point source that emitsthe sound isotropically–that is, with equal intensityin all directions.

(a) (�) The wavefronts spreading from such anisotropic point source S at a particular instantare concentric spheres in three dimensions.

(b) (�) We assume that there is no loss ofenergies during the propagation of sound.Then the time rate at which energy istransferred through the surface by the soundwaves must equal the time rate at whichenergy is emitted by the source (that is, thepower Ps of the source).

(c) (�) The intensity I at the sphere must thenbe

I =Ps

4πr2,

where 4πr2 is the area of the sphere of radiusr.




3. Consider a thin slice of air of thickness dx, area A,and mass dm, oscillating back and forth as thesound wave of

s(x, t) = sm cos(kx− ωt)

passes through it.

Equilibriumposition

Oscillating fluid element

(a) (�) The kinetic energy dK of the slice of air is

dK =1

2dmv2

s ,

where vs is not the speed of the wave but thespeed of the oscillating element of air:

vs =∂s

∂t= ωsm sin(kx− ωt).

(b) (�) Because the mass of the air element is

dm = ρAdx,

the kinetic energy can be expressed as

dK =1

2(ρAdx) [ωsm sin(kx− ωt)]2.

(c) (�) The time rate of the kinetic energytransported is

dK

dt=

1

2ρA

dx

dt[ωsm sin(kx− ωt)]2.

=1

2ρAv ω2s2

m sin2(kx− ωt).

Here, v is the speed of the wave.

(d) (�) The average rate at which kinetic energyis transported is(

dK

dt

)average

=1

4ρAv ω2s2

m,

where we have used the fact that the averagevalue of the square of a sine (or a cosine)function over one full oscillation is 1

2 :

[sin2(kx− ωt)

]average

=1

2.

(e) (�) We assume that potential energy iscarried along with the wave at this sameaverage rate. The wave intensity I, which isthe average rate per unit area at which energyof both kinds is transmitted by the wave, is

I =2(dKdt

)average

A=

1

2ρvω2s2

m.

4. (�) The sound level β in decibels (dB) is defined as

β = (10 dB) log10

I

I0,

where log10 is not the natural logarithm but thecommon logarithm (logarithm of base 10) and

I0 = 10−12 W/m2

is a reference intensity level to which all intensitiesare compared.

5. (�) The standard reference sound level is aboutthe lower limit of the range of human hearing.

6. (�) For every factor-of-10 increase in intensity, 10dB is added to the sound level.

7. Consider two imaginary spherical surfaces ofdifferent radius, both centered on a point soundsource emitting spherical waves.

(a) (�) The power transmitted across the largersphere is the same as the power transmittedacross the smaller one.

(b) (�) The intensity at a point on the largersphere is less than the intensity at a point onthe smaller one.

8. (�) The sound intensity at distance D from a pointsource is I1. The power output of the source is

P = 4πD2I1.




9. The sound level at a point P is 20 dB below thesound level at a point Q with the distance D0 froma point source. Assume that the intensity from apoint source drops off like the inverse square of thedistance. The distance from the source to point Pis D.

(a) (�) The sound levels at P and Q are given by

βP = (10 dB) log10

IPI0,

βQ = (10 dB) log10

IQI0.

(b) (�) The sound level at a point P is 20 dBbelow the sound level at a point Q. Thus

βP − βQ = (10 dB) log10

IPIQ

= −20dB.

(c) (�) The ratio IPIQ

is

IPIQ

= 10−2.

(d) (�) By making use of the inverse-square lawof attenuation of the intensity, we find that

(D0

D

)2

= 10−2.

(e) (�) Thus the distance D from the source topoint P is

D = 10D0.

17-5 Sources of Musical Sound

1. (�) Standing sound wave patterns can be set up inpipes (that is, resonance can be set up), if sound ofthe proper wavelength is introduced in the pipe.

2. Two Open Ends:Consider a pipe open at both ends.

(a) (�) The pipe will resonate if antinodes(maximum oscillation) occur at the open ends.

(b) (�) The fundamental mode or first harmonichas the maximum wavelength for the standingwave pattern.

(c) (�) The fundamental mode has a single nodeat the midpoint of the pipe and both ends areantinodes. Thus λ = 2L:

f =v

λ=

v

2L.

Here, v is the speed of sound in the air in thepipe, L is the length of the pipe, and λ is thewavelength of the sound wave.

(d) (�) The second harmonic has two nodes inthe pipe and both ends are antinodes. Thusλ = L:

f =v

λ=v

L=

2v

2L.

(e) (�) The pipe will resonate at frequencies

fn =v

λn=nv

2L, n = 1, 2, 3, · · · .

3. One Open End:Consider a pipe closed at one end and open at theother.




(a) (�) The pipe will resonate if an antinode(maximum oscillation) occurs at the open endand if a node occurs at the other end which isclosed.

(b) (�) The fundamental mode or first harmonichas the maximum wavelength for the standingwave pattern.

(c) (�) The fundamental mode has a single nodeat the closed end and the other end isantinode. Thus λ = 4L:

f =v

λ=

v

4L.

Here, v is the speed of sound in the air in thepipe, L is the length of the pipe, and λ is thewavelength of the sound wave.

(d) (�) The second harmonic has a node at theclosed end, one more node in the pipe, and anantinode at the open end. Thus λ = 4L

3 :

f =v

λ=

3v

4L.

(e) (�) The pipe will resonate at frequencies

fn =v

λn=nv

4L, n = 1, 3, 5, · · · .

4. Musical Instruments:

(a) (�) The length of a musical instrumentreflects the range of frequencies over whichthe instrument is designed to function.Smaller length implies higher frequencies.

(b) (�) In any oscillating system that gives rise toa musical sound, whether it is a violin stringor the air in an organ pipe, the fundamentaland one or more of the higher harmonics areusually generated simultaneously.

(c) (�) When different instruments are played atthe same note, they produce the samefundamental frequency but differentintensities for the higher harmonics.

(d) (�) Because different instruments producedifferent net waves, they sound different toyou even when they are played at the samenote.

17-6 Beats

1. Let the time-dependent variations of thedisplacements due to two sound waves of equalamplitude sm be

s1 = sm cosω1t,

s2 = sm cosω2t,

where ω1 > ω2.

Time

(a) (�) There are identities:

ω1 = ω + ω′,

ω2 = ω − ω′,ω = 1

2(ω1 + ω2),

ω′ = 12(ω1 − ω2).




(b) (�)

cosω1t = cosωt cosω′t− sinωt sinω′t,

cosω2t = cosωt cosω′t+ sinωt sinω′t.

(c) (�) The resultant displacement s is

s = s1 + s2

= [2sm cosω′t] cosωt.

(d) (�) ω > ω′. Thus

2π

ω<

2π

ω′.

The oscillating factor cosω′t has the periodmuch longer than the period of the factorcosωt.

(e) (�) The amplitude for the fast oscillatingfactor cosωt is

|2sm cosω′t|.

(f) (�) A maximum amplitude will occurwhenever

| cosω′t| = 1.

This implies that the period of beat isπω′ = 2π

ωbeat. Thus the angular frequency ωbeat

at which beats occur is

ωbeat = 2ω′ = (ω1 − ω2).

The corresponding frequency is

fbeat =ω1 − ω2

2π= f1 − f2.

(g) (�) The average of the two combiningfrequencies

f =ω

2π

is the frequency that we hear.

(h) (�) We also hear a striking variation in theintensity of this sound, and it increases anddecreases in slow, wavering beats that repeatat a frequency of fbeat = f1 − f2.

17-7 The Doppler Effect

1. Motion-related frequency changes are examples ofthe Doppler effect. Suppose that a source generatessound waves to the air with the frequency f and vis the speed of sound through the air.

(a) (�) When the motion of detector or source istoward the other, the sign on its speed mustgive an upward shift in frequency. Towardmeans shift up.

(b) (�) When the motion of detector or source isaway from the other, the sign on its speedmust give a downward shift in frequency.Away means shift down.

(c) (�) If either the detector or the source ismoving, or both are moving, the emittedfrequency f and the detected frequency f ′ arerelated by

f ′ = fv ± vdetector

v ± vsource,

where vdetector is the detector’s speed relativeto the air, and vsource is the source’s speedrelative to the air.

2. Detector Moving, Source Stationary :The source generates sound waves with thefrequency f = 1

T . The source is at rest withrespect to the air.

(a) (�) The wavelength is equal in any directions.Thus

v = λf, λ =v

f.

(b) (�) If the detector is moving toward thesource, then the time difference T ′ ofconsecutive wavefronts to meet the source

λ = vT ′ + vdetectorT′.

Thus

vT = λ = (v + vdetector)T′,

→ f ′ =1

T ′=v + vdetector

vf.

(c) (�) If the detector is moving away from thesource, then the time difference T ′ ofconsecutive wavefronts to meet the source

λ+ vdetectorT′ = vT ′.




Thus

vT = λ = (v − vdetector)T′,

→ f ′ =1

T ′=v − vdetector

vf.

3. Source Moving, Detector Stationary :The source generates sound waves with thefrequency f = 1

T . The detector is at rest withrespect to the air.

(a) (�) If the source is moving with respect to theair, then the wavelength depends on thedirection of sound propagation.

(b) (�) Suppose that the source approaches thedetector with the relative speed vsource. Att = 0 a wavefront departs the source. Att = T the wavefront have displaced by λ = vTtoward the detector and the source havedisplaced by vsourceT . At t = T the nextwavefront departs the source and thiswavefront meets the detector after travelingvT − vsourceT with the speed of sound v. Thistime gap T ′ = 1

f ′ satisfies

vT ′ = (v − vsource)T.

As a result,

f ′ =v

v − vsourcef.

(c) (�) If the source is moving away from thedetector, then

f ′ =v

v + vsourcef.

17-8 Supersonic Speeds, Shock Waves

1. (�) If the speed of a source relative to the mediumexceeds the speed of sound in the medium, theDoppler equation no longer applies.

Speed of source = Speed of sound

Speed of source > Speed of sound

(a) (�) If a source is moving toward a stationarydetector at a speed vsource equal to the speedof sound v,

f ′ = limvsource→v

fv ± vdetector

v − vsource→∞.

(b) (�) For supersonic speeds, vsource > v, theformula of Doppler effect breaks down.

(c) (�) The radius of any wavefront is vt, where tis the time that has elapsed since the sourceemitted that wavefront.

(d) (�) A shock wave exists along the surface ofthis cone, because the bunching of wavefrontscauses an abrupt rise and fall of air pressureas the surface passes through any point.

(e) (�) In such a case, shock waves result. Thehalf-angle θ of the Mach cone is given by

sin θ =vt

vsourcet=

v

vsource.

(f) (�) vsourcev is the Mach number.

(g) (�) The shock wave generated by a supersonicaircraft or projectile produces a burst ofsound, called a sonic boom, in which the airpressure first suddenly increases and thensuddenly decreases below normal beforereturning to normal.




(h) (�) When a long bull whip is snapped, its tipis moving faster than sound and produces a

small sonic boom, the crack of the whip.


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