INTRODUCTION
TO WAVES
ISC 234
11/04/2022
AAMUSTED_DIS_S. Y.
1
NOTE
✓ These slides are intended as an addition to the lectures given in class. They are
NOT designed to replace the actual lectures.
✓ Some of the notes will contain less information than in the actual lecture.
✓ Not all equations which will be needed for exams are contained or completed
in these slides.
✓ If you notice any typos - let me know at syunus@aamusted.edu.gh .
✓ The slides will be available both in soft and hard copy.
11/04/2022
AAMUSTED_DIS_S. Y.
2
OUTLINE
✓
✓
Types of Waves
Wave Speed
Reference Books:
Cutnell, J.D., & Johnson, K.W., (2007). Physics (7th
ed.). Hoboken, NJ: John Wiley & Sons, Inc.
✓
✓
✓
Reflection and Transmission
Energy & Power Of Waves
The Linear Wave Equation
✓
✓
✓
✓
Superposition and Interference
Sound Waves
Doppler Effect
Standing Waves & Resonance: With
Boundary Conditions
Structure and Functions of the Human Ear
✓
11/04/2022
Serwey, R. A., & Jewet, J. W. (2000). Physics for
Scientist and Engineers with Modern Physics. (9th ed.).
Cengage Learning.
Hewit, P.G. (2006). Conceptual Physics (10th ed.). San
Francisco: Addison Wesley
Young, H.D. & Freedman, R. A. (2004). University
Physics – Vol. 2 (11th ed.). San Francisco, CA: Pearson
Education, Inc.
AAMUSTED_DIS_S. Y.
3
TOPIC HIGHLIGHTS
11/04/2022
AAMUSTED_DIS_S. Y.
4
✓
✓ The Linear Wave Equation
Types of Waves
•
✓
E.M. waves, Mechanical, and matter waves
Wave Speed
•
Speed of a travelling wave is: 𝑣 = 𝑓𝜆 [PROVE]
•
Wave speed on a stretched string depends on
medium properties not wave properties: 𝑣 =
✓
𝜏
𝜇
Average rate at which energy is transmitted by, a
1
wave (Power) is given by; 𝑃𝑎𝑣𝑔 = 𝜇𝜈𝜔2 𝑦𝑚 2
2
✓
General linear wave eqn. and the linear wave
eqn. on a string
•
Gives a complete description of the wave
motion
•
Helps derive an expression for the wave speed
•
Revision needed: Net forces, Newtons laws,
& PDEs
✓ Superposition and Interference
Energy & Power of Waves
•
•
•
A phenomenon of combining waves
•
Apply the principle of superposition to
show that two overlapping waves add
algebraically to give a resultant (or net)
wave
•
The resultant wave depends on the extent to
which the waves are in phase with respect
to each other
Reflection and Transmission
•
How a traveling wave is affected when it
encounters a change in the medium ( Tight/loose
on a pole; thinner/thicker thread)
11/04/2022
AAMUSTED_DIS_S. Y.
5
✓
Sound Waves
• Any longitudinal wave (can travel thru solid >,
liquid >, or gases). speed of sound depends on
medium properties.
• Motion toward tends to shift freq. up and motion
away tends to shift it down.
𝑓′ = 𝑓
𝑣±𝑣𝑑
𝑣±𝑣𝑠
• Sound intensity: sound power (P) per unit area
• Decibel (dB): a measurement unit when
comparing two sound intensities.
• Beats: periodic and repeating fluctuations heard
in the intensity of sound when 2 waves of similar
frequencies interfere. 𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2
✓
✓ Standing Waves & Resonance: With
Boundary Conditions
•
A standing wave is an oscillation pattern with
a stationary outline that results from the
superposition of two identical waves traveling
in opposite directions. [The interference only
produces a standing wave at certain
frequencies]
•
Th patterns of standing waves do not move left
or right; the locations of the maxima and
minima do not change.
•
Eqn. of a standing wave:
Doppler Effect (Sound)
• Shift in the detected freq. from the emitted freq.
by a source due to the relative motion between
the source and the detector
• Calculating the shift in sound freq. for diff.
scenarios.
11/04/2022
y = 𝑦1 + 𝑦2 ⇒ 𝑦 = 2𝑦𝑚 sin 𝑘𝑥 cos 𝑤𝑡
AAMUSTED_DIS_S. Y.
6
✓ Standing Waves & Resonance: With
Boundary Conditions Cont’d
For waves on strings
•
For a string of length L fixed at both ends. The
natural frequencies are; [T is the tension in the
string and m is its linear mass density]
•
Such a standing wave is said to be produced at
resonance, and the selected frequencies, called
resonant frequencies.
•
Resonance: occurs when one object vibrating at
the same natural frequency as a second object,
forces that second object into vibrational motion.
•
Resonance is important in the use of musical
instruments which rely on air columns.
For sound waves [Standing Wave in Air
Columns]
•
A common model of resonance is a resonance
tube; a hollow cylindrical tube partially filled with
water and forced into vibration by a tuning fork.
•
•
The tuning fork forces the air in the tube into
resonance
11/04/2022
𝑛 𝑇
𝑓𝑛 =
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1,2,3, …
2𝐿 𝜇
For sound waves with speed v in an air column
of length L open at both ends, the natural
frequencies are;
AAMUSTED_DIS_S. Y.
𝑓𝑛 = 𝑛
𝑣
2𝐿
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1,2,3, …
7
TYPES OF WAVES
11/04/2022
AAMUSTED_DIS_S. Y.
8
▪ Governed by Newton’s laws
▪ Only exist within a material medium
▪ Eg: Water, sound, & string waves
Mechanical Waves
▪ Direction of oscillation is Ʇ to
that of the wave. Eg: All E.M.
waves, waves on a string
Matter
Waves
Electromagnetic Waves
▪ Associated with
protons, & electrons
▪ Direction of oscillation is
\ \ to that of the wave. Eg:
Sound & ‘slinky’ waves
Longitudinal Waves
Transverse Waves
Surface Waves
11/04/2022
▪ Require no material medium
▪ Travels at c = 3 × 108 m/s
▪ Eg: UV-VIS, TV, radio, micro waves
Waves
▪ Particles undergo circular motion.
▪ Deals with only particles at the
surface. Eg. Waves on water surface
AAMUSTED_DIS_S. Y.
9
A
B
Fig. 1: Periodic waves; (A) Transverse & (B)
Longitudinal waves
11/04/2022
AAMUSTED_DIS_S. Y.
✓
Both transverse and longitudinal waves are
Periodic / Progressive
✓
Particles vibrate with the same amplitude
and frequency
✓
Frequency: rate at which each crest of the
wave passes through a given point per unit
of time.
✓
Wavelength: the horizontal distance
between two successive crests, two
successive trough or any two successive
equivalent points on the wave.
✓
Amplitude: the distance between a crest
and the undisturbed position
✓
Crest: the highest point on the wave
pattern
✓
Trough: the lowest point on the wave
pattern
𝑣 = 𝑓𝜆
10
✓ To completely describe a wave on a string, imagine a sinusoidal wave like that of
Fig. 1 (A) 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 the
sinusoidal wave function:
Amplitude Oscillating Term
Displacement
𝑦 𝑥, 𝑡 = 𝐴 sin 𝑘𝑥 − 𝜔𝑡 = 𝐴 sin
2𝜋
𝑥
𝜆
−
2𝜋
𝑡
𝑇
= 𝐴 sin 2𝜋
𝑥
𝜆
−
𝑡
𝑇
(1)
Phase
Where k is the wavenumber, x is the position, 𝜔 is the angular frequency, and t is time
[-ve sign depicts a travel in the positive direction.
✓ Equation (1) can be used to determine the shape of the wave at any given time
11/04/2022
AAMUSTED_DIS_S. Y.
11
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. List two main features common to all waves.
[ (a) A wave is a traveling disturbance; (b) A wave carries energy from place to place]
2. _________________ waves travel on the strings of instruments such as guitars.
[ Transverse]
3. All E.M waves are _______________________________. [ Transverse]
4. The transverse and longitudinal waves that we have been discussing are called mechanical or
___________________ waves because they consist of cycles or patterns that are produced over and
over again by the source.
[Periodic Waves]
5. The ___________________________is the horizontal length of one cycle of the wave. [wavelength]
6. The ____________________________ is the time required for one complete up/down cycle. [period]
7. AM and FM radio waves are transverse waves consisting of electric and magnetic disturbances
traveling at a speed of 3 × 108 𝑚/𝑠. A station broadcasts an AM radio wave whose frequency is
1230 × 103 𝐻𝑧 and an FM radio wave whose frequency is 91.9 × 106 𝐻𝑧 . Find the distance
between adjacent crests in each wave.
[𝝀𝒂𝒎 = 𝟐𝟒𝟒 𝒎 𝒂𝒏𝒅 𝝀𝒇𝒎 = 𝟑. 𝟐𝟔 𝒎]
11/04/2022
AAMUSTED_DIS_S. Y.
12
WAVE SPEED
11/04/2022
AAMUSTED_DIS_S. Y.
13
For A Traveling Wave
✓ Consider a wave traveling in the +ve x direction
✓ The entire wave moves a distance Δx in a time interval Δt.
✓ The ratio Δx/Δt or dx/dt is the wave speed v.
✓ As the wave in moves, each point of the moving wave
(not the string) such as pt ‘A’ moves and retains it’s
displacement y.
✓ If point A retains its displacement as it moves, the phase
becomes constant. i.e.
kx - ωt = a constant
(1)
Note: Both x and t are changing such that equation (1)
remains a constant.
Fig. 2: Waves at t = 0 and t = Δt
✓ Derivative of equation (1) gives;
𝑤
𝜆
𝑣 = = = 𝑓𝜆
(2)
𝑘
𝑇
✓ That is; the wave moves a distance of one wavelength in
one period of oscillation.
11/04/2022
AAMUSTED_DIS_S. Y.
14
For A Stretched String
✓ The speed of a wave is not only related to the wavelength and frequency, but mainly depends on the
properties of the medium
✓ As waves travel through a medium, the particles of that medium are forced to oscillate
✓ This requires both mass (for kinetic energy) and elasticity (for potential energy) which affects the
wave speed.
∆𝒍
𝝉
𝝉
∆𝒍
Fig. 3: Wave speed in a stretched string
11/04/2022
AAMUSTED_DIS_S. Y.
15
✓ Consider a pulse moving on a stretched string element of length Δl, under tension force F or 𝜏,with
Ԧ
a
uniform speed v.
✓ The pulse makes an approximate arc of the same length Δl, when it reaches the top of the pulse.
✓ The same magnitude of tension force 𝝉 acts on both ends of the arc.
✓ The arc has extremely small length Δl, so can be taken as circular in a circle of radius R, and therefore
∆𝑙 = R(2θ)
(1)
✓ If μ is the mass per unit length of the string (linear density), the mass of the element of the string is
dm = μ∆𝑙 = μR(2θ)
(2)
✓ The horizontal components of tension 𝜏Ԧ are equal and opposite and cancel each other.
✓ The vertical components sum up to give the centripetal force on the arc:
𝜏sinθ + τsinθ = 2τsinθ = dm
Taking sinθ = θ and replacing dm gives;
11/04/2022
2τθ =
v2 Centripetal
(3)
R acceleration
μR(2θ) v2
R
AAMUSTED_DIS_S. Y.
∴𝑣=
𝜏
𝜇
(4)
16
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. Transverse waves travel on each string of an electric guitar after the string is plucked. The length of
each string between its two fixed ends is 0.628 m, and the mass is 0.208 g for the highest pitched E
string and 3.32 g for the lowest pitched E string. Each string is under a tension of 226 N. Find the
speeds of the waves on the two strings. [High = 286 m/s and Low = 207 m/s]
2. The speed of a transverse wave on a string is 𝑣𝑤𝑎𝑣𝑒 , and the speed at which a string particle moves is
𝑣𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 . Are the speeds 𝑣𝑤𝑎𝑣𝑒 and 𝑣𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 identical Or different.
[Different; 𝒗𝒘𝒂𝒗𝒆 =
𝝉
𝝁
≠ 𝒗𝒑𝒂𝒓𝒕𝒊𝒄𝒍𝒆 =
𝝏𝒚
𝝏𝒕
=
𝝏
𝝏𝒕
𝑨𝒔𝒊𝒏 𝝎𝒕
= 𝑨𝝎𝒔𝒊𝒏𝝎𝒕]
3. The displacement function of a particle caused by a wave traveling in the -x direction, with an
𝟐𝝅
𝟐𝝅
amplitude A, frequency f, and wavelength 𝜆 is given by; [𝒚 𝒙, 𝒕 = 𝑨 𝒔𝒊𝒏
𝒙+ 𝒕 ]
𝝀
4. The phase of this wave function is given by; [
𝟐𝝅
𝟐𝝅
𝒙+ 𝒕
𝝀
𝑻
𝑻
]
5. Light travels at a speed of 3 × 108 𝑚/𝑠. The human eye is most sensitive to yellowish-green light,
which has a wavelength of 5.45 × 107 𝑚 . What is the frequency of this light? [𝟓. 𝟓 × 𝟏𝟎𝟏𝟒 𝑯𝒛]
11/04/2022
AAMUSTED_DIS_S. Y.
17
REFLECTION AND
TRANSMISSION
11/04/2022
AAMUSTED_DIS_S. Y.
18
✓ How is a traveling wave affected when it
encounters a change in the medium? E.g. a
pulse traveling on a string that is rigidly
attached to a support at one end (Fig. 4)
✓ On reaching the support, the pulse
undergoes reflection; that is, the pulse
moves back along the string in the opposite
direction
✓ From Fig. 4 C, the reflected pulse is
inverted, because when the pulse reaches
the fixed end of the string, the string
produces an upward force on the support.
And by Newton’s third law, the support
must exert an equal-magnitude and
oppositely directed (downward) reaction
force on the string.
11/04/2022
Fig. 4: Reflection of a traveling pulse at the fixed
end of a stretched string
AAMUSTED_DIS_S. Y.
19
✓ Consider a pulse traveling on a
string that is loosely attached to a
support at one end such that its free
to move vertically on a smooth post
without friction (Fig. 5)
✓ On reaching the support, the pulse
undergoes reflection
✓ From Fig. 5 C, the reflected pulse is
NOT inverted, because, when it
reaches the post, the pulse exerts a
force on the free end of the string,
causing the ring to accelerate
upward. The ring rises as high as the
incoming pulse, and then the
downward component of the tension
force pulls the ring back down.
11/04/2022
Fig. 5: Reflection of a traveling pulse at the free
end of a stretched string
AAMUSTED_DIS_S. Y.
20
✓ Consider a pulse traveling on a string that is attached to a
support such that the boundary is intermediate between the
extremes; fixed and loose(Fig. 6)
✓ On reaching the support, part of the energy in the incident
pulse is reflected and part undergoes transmission
✓ When a pulse traveling on the light string reaches the
boundary between the light string and a heavier string (Fig.
6.), part of the pulse is reflected and inverted and part is
transmitted to the heavier string.
✓ The reflected pulse has a smaller amplitude than the
incident pulse per the principle of conservation of energy
since the energy carried by a wave is related to its
amplitude.
✓ That is; the sum of the energies of these two pulses
(reflected and transmitted) must equal the energy of the
incident pulse. And since, the reflected pulse contains only
part of the energy of the incident pulse, its amplitude must
be smaller.
11/04/2022
AAMUSTED_DIS_S. Y.
Fig. 6: (a) A pulse traveling from a light string
to a heavier string. (b) The situation after the
pulse reaches the junction.
21
✓ Consider a pulse traveling on a string that is attached
to a support such that the boundary is intermediate
between the extremes; fixed and loose(Fig. 7)
✓ On reaching the support, part of the energy in the
incident pulse is reflected and part undergoes
transmission
✓ When a pulse traveling on a heavy string reaches the
boundary between the heavy string and a lighter string
(Fig. 7), part of the pulse is reflected but NOT
inverted and part is transmitted to the heavier string.
✓ In either case (Fig. 6 or Fig. 7), the relative heights of
the reflected and transmitted pulses depend on the
relative densities of the two strings.
✓ If the strings are identical, there is no discontinuity at
the boundary and no reflection takes place.
11/04/2022
AAMUSTED_DIS_S. Y.
Fig. 7: (a) A pulse traveling from a
heavy string to a lighter string. (b) The
situation after the pulse reaches the junction.
22
✓ According to equation (4),
𝑣=
𝜏
𝜇
(4)
✓ The speed of a wave on a string increases as the mass per unit length of the string
(𝜇) decreases.
✓ In other words, a wave travels more rapidly on a light string than on a heavy string if both
are under the same tension.
✓ The following general rules apply to reflected waves:
•
When a wave or pulse travels from medium A to medium B and vA > vB (that is, when
B is denser than A), it is inverted upon reflection.
•
When a wave or pulse travels from medium A to medium B and vA < vB (that is, when
A is denser than B), it is not inverted upon reflection.
11/04/2022
AAMUSTED_DIS_S. Y.
23
ENERGY & POWER
OF WAVES
11/04/2022
AAMUSTED_DIS_S. Y.
24
✓ The energy provided when a wave is set on a stretched
string is transported as both kinetic energy and elastic
potential energy.
Kinetic Energy
✓ A string element of mass dm, oscillating transversely in
simple harmonic motion as the wave passes through it,
has kinetic energy associated with its transverse
velocity 𝑣.
✓ When the element is rushing through its rest position at
y = 0 (b), its transverse velocity—and thus its kinetic
energy—is a maximum.
✓ When the element is at its maximum position y = 𝑦𝑚
(a), its transverse velocity—and thus its kinetic
energy—is 0.
11/04/2022
AAMUSTED_DIS_S. Y.
Fig. 8: a traveling wave on a string at time
t = 0. a is at displacement y = ym, and b is
at displacement y = 0.
25
Elastic Potential Energy
✓ As a string element of length dx oscillates transversely, its length must increase and decrease in a
periodic way if the string element is to fit the sinusoidal wave form. Elastic potential energy is
associated with these length changes, just as for a spring
✓ At point ‘a’ its length has its normal undisturbed value dx, so its elastic potential energy is zero.
However, when the element is rushing through its y = 0 position, it has maximum stretch and thus
maximum elastic potential energy.
Energy Transport
✓ The oscillating string element has both its maximum kinetic and elastic potential energy at y = 0.
✓ As a wave travels along the string, tension forces in the string continuously do work to transfer
energy from regions with energy to regions with no energy.
✓ As the wave moves into sections that were previously at rest, energy is transferred into those new
sections. That is, the wave transports the energy along the string.
11/04/2022
AAMUSTED_DIS_S. Y.
26
The Rate of Energy Transmission
1
2
✓ Kinetic energy dK associated with a string element of mass dm is given by: 𝑑𝐾 = 𝑑𝑚 𝑣 2
(1)
✓ Differentiating the wave element displacement eqn. with respect to time (x = constant) to get 𝜇 :
𝑑𝑦
𝑣=
= −𝜔𝑦𝑚 cos 𝑘𝑥 − 𝜔𝑡
𝑑𝑡
and
𝑑𝑚 = 𝜇𝑑𝑥
1
2
✓ Eqn (1) now becomes; 𝑑𝐾 = 𝜇𝑑𝑥 −𝜔𝑦𝑚 2 𝑐𝑜𝑠 2 𝑘𝑥 − 𝜔𝑡
(2)
✓ Dividing Eqn. (2) by dt to get the rate at which kinetic energy is carried along by the wave. And
assuming that the average value of the square of a cosine function over an integer number of periods
is ½ gives eqn. (3)
𝑑𝐾 1 𝑑𝑥
= 𝜇
−𝜔𝑦𝑚 2 𝑐𝑜𝑠 2 𝑘𝑥 − 𝜔𝑡
𝑑𝑡 2 𝑑𝑡
𝑑𝐾
𝑑𝑡
11/04/2022
=
2
1
2
𝜇𝑣𝜔 𝑦𝑚 𝑐𝑜𝑠 2
2
𝑘𝑥 − 𝜔𝑡
→
𝑑𝐾
𝑑𝑡 𝑎𝑣𝑔
AAMUSTED_DIS_S. Y.
=
𝑑𝑥
𝑏𝑢𝑡
=𝑣
𝑑𝑡
2
1
2
𝜇𝑣𝜔 𝑦𝑚
4
(3)
27
The Rate of Energy Transmission
✓ Elastic potential energy is also carried along with the wave, at the same average rate as in Eqn. (3)
✓ The average power, is then given by ;
𝑃𝑎𝑣𝑔 = 2
𝑑𝐾
𝑑𝑡 𝑎𝑣𝑔
=
2
1
2
𝜇𝑣𝜔 𝑦𝑚
2
(4)
✓ The factors μ and v in this equation depend on the material and tension of the string.
✓ The factors ω and ym depend on the process that generates the wave.
✓ The dependence of the average power of a wave on the square of its amplitude and the square of its
angular frequency is a general result, true for waves of all types.
TRY:
✓ A string has linear density μ = 525 g/m and is under tension τ = 45 N. We send a sinusoidal wave
with frequency f = 120 Hz and amplitude ym = 8.5 mm along the string. At what average rate does
the wave transport energy?
11/04/2022
AAMUSTED_DIS_S. Y.
28
THE LINEAR WAVE
EQUATION
11/04/2022
AAMUSTED_DIS_S. Y.
30
✓ Deriving the wave equation by applying Newton’s second law to the element’s motion
(Transverse)
✓ Newton’s second law written for y components as;
𝐹𝑛𝑒𝑡,𝑦 = 𝑚𝑎𝑦
(1)
✓ Where the mass: dm can be written in terms of the string’s linear density μ and the element’s
length ∆𝑙 as;
𝑑𝑚 = 𝜇∆𝑙 = μ𝑑𝑥
✓ The acceleration can be written as
𝑎𝑦 =
✓ The magnitude of the tension force;
𝜏=
11/04/2022
(2)
𝑑2 𝑦
𝑑𝑡 2
(3)
𝐹2𝑥 2 + 𝐹2𝑥 2
AAMUSTED_DIS_S. Y.
(4)
31
✓ But we assumed the element is slightly tilted (Fig. 3), hence 𝐹2𝑦 ≪ 𝐹2𝑥 .
✓ Then we rewrite eqn. (3) as ;
𝜏 = 𝐹2𝑥
(5)
✓ But the string slope 𝑆2 is related to the Force components as;
𝐹2𝑦
𝐹2𝑥
𝐹2𝑦 = 𝜏𝑆2 and similarly, 𝐹1𝑦 = 𝜏𝑆1
= 𝑆2 =
𝐹2𝑦
𝜏
(6)
(7)
✓ Hence the net force can be written as;
𝐹𝑛𝑒𝑡,𝑦 = 𝜏𝑆2 − 𝜏𝑆1 = 𝑚𝑎𝑦 = μ𝑑𝑥
𝑆2 −𝑆1
𝑑𝑥
11/04/2022
=
𝜇 𝑑2 𝑦
𝜏 𝑑𝑡 2
𝑑2 𝑦
𝑑𝑡 2
But 𝑆2 − 𝑆1 = 𝑑𝑆 = 𝑑
AAMUSTED_DIS_S. Y.
𝑑𝑦
𝑑𝑥
(8)
32
𝑑𝑦
𝑑 𝑑𝑥
𝑑𝑥
✓ Hence;
∴
=
𝜇 𝑑2 𝑦
𝜏 𝑑𝑡 2
𝑑2 𝑦
𝑑𝑥 2
𝜕2 𝑦
𝜕𝑥 2
since 𝑣 =
=
𝜇 𝑑2 𝑦
𝜏 𝑑𝑡 2
(9)
=
𝜇 𝜕2 𝑦
𝜏 𝜕𝑡 2
(10)
=
1 𝜕2 𝑦
𝑣 2 𝜕𝑡 2
(11)
NOTE THE CHANGE
FROM FULL (d) TO
PARTIAL(𝝏)
DERIVATIVES
𝜏
𝜇
𝜕2 𝑦
𝜕𝑥 2
✓ Eqn. (11) is the general differential equation that governs the travel of waves of all types
11/04/2022
AAMUSTED_DIS_S. Y.
33
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. A hand moves the end of the Slinky up and down through two complete cycles in one second. The
wave moves along the Slinky at a speed of 0.50 m/s. Find the distance between two adjacent crests
on the wave. [ 0.5/2 = 0.25 m]
2. The middle C string on a piano is under a tension of 944 N. The period and wavelength of a wave on
this string are 3.82 ms and 1.26 m, respectively. Find the linear density of the string.
[ 1st find v and use it to find 𝝁 = 𝟖. 𝟔𝟖 × 𝟏𝟎−𝟑 ]
3. The drawing below shows two transverse waves traveling on strings. The linear density of each
string is 0.065 kg/m. The tension is provided by a 26-N block that is hanging from the string. Find
the speed of the wave in part (a) and in part (b) of the drawing. [ a) 20.0 m/s b) 14 m/s]
11/04/2022
AAMUSTED_DIS_S. Y.
34
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1.
A wave traveling in the - ve x direction has an amplitude of 0.35 m, a speed of 5.2 m/s, and a frequency of 14
𝟐𝝅
Hz. Write the equation of the wave in terms of the frequency and wavelength. [𝒚 𝒙, 𝒕 = 𝑨 𝒔𝒊𝒏
𝒙 + 𝟐𝝅𝒇𝒕 ]
𝝀
2.
The tension in a string is 15 N, and its linear density is 0.85 kg/m. A wave on the string travels toward the x
direction; it has an amplitude of 3.6 cm and a frequency of 12 Hz. What are the (a) speed and (b) wavelength of
the wave? (c) Write down a mathematical expression for the wave, substituting numbers for the variables A, f,
and 𝜆.
[a) Speed = 4.2 m/s b) wavelength = 0.35 m
𝟐𝝅
c) 𝒚 𝒙, 𝒕 = 𝟑. 𝟔 × 𝟏𝟎−𝟐 𝒔𝒊𝒏
𝒙 + 𝟐𝝅(𝟏𝟐)𝒕 = 𝟑. 𝟔 × 𝟏𝟎−𝟐 𝒔𝒊𝒏 (𝟏𝟖)𝒙 + (𝟕𝟓 𝒓𝒂𝒅/𝒔)𝒕 ]
𝟎.𝟑𝟓
3.
Give the expression for the average rate at which energy is transmitted by, a sinusoidal wave on a stretched
𝟏
𝟐
𝟐
string. [𝐏 = 𝝁𝒗𝝎𝟐 𝒚𝒎 ]
4.
The general differential equation that governs the travel of waves of all types is given by;
𝝏𝟐 𝒚
[ 𝟐
𝝏𝒙
=
𝟏 𝝏𝟐 𝒚
]
𝒗𝟐 𝝏𝒕𝟐
11/04/2022
AAMUSTED_DIS_S. Y.
35
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. A string along which waves can travel is 2.70 m long and has a mass
of 260 g. The tension in the string is 36.0 N. What must be the
frequency of traveling waves of amplitude 7.70 mm for the average
power to be 85.0 W?
Solution:
1/2
𝟏
𝟏
𝜏
𝟐
𝐏 = 𝝁𝒗𝝎𝟐 𝒚𝒎
𝑯𝒆𝒏𝒄𝒆 𝟖𝟓. 𝟎 𝑾 = 𝝁 1/2 𝟒𝝅𝟐 𝒇𝟐 𝟕. 𝟕 × 𝟏𝟎−𝟑
𝟐
𝟐
𝜇
𝟏 𝟏/𝟐 𝟏/𝟐 𝟐 𝟐
𝟖𝟓. 𝟎 × 𝟐
𝟐
−𝟑
𝟐
= 𝝁 𝝉 𝟒𝝅 𝒇 𝟕. 𝟕 × 𝟏𝟎
⇒
=
𝒇
𝟐
𝝁𝝉 × 𝟒𝝅𝟐 𝟕. 𝟕 × 𝟏𝟎−𝟑 𝟐
11/04/2022
AAMUSTED_DIS_S. Y.
𝟐
36
SUPERPOSITION
AND
INTERFERENCE
11/04/2022
AAMUSTED_DIS_S. Y.
37
✓ When two waves travel simultaneously through
a medium or along the same stretched string,
their combined effect at any point can be
determined by the superposition principle.
✓ Let 𝑦1 (𝑥, 𝑡) and 𝑦2 𝑥, 𝑡 , be the displacements
that the string would experience if each wave
traveled alone. The displacement of the string
when the waves overlap is then the algebraic
sum 𝑦 ′ (𝑥, 𝑡) = 𝑦1 (𝑥, 𝑡) + 𝑦2 (𝑥, 𝑡)
✓ Hence in other words, the principle of
superposition says that when several effects
occur simultaneously, their net effect is the sum
of the individual effects.
NB: Overlapping waves do not in any way alter the
travel of each other.
11/04/2022
AAMUSTED_DIS_S. Y.
Fig. 8: Two pulses traveling in opposite
directions along a stretched string
38
✓ After the superposition of two waves, the resultant wave depends on the extent to which the
waves are in phase with respect to each other.
✓ If the waves are exactly in phase (so that the peaks and valleys of one are exactly aligned
with those of the other), they combine to double the displacement of either wave acting
alone.
✓ If they are exactly out of phase (the peaks of one are exactly aligned with the valleys of the
other), they combine to cancel everywhere, and the string remains straight.
✓ This phenomenon of combining waves interference, which only affects the wave
displacement not its travel.
✓ Consider two waves traveling along a stretched string given by;
𝑦1 𝑥, 𝑡 = 𝑦𝑚 sin(𝑘𝑥 − 𝜔𝑡)
𝑦2 𝑥, 𝑡 = 𝑦𝑚 sin(𝑘𝑥 − 𝜔𝑡 + 𝜙)
11/04/2022
AAMUSTED_DIS_S. Y.
(1)
(2)
39
✓ Both waves in eqn. (1) and (2) travel in the positive direction of the x axis, with the same
speed. They differ only by the phase constant. These waves are said to be out of phase by ϕ.
✓ Applying the superposition principle, the resultant wave is;
′
𝑦 𝑥, 𝑡 =
1
2𝑦𝑚 𝑐𝑜𝑠 𝜙
2
✓ Hint; Apply sinA + 𝑠𝑖𝑛𝐵
=
sin 𝑘𝑥 − 𝜔𝑡 +
1
2𝑠𝑖𝑛
2
𝐴+𝐵
1
𝜙
2
1
𝑐𝑜𝑠 (𝐴
2
(3)
− 𝐵)
(a)
✓ The
resultant wave differs from the interfering waves in two respects: (1) its phase constant
1
is 𝜙, and (2) its amplitude;
2
′
𝑦𝑚
11/04/2022
=
1
2𝑦𝑚 𝑐𝑜𝑠 𝜙
2
AAMUSTED_DIS_S. Y.
(4)
40
✓ If 𝜙 = 0, then the two interfering waves are exactly in phase, and hence eqn. (3) becomes;
𝑦 ′ 𝑥, 𝑡 = 2𝑦𝑚 sin 𝑘𝑥 − 𝜔𝑡
(5)
✓ Interference that produces the greatest possible amplitude is called fully constructive
interference.
✓ If 𝜙 = π rad (or 180°) , then the two interfering waves are exactly out of phase, and hence
eqn. (3) becomes;
𝜋
𝜋
′
𝑦 𝑥, 𝑡 = 2𝑦𝑚 𝑐𝑜𝑠 sin 𝑘𝑥 − 𝜔𝑡 + = 0
(6)
2
2
✓ Interference that produces no amplitude is called fully destructive interference
✓ Two waves with the same wavelength are in phase if their phase difference is zero or any
integer number of their wavelengths.
11/04/2022
AAMUSTED_DIS_S. Y.
41
11/04/2022
AAMUSTED_DIS_S. Y.
42
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. Two sinusoidal waves on the same string exhibit interference, adding or canceling according to the
principle of superposition. If the two are traveling in the same direction and have the same amplitude
ym and frequency (hence the same wavelength) but differ in phase by a phase constant ϕ, the resultant
𝟏
𝟏
wave is given by ___________. [𝒚′ 𝒙, 𝒕 = 𝟐𝒚𝒎 𝒄𝒐𝒔 𝝓 𝐬𝐢𝐧 𝒌𝒙 − 𝝎𝒕 + 𝝓 ]
𝟐
𝟐
2. For ϕ = 0, the amplitude of the resultant wave is _________________. [𝟐𝒚𝒎 ]
3. Two identical sinusoidal waves, moving in the same direction along a stretched string, interfere with
each other. The amplitude ym of each wave is 9.8 mm, and the phase difference ϕ between them is
100°. (a) What is the amplitude y′m of the resultant wave due to the interference, and what is the type
of this interference?[= |(2)(9.8 mm) cos(100°/2)= 13 mm ]
4. Two identical traveling waves, moving in the same direction, are out of phase by π/2 rad. What is the
amplitude of the resultant wave in terms of the common amplitude ym of the two combining waves?
1
′
[𝑦𝑚 = 2𝑦𝑚 𝑐𝑜𝑠 𝜙 ]
2
11/04/2022
AAMUSTED_DIS_S. Y.
43
SOUND WAVES
11/04/2022
AAMUSTED_DIS_S. Y.
44
✓ Sound waves are vibrations that propagate through a medium in the form of longitudinal
waves. Speed of sound in solid > in liquid > in gasses
✓ Most commonly experienced as waves traveling through air which aid in human hearing
✓ The travel of sound waves through air, disturbs the air elements/particles from their
equilibrium positions.
✓ These movements are accompanied by changes in density and pressure of the air along the
direction of wave motion. Sound can be reflected and refracted.
✓ Sound waves are divided into three categories based on frequency ranges;
• Audible waves: frequencies within the range of sensitivity of the human ear (Fig. a) (20 –
20k Hz). Mostly generated by musical instruments, human voices, or loudspeakers.
• Infrasonic waves: frequencies below the audible range. Elephants can use infrasonic
waves to communicate with one another, even when separated by many kilometers.
• Ultrasonic waves: frequencies above the audible range. Dogs easily hear ultrasonic sound
although humans cannot detect it at all. Ultrasonic waves are also used in medical
imaging.
11/04/2022
AAMUSTED_DIS_S. Y.
45
Fig. a: The human ear
11/04/2022
AAMUSTED_DIS_S. Y.
46
✓ Considering the sound wave in Fig. 9, point S
represents a tiny sound source, called a point source,
that emits sound waves in all directions.
✓ The wavefronts and rays indicate the direction of
travel and the spread of the sound waves.
NB: For a traveling sound wave, the
frequency remains the same from
the source but the wavelength
changes for changing speed.
𝒗 = 𝒇𝝀
✓ Wavefronts are surfaces over which the oscillations
due to the sound wave have the same value; such
surfaces are represented by whole or partial circles in
a two-dimensional drawing for a point source.
✓ Rays are directed lines perpendicular to the
wavefronts that indicate the direction of travel of the
wavefronts.
✓ Near S (Fig. 9), the wavefronts are spherical and
spread out. But as the wavefronts move outward and
their radii become larger, their curvature decreases.
These are said to be planar.
11/04/2022
AAMUSTED_DIS_S. Y.
Fig. 9: A sound wave travels from
a point source S through a 3D medium.
47
Speed of Sound Waves
✓ The speed of any mechanical wave, transverse or longitudinal, depends on both an inertial
and an elastic property of the medium.
✓ Generalizing the wave speed eqn. for a transverse wave along a stretched string (Slide 16),
gives;
𝑣=
𝜏
𝜇
=
𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
𝑖𝑛𝑒𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑜𝑝𝑒𝑟𝑡𝑦
(1)
✓ For longitudinal waves in air; the inertial property, corresponding to μ, becomes the volume
density ρ of air.
✓ As sound wave passes through air, the inertial property is associated with the periodic
compressions and expansions of small volume elements of the air.
✓ The property that determines the extent to which an element of a medium changes in volume
when the pressure on it changes is the bulk modulus B, given by;
11/04/2022
AAMUSTED_DIS_S. Y.
48
𝐵=
∆𝑝
− ∆𝑉
ൗ𝑉
(2)
✓ Hence, the speed of sound in a medium with bulk modulus B and density ρ is given by;
𝑣=
𝐵
𝜌
(3)
✓ For longitudinal sound waves in a solid rod, for example, the speed of sound depends on
Young’s modulus Y and the density 𝛒.
✓ For sound traveling through air, its speed also depends on the temperature of the medium.
Relationship between wave speed and air temperature is;
𝑣 = 331 1
𝑇𝑐
+
273
(4)
✓ Where v is in meters/second, speed of sound = 331 m/s in air at 0𝑜 𝐶, and 𝑇𝑐 is the air
temperature in degrees Celsius.
11/04/2022
AAMUSTED_DIS_S. Y.
49
So what? Why do we need this?
✓ To estimate the distance to a thunderstorm. First count the number of seconds between seeing the
flash of lightning and hearing the thunder. [Light travels faster, remember?] Divide this by 5 to get the
𝑚
1
approximate distance to the lightning in miles. v at 20𝑜 𝐶 = 343 ≅ 𝑚𝑖/𝑠.
𝑠
3
Intensity and Sound Level
✓ Intensity (I); the average rate per unit area at which energy is transferred by the wave through
or onto the surface.
𝑃
𝐴
𝐼=
(5)
P = power (time rate of energy transfer) of the sound wave and A = surface area intercepting the
sound.
✓ Intensity I is related to the displacement amplitude 𝑦𝑚 of the sound by the relation;
𝐼=
NOTE: 𝑃𝑎𝑣𝑔 =
11/04/2022
2
1
2
𝜇𝑣𝜔 𝑦𝑚
2
1
𝜌𝑣𝜔2
2
⇒ 𝐼=
𝑃
𝐴
=
𝑦𝑚 2
2
1
2
𝜇𝑣𝜔 𝑦𝑚
2𝐴
AAMUSTED_DIS_S. Y.
(6)
𝐵𝑢𝑡
𝜇
𝐴
=𝜌
50
✓ Hence, the intensity of a periodic sound wave is proportional to the square of the
displacement amplitude and to the square of the angular frequency (Eqn, 6).
✓ But the relationship between pressure amplitude and displacement amplitude for a sound
wave is given by;
∆𝑃𝑚𝑎𝑥 = 𝜌𝑣𝜔𝑆𝑚𝑎𝑥
(7)
[LOOKUP: Pressure variations in sound]
✓ Hence eqn. (6) can be written as;
𝐼=
∆𝑃𝑚𝑎𝑥 2
2𝜌𝑣
(8)
✓ Ignoring echoes and assuming that the sound source is a point source that emits the sound
isotropically—that is, with equal intensity in all directions (as in Fig. 9). The intensity at the
sphere becomes; (Ps is the power of the source)
𝐼
11/04/2022
𝑃
𝑃𝑠
= =
𝐴
4𝜋𝑟 2
AAMUSTED_DIS_S. Y.
(9)
51
✓ Displacement amplitude at the human ear is 10−5 − 10−11 (loudest tolerable sound to the faintest
detectable sound).
✓ Sound level 𝛽, defined as;
𝛽 = 10
𝐼
𝑑𝐵 log
𝐼𝑜
(10)
✓ dB = decibel, and 𝐼𝑜 = standard reference intensity (10−12 𝑊/𝑚2 ) = Threshold Intensity = 0 dB
𝑾
✓ 𝟏𝟎−𝟏𝟐 𝟐 = 𝟎 𝒅𝑩 corresponds to a sound of frequency 1000 Hz, which is a standard reference
𝒎
frequency in acoustics.
✓ There is however, no simple relationship between physical measurements and psychological
“measurements” of sound
✓ A 100-Hz, 30-dB sound is psychologically “equal” in loudness to the 1 000-Hz, 0-dB sound
✓ Both are just barely audible, but are not physically equal in sound level.
✓ Humans are sensitive to frequencies ranging from about 20 Hz to about 20 000 Hz.
✓ The human ears are particularly insensitive to low frequencies and low intensity levels.
11/04/2022
AAMUSTED_DIS_S. Y.
52
Beats
✓ If two sounds with similar frequencies (say 552 and 564 Hz), reaches the ears simultaneously, the
sound heard will have a frequency of 558 Hz, (average of the two combining frequencies).
✓ A striking variation in the intensity of the resulting sound is also heard, increasing and decreasing in
slow, wavering beats repeated at a frequency of 12 Hz, (difference between the two original
frequencies).
✓ From this beat phenomenon, it can be said that beats arise when two waves having slightly different
frequencies, f1 and f2, are detected together. The beat frequency is derived as;
𝑓𝑏𝑒𝑎𝑡 = 𝑓1 − 𝑓2
(11)
Applications of Sound in Medicine
✓ Ultrasonic Imaging: Ultrasonic imaging is mostly employed to; examine the developing fetus, detect tumors in
the liver, kidney, brain, and pancreas, monitor the movement of the heart valves (“echocardiography”) and large
blood vessels in real-time.
✓ Cavitron Ultrasonic Surgical Aspirator (CUSA): Used to remove brain tumors once thought to be inoperable.
✓ Doppler flow meter: A medical application of the Doppler effect that measures the speed of blood flow, using
transmitting and receiving elements placed directly on the skin
11/04/2022
AAMUSTED_DIS_S. Y.
53
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. A 12 × 10−5 W of sound power passes perpendicularly through surfaces 1 and 2. These surfaces have
areas of A1 4.0𝑚2 and A2 12 𝑚2 . Determine the sound intensity at each surface and discuss why
listener 2 hears a quieter sound than listener 1. [The intensity is 𝟑 × 𝟏𝟎−𝟓 𝑾/𝒎𝟐 for A1 and 𝟏 ×
𝟏𝟎−𝟓 𝑾/𝒎𝟐 for A2 respectively. A2 is quieter due to its larger area]
2. Audio system 1 produces an intensity level of 𝛽1 = 90.0 𝑑𝐵 and system 2 produces an intensity level
𝐼
of 𝛽2 = 93.0 𝑑𝐵 . The corresponding intensities (in 𝑊/𝑚2 ) are 𝐼1 and 𝐼2 . Determine the ratio 2 .
[𝜷𝟐 − 𝜷𝟏 = 𝟗𝟑 − 𝟗𝟎 = 𝟏𝟎 𝒅𝑩 𝐥𝐨𝐠
𝑰𝟐
𝑰𝟏
𝐼1
= 𝟐. 𝟎 ]
3. A point source emits sound waves with an average power output of 80.0 W. Find the intensity 3.00 m
from the source and the distance at which the intensity of the sound is 1 × 10−8 𝑊/𝑚2 . [0.707
𝑾/𝒎𝟐 and 𝟐. 𝟓𝟐 × 𝟏𝟎𝟒 𝒎 ]
4. Two students with identical voices are seated the same distance from the lecturer. The intensity of
sound delivered by each student at the lecturer's location is 2 × 10−7 𝑊/𝑚2 . Find the sound level
heard by the lecturer when one student is talking and when both students are talking. [ 53 dB and 56
dB ]
11/04/2022
AAMUSTED_DIS_S. Y.
54
ASSIGNMENT_1
1.
Show that 𝑣 =
𝜏
𝜇
by the direct application of Newton’s laws. Where v is the speed of waves, 𝜏 is
the tension, and 𝜇 is the linear density of the string. [8 marks]
2.
Calculate the sound level (in decibels) of a sound wave that has an intensity of 4.00 𝜇𝑊/𝑚2 . [3
marks]
3.
The area of a typical eardrum is about 5 × 10−5 𝑚2 . (a) Calculate the average sound power incident
on an eardrum at the threshold of pain, which corresponds to an intensity of 1.00 𝑊/𝑚2 . (b) How
much energy is transferred to the eardrum exposed to this sound for 1.00 min?
[4 marks]
4
At a distance of 3.8 m from a siren, the sound intensity is 3.6 × 10−2 𝑊/𝑚2 . Assuming that the siren
radiates sound uniformly in all directions, find the total power radiated. [4 marks]
5.
Calculate the beat frequency for the two Frequencies of Waves given by 720Hz and 280 Hz
respectively. [1 marks]
11/04/2022
AAMUSTED_DIS_S. Y.
55
DOPPLER EFFECT
11/04/2022
AAMUSTED_DIS_S. Y.
56
✓ An ambulance parked by the hospital is sounding its 1000 Hz siren. If you are also parked by the
hospital, you will hear that same frequency.
✓ However, if the ambulance starts to move, and there is relative motion between you and the
ambulance, either toward or away from each other, you will hear a different frequency.
✓ Consider the following cases:
• driving toward the ambulance at 75 mi/h; a higher frequency (1096 Hz, an increase of 96 Hz)
will be heard.
• driving away from the ambulance at that same speed, a lower frequency (904 Hz, a decrease of
96 Hz) will be heard.
✓ Motion-related frequency changes are examples of Doppler effect. Doppler effect holds for sound
and electromagnetic waves
✓ Doppler shift: the difference between the observed f ′ and source 𝑓 frequencies.
✓ Considering only sound waves, and taking air as a reference frame through which these waves travel,
let’s measure the speeds of a source S of sound waves (Ambulance) and a detector D of those waves
(You) relative to that body of air
11/04/2022
AAMUSTED_DIS_S. Y.
57
✓ Assume the following two cases: 1. Detector (You) moving, Source (ambulance) stationary and 2.
Source (ambulance) moving, Detector (You) stationary
✓ Generally, if either the detector or the source is moving, or both are moving, the emitted frequency f
and the detected frequency f ′ are related by;
′
f =
v±vD
f
v±vS
(1)
✓ v = speed of sound through the air, 𝑣𝐷 = detector’s speed relative to the air, and 𝑣𝑆 = source’s speed
relative to the air. The choice of plus or minus signs is set by either case 1 or 2.
Detector (You) Moving, Source (ambulance) Stationary
✓ When D is also stationary, there will be no Doppler effect;𝑣𝐷 = 𝑣𝑆 = 0 &
𝑓 ′ = 𝑓 = 𝜆𝑣
(2)
✓ If D is moving at speed 𝑣𝐷 towards a stationary source S that emits spherical wavefronts, of
wavelength λ and frequency f, moving at the speed v of sound in air, 𝑣𝑆 = 0 & 𝑓 ′ > 𝑓
That is;
11/04/2022
𝑓′
=
𝑣+𝑣𝐷
𝑣+𝑣𝐷
′
→𝑓 =𝑓
𝜆
𝑣
AAMUSTED_DIS_S. Y.
(3)
58
′
𝑣−𝑣𝐷
𝑓
𝑣
𝑣±𝑣𝐷
𝑓
𝑣
(5)
′
✓ If D (You) is moving at speed vD away from a stationary source S; f < f
✓ Hence, the general eqn. for “ moving D and stationary S is;
𝑓′
=
𝑓 =
(4)
Source (ambulance) Moving, Detector (You) Stationary
✓ When D is also stationary, there will be no Doppler effect; 𝑣𝐷 = 𝑣𝑆 = 0 &
𝑣
𝑓 ′ = 𝑓 = 𝜆 (6)
✓ If S is moving at speed 𝑣𝑆 , emitting spherical wavefronts, of wavelength λ and frequency f, moving
at the speed v of sound in air, towards a stationary detector D (You), 𝑣𝐷 = 0 & 𝑓 ′ > 𝑓
That is;
𝑓′
=
𝑣
𝑓
𝑣−𝑣𝑆
(7)
✓ If S is moving at speed 𝑣𝑠 away from a stationary detector D; 𝑓 ′ < 𝑓
𝑓′
=
𝑣
𝑓
𝑣+𝑣𝑆
✓ Hence, the general eqn. for “ moving S and stationary D is;
11/04/2022
AAMUSTED_DIS_S. Y.
(8)
𝑓′
=
𝑣
𝑓
𝑣±𝑣𝑆
(9)
59
Getting Back the General Doppler Effect Equation
✓ To get the general Doppler equation in eqn. (1), replace eqn. (5) with the 𝑓 ′ in eqn. (9)
𝑓′
=
𝑣±𝑣𝐷
𝑓
𝑣
×
𝑣
𝑣±𝑣𝑆
=
𝑣±𝑣𝐷
𝑓
𝑣±𝑣𝑆
Note: Equation (10) is the
(10)
eqn. to remember ☺
✓ Eqn. (10) holds when both D and S are moving, as well as in the two specific situations
discussed above.
✓ In a nutshell; 1. when D is moving and S is stationary, substitute 𝒗𝑺 = 𝟎 into eqn. (10) and 2.
when S is moving and D is stationary, substitute 𝒗𝑫 = 𝟎 into eqn. (10) .
TRY:
1.
An ambulance is traveling at a speed of 44.7 m/s when the driver sounds the 415-Hz horn. The speed of sound
is 343 m/s. What are the frequency and wavelength of the sound, as perceived by a person standing by the
hospital as the ambulance is;
a) Approaching (moving towards) the observer b) Leaving / moving away from the observer?
11/04/2022
AAMUSTED_DIS_S. Y.
60
Solution
As the ambulance approaches (moves towards), the person hears a sound with a frequency
greater than 415 Hz because of the Doppler effect. But as it moves away, the person hears a
frequency that is less than 415 Hz .
a) When the ambulance approaches (moves towards), the observed frequency is;
𝑓′ = 𝑓
𝑣
𝑣−𝑣𝑆
= 415 𝐻𝑧
343 𝑚/𝑠
343 𝑚/𝑠 −44.7 𝑚/𝑠
∴ 𝑓 ′ = 477.2 𝐻𝑧 ≈ 𝟒𝟕𝟕 𝑯𝒛But
= 415 𝐻𝑧
𝜆′ =
𝑣
𝑓′
=
343 𝑚/𝑠
298.3 𝑚/𝑠
343 𝑚/𝑠
477 𝐻𝑧
= 𝟎. 𝟕𝟏𝟗 𝒎
b) When the ambulance leaves, the observed frequency is
𝑓′
𝑣
343 𝑚/𝑠
343 𝑚/𝑠
=𝑓
= 415 𝐻𝑧
= 415 𝐻𝑧
𝑣 + 𝑣𝑆
343 𝑚/𝑠 + 44.7 𝑚/𝑠
387.7 𝑚/𝑠
∴ 𝑓 ′ = 367.15 𝐻𝑧 ≈ 𝟑𝟔𝟕 𝑯𝒛
11/04/2022
But
𝜆′ =
AAMUSTED_DIS_S. Y.
𝑣
𝑓′
=
343 𝑚/𝑠
367 𝐻𝑧
= 𝟎. 𝟗𝟑𝟓 𝒎
61
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. When a car is at rest, its horn emits a frequency of 600 Hz, A lady standing in the middle of
the street with this car behind her hears the horn at a frequency of 580 Hz. Does she really
have to run out of the way? Why? [No, because the observed frequency is less than the
source frequency 𝒇′ < 𝒇 , hence the car is moving away from her NOT towards her]
2. Two submarines are under water and approaching each other head-on. Sub A
has a speed of 12 m/s and sub B has a speed of 8 m/s. Sub A sends out a
1550-Hz sonar wave that travels at a speed of 1522 m/s. What is the
𝒗+𝒗𝑫
′
frequency detected by sub B? [𝒇 = 𝒇
= 𝟏𝟓𝟕𝟎 𝑯𝒛]
𝒗−𝒗𝑺
11/04/2022
AAMUSTED_DIS_S. Y.
62
STANDING WAVES &
RESONANCE: WITH
BOUNDARY CONDITIONS
11/04/2022
AAMUSTED_DIS_S. Y.
63
✓ A standing wave is an oscillation pattern with a stationary outline that results from the superposition
of two identical waves traveling in opposite directions.
✓ This interference only produces a standing wave at certain specific frequencies
✓ The patterns of standing waves do not move left or right; the locations of the maxima and minima
do not change.
✓ Such a standing wave is said to be produced at resonance, and the selected frequencies, called
resonant frequencies.
✓ Resonance: occurs when one object vibrating at the same natural frequency as a second object,
forces that second object into vibrational motion.
✓ Resonance is important in the use of musical instruments which rely on air columns.
✓ A common model of resonance is a resonance tube; a hollow cylindrical tube partially filled with
water and forced into vibration by a tuning fork.
✓ The tuning fork forces the air in the tube into resonance.
11/04/2022
AAMUSTED_DIS_S. Y.
64
✓ Consider sound waves from two loudspeakers, facing each other and emitting sound of the same
frequency and amplitude
✓ This gives two identical waves travelling in opposite directions in the same medium which will
combine in accordance with waves in the interference model.
✓ This situation can be analyzed by considering wave functions;
𝑦1 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡) 𝑎𝑛𝑑
𝑦2 = 𝐴 sin(𝑘𝑥 + 𝜔𝑡)
(1)
✓ 𝑦1 traveling in the +ve x direction and 𝑦2 traveling in the -ve x direction.
𝑦 = 𝑦1 + 𝑦2 = 𝐴 sin(𝑘𝑥 − 𝜔𝑡) + 𝐴 sin(𝑘𝑥 + 𝜔𝑡)
(2) [Superposition Principle]
✓ Using the trigonometric identity sin 𝑎 ± 𝑏 = 𝑠𝑖𝑛𝑎 𝑐𝑜𝑠𝑏 ± cos 𝑎 sin 𝑏, eqn. (2) reduces to;
y = (2𝐴 sin 𝑘𝑥)𝑐𝑜𝑠 𝜔𝑡
(3)
✓ Eqn. (3) is not an expression for a single traveling wave since it does not contain a function of
𝑘𝑥 ± 𝜔𝑡
✓ Eqn. (3) also shows that the amplitude of the simple harmonic motion of an element of the medium
has a minimum value of zero when x satisfies the condition sin kx = 0, that is, when;
11/04/2022
AAMUSTED_DIS_S. Y.
65
𝜆
𝜆
𝜆
𝜆
𝜆 3𝜆
𝑘𝑥 = 0, 𝜋, 2𝜋, 3𝜋, … → 𝑥 =
× 0,
× 𝜋,
× 2𝜋,
× 3𝜋 = 0, , 𝜆,
....
2𝜋
2𝜋
2𝜋
2𝜋
2
2
𝑛𝜆
=
for 𝑛 = 0,1,2,3, . . .
2
✓ These points of zero amplitude are called nodes.
✓ The positions in the medium at which an element experiences the greatest maximum
displacement occurs are called antinode
✓ This element has an amplitude of 2A (Amplitude of a standing wave)
✓ Antinodes are located at positions for which the coordinate x satisfies the condition
sin 𝑘𝑥 = ±1 , that is, when
𝑘𝑥 =
𝜋 3𝜋 5𝜋
, , ,...
2 2 2
∴𝑥=
𝜆 3𝜆 5𝜆
𝑛𝜆
, , ,...
4 4 4
4
𝑓𝑜𝑟 𝑛 = 1,3,5, . . .
11/04/2022
AAMUSTED_DIS_S. Y.
66
Inverted Relative
Pattern in (a)
(A)
In Phase
(B)
Out of Phase
(C)
In Phase
Fig. 10: Standing-wave patterns produced at various times by two waves traveling in opposite directions. The resultant wave
y, the nodes (N) are points of zero displacement and the antinodes (A) are points of maximum displacement.
11/04/2022
AAMUSTED_DIS_S. Y.
67
𝜆
✓ Distance between adjacent antinodes is equal to
2
𝜆
✓ Distance between adjacent nodes is equal to
2
𝜆
✓ Distance between a node and an adjacent antinode is .
4
TRY
1. Two waves traveling in opposite directions produce a standing wave. The individual wave
functions are 𝑦1 = 4.0 sin(3.0 𝑥 − 2.0 𝑡) and 𝑦2 = 4.0 sin(3.0 𝑥 + 2.0 𝑡) where x and y
are measured in centimeters and t is in seconds. (A) Find the amplitude of the simple
harmonic motion of the element of the medium located at x = 2.3 cm. (B) Find the positions
of the nodes and antinodes if one end of the string is at x = 0.
11/04/2022
AAMUSTED_DIS_S. Y.
68
Solution
𝑦1 and 𝑦2 are identical except for their directions of travel; A = 4.0 , k =
rad
3.0
,
cm
&ω=
rad
2.0
cm
(A) For x = 2.3 at t =0
y = 𝑦1 + 𝑦2 = (2𝐴 sin 𝑘𝑥)𝑐𝑜𝑠 𝜔𝑡 = 8.0 𝑠𝑖𝑛3.0𝑥 cos 2.0 𝑡 = 8.0 𝑠𝑖𝑛6.9 𝑟𝑎𝑑 = 4.6
(B) Finding the wavelength; k =
2𝜋
𝜆
→𝜆=
2𝜋
𝑘
=
2𝜋
3.0
For the nodes;
𝜆 3𝜆
𝑛𝜆 𝑛 2𝜋ൗ3
𝜋
x = 0, , 𝜆,
.... =
=
=𝑛
2
2
2
2
3
For antinodes;𝑥 =
11/04/2022
𝜆 3𝜆 5𝜆
𝑛𝜆
, , ,...
4 4 4
4
=𝑛
2𝜋Τ
3
4
=𝑛
𝜋
6
for 𝑛 = 0,1,2,3, . . .
𝑓𝑜𝑟 𝑛 = 1, 3, 5, 7, . . .
AAMUSTED_DIS_S. Y.
69
✓ For a string of length L fixed at both ends; the normal modes of oscillation can be described
by imposing the boundary conditions that the ends be nodes and that the nodes be separated
by a specific wavelength with antinodes in between the nodes as shown in Fig. 11.
A
B
C
Fig. 11: Harmonic series formed from the normal modes of vibration for a string fixed at both ends
11/04/2022
AAMUSTED_DIS_S. Y.
70
✓ First normal mode (Fig. 11 A);
• occurs when the wavelength 𝜆1 = 2𝐿
• the string is vibrating in one loop (The section from one node to the next node)
✓ Second normal mode (Fig. 11 B);
• occurs when the wavelength 𝜆2 = 𝐿
• the string is vibrating in two loops
✓ Third normal mode (Fig. 11 C);
•
occurs when the wavelength 𝜆3 =
•
the string is vibrating in three loops
2𝐿
3
So generally;
2𝐿
𝜆𝑛 =
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, … .
𝑛
n refers to the nth normal mode of
oscillation
✓ The natural frequencies associated with the modes of oscillation are obtained from the
𝑣
relationship f = , where the wave speed v is the same for all frequencies.
𝜆
✓ Hence the natural frequencies 𝑓𝑛 of the normal modes are; 𝑓𝑛 = 𝑛
11/04/2022
AAMUSTED_DIS_S. Y.
𝑣
2𝐿
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, …
71
✓ But for waves on a string, 𝑣 =
𝜏
𝜇
✓ The fundamental frequency 𝑓1 =
, hence 𝑓𝑛 =
1
2𝐿
𝜏
𝜇
𝑛
2𝐿
𝜏
𝜇
This means, the string’s frequency is
directly proportional to it’s tension and
inversely proportional to its length
is the frequency of the first harmonic. The second
harmonic will follow as 𝑓2 = 2𝑓1 . So generally, 𝑓𝑛 = 𝑛𝑓1 is that of the nth harmonic
✓ The waves under boundary conditions model can also be applied to sound waves in an air
column
✓ The resulting wave in this case is due to interference between longitudinal sound waves
traveling in opposite directions.
✓ Consider a pipe closed at one end, and another pipe open at both ends (Fig. 12); the closed
end is a displacement node or a pressure antinode
✓ The open ends of an air column are approximately displacement antinodes (end correction =
0.6R) or pressure nodes.
11/04/2022
AAMUSTED_DIS_S. Y.
72
A
B
Fig. 12: Standing longitudinal waves in (a) a column open at both ends and (b) a column closed at one end.
11/04/2022
AAMUSTED_DIS_S. Y.
73
✓ From Fig. 12 A [Open at both ends];
• For the first mode, the standing wave extends between two adjacent antinodes which is
half a wavelength; 𝜆1 = 2𝐿 where L is the length of the pipe.
✓ The general relation;
𝜆𝑛 =
2𝐿
𝑛
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, … . also works in this case of
sound (standing) waves in air columns. Where n refers to the nth normal mode of oscillation
𝑣
𝑛
2𝐿
✓ The fundamental frequency also follows suite; 𝑓𝑛 =
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, … and
hence 𝑓𝑛 = 𝑛𝑓1 , giving the frequencies of the higher harmonics as; 2𝑓1 , 3𝑓1 , . . .
Note: 𝑣 here refers to speed of sound waves in air NOT the speed of waves on strings
✓ From Fig. 12 B [Closed at one end];
• For the first mode, the standing wave extends from an antinode to an adjacent node
which is one-fourth of a wavelength; 𝜆1 = 4𝐿 where L is the length of the pipe.
11/04/2022
AAMUSTED_DIS_S. Y.
74
✓ The general relation;
𝜆𝑛 =
4𝐿
𝑛
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 2, 3, … . also works in this case of sound
(standing) waves in air columns. Where n refers to the nth normal mode of oscillation
𝑣
𝑛
4𝐿
✓ The fundamental frequency also follows suite; 𝑓𝑛 =
𝑤ℎ𝑒𝑟𝑒 𝑛 = 1, 3, 5, … and
hence 𝑓𝑛 = 𝑛𝑓1 , giving the frequencies of the higher harmonics as; 3𝑓1 , 5𝑓1 , . . .
✓ Musical instruments based on air columns are generally excited by resonance.
11/04/2022
AAMUSTED_DIS_S. Y.
75
𝐄𝐧𝐝 𝐨𝐟 𝐓𝐨𝐩𝐢𝐜 𝐓𝐫𝐢𝐚𝐥𝐬
1. The middle C string on a piano has a fundamental frequency of 262 Hz, and the string for the
first A above middle C has a fundamental frequency of 440 Hz. Calculate the frequencies of
the next two harmonics of the C string. [𝒇𝟐 = 𝟐𝒇𝟏 = 𝟐 𝟐𝟔𝟐 = 𝟓𝟐𝟒 𝑯𝒛, 𝒂𝒏𝒅 𝒇𝟑 =
𝟑𝒇𝟏 = 𝟕𝟖𝟔 𝑯𝒛 ]
2. Two sinusoidal waves traveling in opposite directions interfere to produce a standing wave
with the wave function y = 1.50 sin 0.400𝑥 cos(200𝑡). where x and y are in meters and t
is in seconds. Determine (a) the wavelength, (b) the frequency, and (c) the speed of the
interfering waves. [𝝀 = 𝟏𝟓. 𝟕 𝒎, 𝒇 = 𝟑𝟏. 𝟖 𝑯𝒛, 𝒂𝒏𝒅 𝒗 = 𝟓𝟎𝟎 𝒎/𝒔 ]
11/04/2022
AAMUSTED_DIS_S. Y.
76
ASSIGNMENT 2
1. An ambulance moving at 42 m/s sounds its siren whose frequency is 450 Hz. A car is
moving in the same direction as the ambulance at 25 m/s. What frequency does a person in
the car hear (a) as the ambulance approaches the car? (b) After the ambulance passes the
car? [6 Marks]
2. A section of drainage culvert 1.23 m in length makes a howling noise when the wind blows
across its open ends.. Determine the frequencies of the first three harmonics (𝑓1 , 𝑓2 , 𝑓3 ) of
the culvert if it is cylindrical in shape and open at both ends. Take v = 343 m/s as the speed
of sound in air. [6 𝐌𝐚𝐫𝐤𝐬]
3. Verify by direct substitution that the wave function for a standing wave given by; y =
(2𝐴 sin 𝑘𝑥)𝑐𝑜𝑠 𝜔𝑡 is a solution of the general linear wave equation;
𝜕2 𝑦
𝜕𝑥 2
=
1 𝜕2 𝑦
𝑣 2 𝜕𝑡 2
[8 Marks]
11/04/2022
AAMUSTED_DIS_S. Y.
77
READING ASSIGNMENT_3
1.Explain standing waves in rods and membranes
2.Explain the structure and functions of the human ear
11/04/2022
AAMUSTED_DIS_S. Y.
78