General Physics I Quiz Samples for Chapter 16 Waves - I May 18, 2020 Name: Department: Student ID #: Notice +2 (-1) points per correct (incorrect) answer. No penalty for an unanswered question. Fill the blank ( ) with (✘) if the statement is correct (incorrect). Textbook: Walker, Halliday, Resnick, Principles of Physics, Tenth Edition, John Wiley & Sons (2014). 16-1 Transverse Waves 1. Waves are of three main types: (a) () Mechanical waves. Common examples include water waves, sound waves, and seismic waves. All these waves have two central features: They are governed by Newton’s laws, and they can exist only within a material medium, such as water, air, and rock. (b) () Electromagnetic waves. Common examples include visible and ultraviolet light, radio and television waves, microwaves, x rays, and radar waves. These waves require no material medium to exist. Light waves from stars, for example, travel through the vacuum of space to reach us. All electromagnetic waves travel through a vacuum at the same speed c = 299 792 458 m/s. (c) () Matter waves. These waves are associated with electrons, protons, and other fundamental particles, and even atoms and molecules. Because we commonly think of these particles as constituting matter, such waves are called matter waves. 2. Transverse Waves and Longitudinal Waves: (a) () We would find that the displacement of every such oscillating string element is perpendicular to the direction of travel of the wave. This motion is said to be transverse, and the wave is said to be a transverse wave. (b) () If you push and pull on the piston in simple harmonic motion, a sinusoidal wave travels along the pipe. Because the motion of the elements of air is parallel to the direction of the wave’s travel, the motion is said to be longitudinal, and the wave is said to be a longitudinal wave. 3. Description of a Wave : Consider a sinusoidal wave traveling in the positive direction of an x axis. At time t, the displacement y of the element located at position x is given by y(x, t)= y m sin(kx - ωt + φ). (a) () y(x, t) : displacement y m : amplitude sin(kx - ωt + φ) : oscillating term kx - ωt + φ : phase k : wave number x : position ω : angular frequency t : time φ : phase constant (b) () The wavelength λ of a wave is the distance (parallel to the direction of the wave’s travel) between repetitions of the shape of the wave (or wave shape). The angular wave number can be expressed in terms of λ as k = 2π λ . 2020 KPOPE All rights reserved. Korea University Page 1 of 10
Transcript
General Physics IQuiz Samples for Chapter 16
Waves - IMay 18, 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).
(a) (�) Mechanical waves. Common examplesinclude water waves, sound waves, and seismicwaves. All these waves have two centralfeatures: They are governed by Newton’slaws, and they can exist only within amaterial medium, such as water, air, and rock.
(b) (�) Electromagnetic waves. Commonexamples include visible and ultraviolet light,radio and television waves, microwaves, xrays, and radar waves. These waves require nomaterial medium to exist. Light waves fromstars, for example, travel through the vacuumof space to reach us. All electromagneticwaves travel through a vacuum at the samespeed c = 299 792 458 m/s.
(c) (�) Matter waves. These waves areassociated with electrons, protons, and otherfundamental particles, and even atoms andmolecules. Because we commonly think ofthese particles as constituting matter, suchwaves are called matter waves.
2. Transverse Waves and Longitudinal Waves:
(a) (�) We would find that the displacement ofevery such oscillating string element isperpendicular to the direction of travel of thewave. This motion is said to be transverse,and the wave is said to be a transverse wave.
(b) (�) If you push and pull on the piston insimple harmonic motion, a sinusoidal wavetravels along the pipe. Because the motion ofthe elements of air is parallel to the directionof the wave’s travel, the motion is said to belongitudinal, and the wave is said to be alongitudinal wave.
3. Description of a Wave: Consider a sinusoidalwave traveling in the positive direction of an xaxis. At time t, the displacement y of the elementlocated at position x is given by
(b) (�) The wavelength λ of a wave is thedistance (parallel to the direction of thewave’s travel) between repetitions of the shapeof the wave (or wave shape). The angularwave number can be expressed in terms of λ as
(c) (�) We define the period of oscillation T of awave to be the time any string element takesto move through one full oscillation. Theangular frequency can be expressed in termsof T as
ω =2π
T.
(d) (�) The frequency f of a wave is defined as 1T
and is related to the angular frequency ω by
f =1
T=
ω
2π.
4. Wave Speed v:
(a) (�) If point A retains its displacement as itmoves, the phase iny(x, t) = ym sin(kx− ωt+ φ) giving it thatdisplacement must remain a constant:
kx− ωt+ φ = constant.
(b) (�) The wave speed can be computed bytaking the time derivative of kx− ωt+ φ:
dx
dt= v =
ω
k.
(c) (�) Because
ω =2π
T, k =
2π
λ,
dx
dt= v =
ω
k=
2π
T· λ
2π=λ
T= λf.
(d) (�) y(x, t) = ym sin(kx+ ωt+ φ) is travelingalong the −i direction whiley(x, t) = ym sin(kx− ωt+ φ) is traveling alongthe +i direction.
5. Consider a function y(x, t) = f(kx− ωt).(a) (�) The graph for y(x, t) is translated by +ωt
kin comparison with y(x, 0):
v =ω
k.
Thus the wave is traveling along the +i
direction.
(b) (�) y′(x, t) = f(kx+ ωt) is translated by −ωtk
in comparison with y′(x, 0):
v = −ωk.
Thus the wave is traveling along the −i
direction.
6. The displacement of a string carrying a travelingsinusoidal wave is given by
y(x, t) = ym sin(kx− ωt+ φ).
At time t = 0 the point at x = 0 has adisplacement of 0 and is moving in the positive ydirection. We assume that φ ∈ [0, 2π).
(a) (�) The conditions are summarized as
y(0, 0) = ym sinφ = 0,
∂y
∂t
∣∣∣(x,t)=(0,0)
= −ωym cosφ > 0.
(b) (�) The solutions of the equation are
φ = nπ, n = 0, 1.
(c) (�) The solutions of the inequality are
π
2< φ <
3
2π.
(d) (�) The phase constant is given by φ = π.
16-2 Wave Speed on a Stretched String
1. (�) The speed of a wave on a stretched string isset by properties of the string, not properties ofthe wave such as frequency or amplitude. Thespeed of a wave on a string with tension τ andlinear density µ is
v =
√τ
µ.
2. Consider a symmetrical pulse traveling to the right.Suppose that we view the pulse from a referenceframe in which the pulse is stationary and thestring appears to move right to left with speed v.We find speed v by applying Newton’s second lawto a string element of length ∆l, located at the topof the pulse. The string tension is τ .
(a) (�) Consider a small string element of length∆l within the pulse, an element that forms anarc of a circle of radius R and subtending anangle 2θ at the center of that circle. Therestoring force F acting on the the central linesegment of infinitesimal length ∆l is
F = 2τ sin θ.
Here, τ sin θ is the vertical restoring force onthe right and on the left each.
(b) (�) If ∆l is small, then θ is small and
sin θ = θ − θ3
3!+θ5
5!− · · · ≈ θ.
Then,
F ≈ 2τθ = τ∆l
R.
Here, ∆l = R(2θ) is the length of the arc of acircle of radius R.
(c) (�) The mass of the string of length ∆l is
∆m = µ∆l ∆l =∆m
µ,
where µ is the mass per unit length of thestring. Thus
F ≈ τ∆l
R=τ
µ
∆m
R.
(d) (�) At the moment shown in the figure, thestring element ∆l is moving in an arc of acircle. Thus, it has a centripetal force towardthe center of that circle, given by
F = (∆m)a = v2 ∆m
R=τ
µ
∆m
R.
Therefore,
v =
√τ
µ.
(e) (�) By carrying out the dimensional analysis,we find that[√
τ
µ
]=
√[M ][L][T ]−2
[M ][L]−1= [L][T ]−1.
Thus the dimensions are the same for those ofthe velocity.
3. A stretched string, clamped at its ends, vibrates ata particular frequency.
(a) (�) If the tension and the linear mass densityof the string are τ and µ, then
v =
√τ
µ.
(b) (�) The frequency is
f =1
T=λ
T
1
λ=v
λ.
(c) (�) If we change the string tension by a factorof n, then the speed of the wave varies as
v′ =
√nτ
µ=√nv.
Thus the frequency varies as
f ′ =v′
λ=√nf.
16-3 Energy and Power of a Wave Travellingalong a String
1. (�) The average power of, or average rate at whichenergy is transmitted by, a sinusoidal wave on astretched string is
Paverage =1
2µω2y2
mv.
2. When we set up a wave on a stretched string, weprovide energy for the motion of the string. As thewave moves away from us, it transports energy asboth kinetic energy and elastic potential energy.Let the wave be
(a) (�) A string element of mass dm, oscillatingtransversely in simple harmonic motion as thewave passes through it, has kinetic energyassociated with its transverse velocity u = uj.
u =∂y
∂t= −ωym cos(kx− ωt).
The kinetic energy of the mass elementdm = µdx is given by
dK =1
2dmu2 =
1
2(µdx)(−ωym)2 cos2(kx−ωt).
(b) (�) Dividing dK by dt gives the rate at whichkinetic energy passes through a stringelement, and thus the rate at which kineticenergy is carried along by the wave.
dK
dt=
1
2µdx
dtω2y2
m cos2(kx− ωt)
=1
2µω2y2
m cos2(kx− ωt)v,
where
v =dx
dt.
(c) (�) The average rate at which kinetic energyis transported is(dK
dt
)average
=1
2µω2y2
m[cos2(kx− ωt)]averagev
=1
4µω2y2
mv,
where we have taken the average over aninteger number of wavelengths and have usedthe fact that the average value of the squareof a cosine function over an integer number ofperiods is
[cos2(kx− ωt)]average =1
2.
(d) (�) Elastic potential energy is also carriedalong with the wave, and at the same averagerate (
dU
dt
)average
=
(dK
dt
)average
=1
4µω2y2
mv.
(e) (�) The average power, which is the averagerate at which energy of both kinds istransmitted by the wave, is then
Paverage =
(dK
dt
)average
+
(dU
dt
)average
= 2
(dK
dt
)average
=1
2µω2y2
mv.
16-4 The Wave Equation
1. (a) (�) The partial derivative of y(x, t) withrespect to x is defined by
∂y(x, t)
∂x= lim
∆x→0
y(x+ ∆x, t)− y(x, t)
∆x.
(b) (�) The second-order partial derivative ofy(x, t) with respect to x is defined by
∂2y(x, t)
∂x2= lim
∆x→0
∂y(x+∆x,t)∂x − ∂y(x,t)
∂x
∆x.
(c) (�) The partial derivative of y(x, t) withrespect to t is defined by
∂y(x, t)
∂t= lim
∆t→0
y(x, t+ ∆t)− y(x, t)
∆t.
(d) (�) The second-order partial derivative ofy(x, t) with respect to t is defined by
∂2y(x, t)
∂t2= lim
∆t→0
∂y(x,t+∆t)∂t − ∂y(x,t)
∂t
∆t.
2. (�) The general differential equation that governsthe travel of waves of all types is
∂2
∂x2y(x, t) =
1
v2
∂2
∂t2y(x, t).
Here the waves travel along an x axis and oscillateparallel to the y axis, and they move with speed v,in either the positive x direction or the negative xdirection.
3. By applying Newton’s second law to the element’smotion, we can derive a general differentialequation, called the wave equation, that governsthe travel of waves of any type. In this problem,we derive the wave equation for the transversewave and the transverse displacement is smallenough to make an approximation ∆` ≈ ∆x, where∆` is the length of the string element within theinterval [x, x+ ∆x].
(a) (�) A string element of mass ∆m and length∆` ≈ ∆x as a wave travels along a string oflinear density µ that is stretched along ahorizontal x axis. The mass of that element is
∆m = µ∆` ≈ µ∆x.
(b) (�) F2 is the force on the right end of theelement and F1 is the force on the left end ofthe element.
F2 = τ [i cos θ + j sin θ]∣∣∣x+∆x
,
F1 = −τ [i cos θ + j sin θ]∣∣∣x.
Note that the tension on the right and that onthe left are opposite.
(c) (�) Because we neglect the horizontaloscillation, The force along the i directioncancels in the net force. If θ is small, then thevertical component along j can be simplifiedby applying the following approximation
sin θ ≈ θ ≈ tan θ.
As a result, the net force on the mass element
∆m is
Fnet = F1 + F2
= τ j[sin θ|x+∆x − sin θ|x]
≈ τ j[tan θ|x+∆x − tan θ|x]
= τ j
[∂y
∂x
∣∣∣∣x+∆x
− ∂y
∂x
∣∣∣∣x
].
(d) (�) The equation of motion for the element is
∆m∂2
∂t2y = τ
[∂y
∂x
∣∣∣∣x+∆x
− ∂y
∂x
∣∣∣∣x
].
Substituting ∆m = µ∆x and dividing bothsides by τ∆x, we find that
µ
τ
∂2
∂t2y =
1
∆x
[∂y
∂x
∣∣∣∣x+∆x
− ∂y
∂x
∣∣∣∣x
].
Taking the limit ∆x→ 0, we find that
1
v2
∂2y
∂t2=∂2y
∂x2,
where
v =
√τ
µ.
16-5 Interference of Waves
1. Principle of Superposition for Waves:
(a) (�) When two or more waves transverse thesame medium, the displacement of anyparticle of the medium is the sum of thedisplacements that the individual waves wouldgive it. Let y1(x, t) and y2(x, t) be thedisplacements that the string wouldexperience if each wave traveled alone. Thedisplacement of the string when the wavesoverlap is then the algebraic sum
y′(x, t) = y1(x, t) + y2(x, t).
Here, y′(x, t) is the resultant wave or netwave.
(b) (�) Overlapping waves do not in any wayalter the travel of each other.
2. Consider two waves
y1(x, t) = ym sin(kx− ωt),y2(x, t) = ym sin(kx− ωt+ φ).
We can make use of the addition formula forthe sine function
sinα = sin[12(α+ β) + 1
2(α− β)]
= sin[12(α+ β)] cos[1
2(α− β)]
+ cos[12(α+ β)] sin[1
2(α− β)],
sinβ = sin[12(α+ β)− 1
2(α− β)]
= sin[12(α+ β)] cos[1
2(α− β)]
− cos[12(α+ β)] sin[1
2(α− β)].
Then, we find that
sinα+ sinβ = 2 sin[12(α+ β)] cos[1
2(α− β)]
= 2 sin(kx− ωt+ 12φ) cos(1
2φ).
(b) (�) The resultant wave is
y′(x, t) = 2ym sin[12(α+ β)] cos[1
2(α− β)]
= [2ym cos(12φ)] sin(kx− ωt+ 1
2φ).
(c) (�) The amplitude of the resultant wave is∣∣2ym cos(12φ)∣∣ .
(d) (�) The oscillating term is
sin(kx− ωt+ 12φ).
3. Consider the resultant wave
y′(x, t) = [2ym cos(12φ)] sin(kx− ωt+ 1
2φ)
of the following two waves
y1(x, t) = ym sin(kx− ωt),y2(x, t) = ym sin(kx− ωt+ φ).
We investigate the effect of the phase constant 12φ.
(a) (�) If φ = 0, then the two waves y1 and y2 areexactly in phase. The resultant wave is
y′(x, t) = 2ym sin(kx− ωt).
The amplitude is doubled. Interference thatproduces the greatest possible amplitude iscalled fully constructive interference.
(b) (�) If φ = π, then the two waves y1 and y2 areexactly out of phase. The resultant wave is
y′(x, t) = 0.
The amplitude is 0. This type of interferenceis called fully destructive interference.
16-6 Phasors
1. (�) A phasor is a vector that rotates around itstail, which is pivoted at the origin of a coordinatesystem. The magnitude of the vector is equal tothe amplitude ym of the wave that it represents.
2. We derive the Taylor’s series expansion of ex aboutx = 0.
(c) (�) The real and imaginary parts of the sumz1 + z2 are
x1 + x2 = Re[z1 + z2],
y1 + y2 = Im[z1 + z2].
(d) (�) Thus the addition of two complexnumbers is very similar to the addition of twotwo-dimensional vectors:
r1 = x1i + y1j,
r2 = x2i + y2j,
r1 + r2 = (x1 + x2)i + (y1 + y2)j.
(e) (�) For a given complex number z,
z = r cos θ + ir sin θ = reiθ,
where
r ≡ |z| =√
[Re(z)]2 + [Im(z)]2, tan θ =Im(z)
Re(z).
(f) (�) For a given complex number z = x+ iy,
|z| =√zz∗,
where z∗ is the complex conjugate of z:
z∗ ≡ x− iy = Re(z)− iIm(z).
(g) (�) For a given complex number z = x+ iy,
z + z∗ = 2Re[z],
z − z∗ = 2i Im[z].
5. [optional: This problem does not appear inthe quiz. Just for advanced students.]Consider r1 sin θ1 + r2 sin θ2. We introduce twocomplex numbers z1 and z2:
z1 = r1(cos θ1 + i sin θ1) = r1eiθ1 ,
z2 = r2(cos θ2 + i sin θ2) = r2eiθ2 ,
z = z1 + z2 = reiθ.
(a) (�) r1 sin θ1 + r2 sin θ2 = Im[z1 + z2].
(b) (�) |z|2 is
|z|2 = zz∗
= (z1 + z2)(z1 + z2)∗
= |z1|2 + |z2|2 + z1z∗2 + z∗1z2
= |z1|2 + |z2|2 + 2Re[z1z∗2 ]
= r21 + r2
2 + 2r1r2 cos(θ1 − θ2).
(c) (�)|z1 + z2|2 = (r1 + r2)2,
where
r1 = iRe[z1] + j Im[z1],
r2 = iRe[z2] + j Im[z2].
6. Consider the resultant wave of the following twowaves of different amplitudes:
1. If two sinusoidal waves of the same amplitude andwavelength travel in opposite directions along astretched string, their interference with each otherproduces a standing wave. To analyze a standingwave, we represent the two waves with theequations
y1(x, t) = ym sin(kx− ωt),y2(x, t) = ym sin(kx+ ωt).
(a) (�) y1 travels to the +i direction and y2
travels to the −i direction. The speed isequally v = ω
k .
(b) (�) The resultant wave is
y′(x, t) = y1(x, t) + y2(x, t)
= ym sin(kx− ωt) + ym sin(kx+ ωt)
= [2ym sin kx] cosωt.
(c) (�) The resultant wave does not travel. Thusit is called the standing wave. The quantity|2ym sin kx| can be viewed as the amplitude ofoscillation of the string element that is locatedat position x.
(d) (�) For a standing wave, the amplitude varieswith position. the amplitude is zero for valuesof kx that give sin kx = 0:
kx = nπ, n = 0, 1, 2, · · · .
Substituting k = 2πλ in this equation and
rearranging, we get
x = nλ
2.
The positions of zero amplitude are called thenodes.
(e) (�) The amplitude of the standing wave has amaximum value of 2ym, which occurs forvalues of kx that give sin kx = ±1. Thosevalues are
(a) (�) A pulse incident from the right isreflected at the left end of the string, which istied to a wall (fixed end). Note that thereflected pulse is inverted from the incidentpulse. In a hard reflection of this kind,there must be a node at the supportbecause the string is fixed there. The reflectedand incident pulses must have opposite signs,so as to cancel each other at that point.
(b) (�) Here the left end of the string is tied to aring that can slide without friction up anddown the rod (free end). Now the pulse isnot inverted by the reflection. Thus, in such asoft reflection, the incident and reflectedpulses reinforce each other, creating anantinode at the end of the string . Themaximum displacement of the ring is twicethe amplitude of either of these two pulses.
3. Resonance
The firstharmonic
The secondharmonic
The thirdharmonic
Standing waves on a string can be set up byreflection of traveling waves from the ends of thestring.
(a) (�) If an end is fixed, it must be the positionof a node. This limits the frequencies at whichstanding waves will occur on a given string.
(b) (�) Each possible frequency is a resonantfrequency, and the corresponding standingwave pattern is an oscillation mode. For astretched string of length L with fixed ends,the resonant frequencies are
fn =v
λn= n
v
2L, n = 1, 2, · · · .
(c) (�) The oscillation mode corresponding ton = 1 is called the fundamental mode orthe first harmonic.
(d) (�) The mode corresponding to n = 2 is thesecond harmonic, and so on.
(e) (�) n is called the harmonic number of thenth harmonic.
