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1. Mechanical Waves
(University Physics, 13th edition, Chapter 15 )
❑ 4 lectures (4 hours) for this chapter:
Lecture 18 & 19 : Different kinds of wave, types and properties of
Mechanical waves, Sinusoidal waves, Mathematical description of a
sinusoidal wave, Graphs of wave function, Speed of a sinusoidal
wave and related problems, the wave equation, Speed of wave
along a stretched string and related problems.
Lecture 20 & 21: Reflection of Waves and Superposition Principle,
Interference of waves and related problems, Standing waves and
related problems, Normal modes of vibration of a string and
related problem, Review and Discussion on Problems for Practice.
Quiz 4 will be on this Chapter.
Final, Spring 2023-24

2. Waves:
• Waves can occur whenever a system is disturbed from equilibrium and
when the disturbance can travel, or propagate from one region of the
system to another. Ripples on a pond, musical sounds, light from the
sun- all these are waves
• As a wave propagates it carries energy. The energy of light waves from
the sun warms the surface of our planet; the energy of seismic waves
can crack our planet’s crust.
• Electromagnetic Waves: Can propagate in empty space, where there
is no medium. Light, Radio waves, x rays, infrared and ultraviolet
radiations, etc.
• Mechanical Waves: Waves that travels within some material called
medium. Waves through a stretched string, sound, water waves, etc.

3. 15.1 Types of Mechanical Waves:

4. ❑ Properties of Mechanical Waves:
1. The disturbance travels with a definite
speed called wave speed through the
medium.
2. The medium itself does not travel through
space, its individual particle undergoes
back-and-forth or up-and-down motions
around their equilibrium positions.
3. The wave motion transport energy from
one region of the medium to other.
➢Wave transport energy, but no matter, from one region to another.

5. 15.2 Sinusoidal Waves
• Each particle of the string
oscillates with SHM. Particles
of the string oscillates with
amplitude A, frequency f, time
period T, and angular
frequency 𝜔 like the mass of
the mass-spring system.
• Periodic waves with SHM are
called sinusoidal waves and
easy to analyze.
• Any periodic motion can be
represented as combination of
sinusoidal waves. So this kind
of wave is important to
understand.

6. Do you
understand?
• The wave pattern travels with constant
speed v and advances a distance of one
wavelength 𝜆 in a time interval of one time
period T. So the wave speed is given by
𝑣 =
𝜆
𝑇
= 𝜆𝑓 [as 𝑓 =
1
𝑇
]

9. Do you
understand?
➢Every particles of the string oscillates with SHM with
the same amplitude and frequency. But the
oscillations of particles at different positions on the
string are not all in step with each other.

10. 15.3 Mathematical description of a wave:
The transverse displacement of the particles of the propagating
medium is considered as the wave function of a sinusoidal wave that is
a function of both position and time:
𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡) [wave moves in +x direction]
𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 + 𝜔𝑡) [wave moves in -x direction]
where k is the wave number k, defines as 𝑘 =
2𝜋
𝜆
, unit: rad/m.
• For a travelling sinusoidal wave:
𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 ∓ 𝜔𝑡)
The quantity (𝑘𝑥 ∓ 𝜔𝑡) is called the phase, it plays the role of an
angular quantity (always measured in radians).

11. Graphs of Wave Function: 𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡)
Figure shows two graphs of wave function y(x,t):
(a) displacement y vs coordinate x at time t=0,
(b) displacement y vs time t at position x=0.
(a) (b)

12. Speed of a Sinusoidal Wave:
• The wave speed is the speed with which a phase of the wave
moves. It is also called phase speed.
• In the fig., a phase A moves a distance Δx in time Δt. The ratio
Δx/Δt (or, in the differential limit, dx/dt) is the wave speed, v.
• For a particular phase the transverse displacement remains
always constant and its possible only if the phase angle
remains constant. Thus, for a sinusoidal wave traveling in the
+x direction
𝑘𝑥 − 𝜔𝑡 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑑
𝑑𝑡
𝑘𝑥 − 𝜔𝑡 = 0
𝑑𝑥
𝑑𝑡
=
𝜔
𝑘
= 𝑣
[Also, 𝑣 =
𝜔
𝑘
=
2𝜋𝑓
2𝜋
𝜆
= 𝑓𝜆]
• In the same way, for a sinusoidal wave traveling in the -x
direction we can show that
𝑣 = −
𝜔
𝑘
Figure: Two snap shots of
a sinusoidal wave at t=0
and then at time t=Δt.
𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡)

13. ❑ A sinusoidal wave travels along a string. The time for a particular point to
move from maximum displacement to zero is 0.170 s. What are (a) the
period and (b) frequency? (c) The wavelength is 1.40 m; what is the wave
speed?
(a) t1 –t2 = T/4 = 0.170 s
T = 4 (0.170) s = 0.680 s
(b) f =1/T = (1/ 0.680) Hz = 1.47 Hz
(c) λ = 1.40 m
v = fλ = 1.47 (1.40) m/s = 2.06 m/s

14. The Wave Equation:
For any particle of a sinusoidal travelling wave along the + x axis:
• Transverse Displacement: 𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡)
• Acceleration:
𝜕2𝑦 𝑥,𝑡
𝜕𝑡2 = −𝜔2
𝐴𝑐𝑜𝑠 𝑘𝑥 − 𝜔𝑡 = −𝜔2
𝑦 𝑥, 𝑡
• Second partial derivatives with respect to x:
𝜕2
𝑦 𝑥, 𝑡
𝜕𝑥2
= −𝑘2𝐴𝑐𝑜𝑠 𝑘𝑥 − 𝜔𝑡 = −𝑘2𝑦 𝑥, 𝑡
• Thus
𝜕2𝑦 𝑥,𝑡
𝜕𝑡2
𝜕2𝑦 𝑥,𝑡
𝜕𝑥2
=
𝜔2
𝑘2 = 𝑣2 [𝑣 =
𝜔
𝑘
]

15. Wave speed on a Stretched String:
• A small string element Δl (of mass Δm) of the pulse makes an angle 2ϴ
at the center of the circle.
• The tension 𝜏 acts along the tangent on both sides of the pulse.
• The horizontal components (𝜏𝑐𝑜𝑠𝜃 and −𝜏𝑐𝑜𝑠𝜃) of 𝜏 cancels each
others.
• The resultant restoring force
𝐹 = 2𝜏𝑠𝑖𝑛𝜃 ≈ 𝜏 2𝜃 = 𝜏
∆𝑙
𝑅
[if 𝜃 is small]
• The centripetal acceleration acts towards the center with the
magnitude
𝒂 = 𝒗𝟐
/𝑹
and the mass of the element (Δl) is
∆𝑚 = 𝜇∆𝑙
where 𝜇 (= ∆𝑚/ ∆𝑙) is the mass per unit length of the string.
ϴ
ϴ
𝜏
𝜏
𝜏𝑐𝑜𝑠𝜃
𝜏𝑐𝑜𝑠𝜃
𝜏𝑠𝑖𝑛𝜃
𝜏𝑠𝑖𝑛𝜃

16. Wave speed on a Stretched String:
• Applying Newton’s second law of motion (𝐹 = 𝑚𝑎):
𝜏
∆𝑙
𝑅
= 𝜇∆𝑙
𝑣2
𝑅
𝒗 =
𝝉
𝝁
➢This equation gives the wave speed for only the special case of mechanical waves
on a stretched string or rope.
➢For many types of mechanical waves:
v =
Restoring force returning the system to equilibrium
Inertia resisting the return to equilibrium

17. ❑ The equation of a transverse wave on a string is y = (2.0 m)cos[(20 m-1)x - (600 s-1)t].
The tension in the string is 15 N. (a) What is the wave speed? (b) Find the linear density of
this string in grams per meter.
Solution: Given, ym = 2.0 m
k = 20 rad/m
𝜔 = 600 rad/s
τ = 15 N
(a) v =
𝜔
𝑘
=
600
20
= 30 m/s
𝑏 𝑣 = √
τ
𝜇
𝑣2 =
τ
𝜇
𝜇 =
τ
𝑣2 =
15
(30)2 = 1.67x10-2 kg/m = 16.7 gm/m
y = (2.0 m)cos[(20 m-1)x - (600 s-1)t]
𝑦 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡)

20. 15.6 Wave Interference, boundary condition
and superposition:
• When a wave strikes the
boundaries of its medium, all or
part of the wave is reflected.
• The initial and reflected waves
overlaps in the same region of
the medium. The overlapping of
the waves is called the
interference.

21. Superposition
Principle:
The transverse displacement of
the medium is the algebraic sum
of the transverse displacement
of the individual waves.
𝒚 𝒙, 𝒕 = 𝒚𝟏 𝒙, 𝒕 + 𝒚𝟐 𝒙, 𝒕

22. Interference of Waves:
Suppose we send two sinusoidal waves of the same wavelength and amplitude in the
same direction along a stretched string.
y1(x, t) = Acos(kx - 𝜔t)
y2(x, t) = Acos(kx - 𝜔t +𝜑)
𝜔 (f ), k (𝜆 ), A , v are same
Superposition principle, y(x, t) = y1(x, t) + y2(x, t)
y(x, t) = Acos(kx - 𝜔t)+ Acos(kx - 𝜔t +𝜑)
= A{cos(kx - 𝜔t)+ cos(kx - 𝜔t +𝜑)}
= A{2 cos ( 𝑘𝑥 − 𝜔𝑡+ 𝑘𝑥 − 𝜔𝑡 +𝜑
2
) cos ( 𝑘𝑥 − 𝜔𝑡− 𝑘𝑥+ 𝜔𝑡 − 𝜑
2
)}
= 2Acos{2( 𝑘𝑥 − 𝜔𝑡) +𝜑
2
} cos ( 𝑘𝑥 − 𝜔𝑡− 𝑘𝑥+ 𝜔𝑡 − 𝜑
2
)
= 2Acos{ 𝑘𝑥 − 𝜔𝑡 + 𝜑
2
} cos (− 𝜑
2
)
y(x, t) = [2Acos (𝜑
2
)] cos (𝑘𝑥 − 𝜔𝑡 + 𝜑
2
) [traveling wave]
Resultant displacement = y(x, t) Amplitude = [2A cos (𝜑
2
)]
Oscillating term = cos (𝑘𝑥 − 𝜔𝑡 + 𝜑
2
)
[c𝒐𝒔𝑨 + 𝒄𝒐𝒔𝑩 = 𝟐𝒄𝒐𝒔
𝑨+𝑩
𝟐
𝒄𝒐𝒔
𝑨−𝑩
𝟐
]

23. If two sinusoidal waves of the same amplitude and wavelength travel in the same direction along a
stretched string, they interfere to produce a resultant sinusoidal wave traveling in that direction.
y(x, t) = [2Acos (0
2
)] cos(𝑘𝑥 − 𝜔𝑡+
0
2
) }
The resultant wave differs from the interfering waves in two respects: (1)
its phase constant is
𝜑
2
and (2) its amplitude is [2ym cos (𝜑
2
)]
Interfering waves: y1(x, t) = Acos(kx - 𝜔t)
y2(x, t) = Acos(kx - 𝜔t +𝜑)
Resultant wave : y(x, t) = [2Acos (𝜑
2
)] cos (𝑘𝑥 − 𝜔𝑡 + 𝜑
2
)
(1) If 𝜑 = 0 rad (00): fully constructive interference
= [2Acos0] cos 𝑘𝑥 − 𝜔𝑡
y(x, t) = 2Acos 𝑘𝑥 − 𝜔𝑡 [greatest amplitude]
(2) If 𝜑 = 𝜋 rad (1800): fully destructive interference
y(x, t) = [2Acos (𝜋
2
)] cos(𝑘𝑥 − 𝜔𝑡 + 𝜋
2
)
= [2A (0)] cos(𝑘𝑥 − 𝜔𝑡 + 𝜋
2
)
y(x, t) = 0

24. (3) If 𝜑 =
2𝜋
3
rad (1200): intermediate interference
y(x, t) = [2A cos(
2𝜋
3
2
)] cos{𝑘𝑥 − 𝜔𝑡 + (
2𝜋
3
2
)}
= 2A cos (𝜋
3
) cos (𝑘𝑥 − 𝜔𝑡 + 𝜋
3
)
= 2A(1
2
) cos(𝑘𝑥 − 𝜔𝑡 + 𝜋
3
)
y(x, t) = Acos (𝑘𝑥 − 𝜔𝑡 + 𝜋
3
)

25. ❑ What phase difference between two identical traveling waves, moving in the same
direction along a stretched string, results in the combined wave having an
amplitude 1.50 times that of the common amplitude of the two combining waves?
Express your answer in (a) degrees, (b) radians, and (c) wavelengths.
y(x, t) = [2A cos (𝜑
2
)] cos(𝑘𝑥 − 𝜔𝑡 + 𝜑
2
)
(a) [2A cos (𝜑
2
)] = 1.50 A
cos (𝜑
2) =
1.50
2
= 0.75
Solution:
𝜑
2 = cos−1
(0.75) = 41.41
𝜑 = 2(41.41)= 82.820
(b) φ = 82.820 (
𝜋 𝑟𝑎𝑑
1800 )= 1.45 rad
c 2π rad = λ
1 rad = (
λ
2π
)
φ = 1.45 rad = 1.45
λ
2π
= 0.23λ

26. 15.7 Standing Wave
along a String
• An incident wave travelling to the left (pink one):
𝑦1 𝑥, 𝑡 = −𝐴𝑐𝑜𝑠 𝑘𝑥 + 𝜔𝑡
• Reflected wave (inverted, reflected from fixed end) travelling to the right (blue one):
𝑦2 𝑥, 𝑡 = 𝐴𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡)
• Applying superposition principle:
𝑦 𝑥, 𝑡 = 𝑦1 + 𝑦2 = A −cos 𝑘𝑥 + 𝜔𝑡 + 𝑐𝑜𝑠(𝑘𝑥 − 𝜔𝑡
= 2𝐴𝑠𝑖𝑛 𝑘𝑥 𝑠𝑖𝑛 𝜔𝑡 = 2𝐴𝑠𝑖𝑛𝑘𝑥 𝑠𝑖𝑛𝜔𝑡
[we know, 𝑐𝑜𝑠 𝑎 ∓ 𝑏 = 𝑐𝑜𝑠𝑎𝑐𝑜𝑠𝑏 ∓ 𝑠𝑖𝑛𝑎𝑠𝑖𝑛𝑏]
• Thus, the equation of standing wave (golden one): 𝑦 𝑥, 𝑡 = 2𝐴𝑠𝑖𝑛𝑘𝑥 𝑠𝑖𝑛𝜔𝑡, where
• Amplitude at position x: 2𝐴𝑠𝑖𝑛𝑘𝑥
• The factor 2A𝑠𝑖𝑛 𝑘𝑥 shows that at each instant the shape of the string is a sine curve.
• But unlike a wave travelling along the string, the wave shape stays in the same position,
oscillating up and down as described by the 𝑠𝑖𝑛 𝜔𝑡 factor, as shown in Fig. 5.24.

27. Standing Waves
on a String:
• Some points of the string never
oscillates, fully destructive interference
occurs, (nodes, marked as N), some
points oscillates with amplitudes 2A,
fully constructive interference occurs,
(antinodes, marked as A).
• Each point in the string still undergoes
simple harmonic motion but all the
points between any successive pair of
nodes oscillate in phase. For a travelling
wave we see a phase difference
between adjacent particles.
• Standing wave, unlike a traveling wave,
does not transfer energy from one end
to the other.

28. Positions of Nodes and Antinodes:
• The equation of standing wave: 𝑦 𝑥, 𝑡 = 2𝐴𝑠𝑖𝑛 𝑘𝑥 𝑠𝑖𝑛 𝜔𝑡
• For nodes: For any time t, y(x,t) = 0, it’s possible only when
𝑠𝑖𝑛 𝑘𝑥 = 0
This occur only when 𝑘𝑥 = 0, 𝜋, 2𝜋, 3𝜋, … . . 𝑛𝜋, and using 𝑘 =
2𝜋
𝜆
The positions of the nodes: 𝑥 = 0,
𝜆
2
,
2𝜆
2
,
3𝜆
2
, … … …
𝑛𝜆
2
, n=0, 1, 2, 3, ….
• For antinodes: Between every two nodes, at the midway, we get one
antinode.
Thus, the positions of the antinodes: 𝑥 =
𝜆
4
,
3𝜆
4
,
5𝜆
4
, … … … (𝑛 +
1
2
)
𝜆
2
,
n=0, 1, 2, 3, …

29. ❑ A string oscillate according to the equation y= (0.50 cm) sin [(π/3 cm-1)x] sin[(40π s-1)t].
What are (a) the amplitude and (b) the speed of the two waves (identical except for direction of travel) whose
superposition gives this oscillation? (c) What is the distance between nodes?
y= [(0.50 cm) sin {(π/3 cm-1)x}] sin[(40π s-1)t]
y = [2Asin kx] sinωt
(a) 2A = 0.50 cm
2A = 0.0050 m
A = 0.0050/2
A = 0.0025 m Ans.
(b) v =
ω
k
=
40π
π
0.03
= 0.03(40) = 1.20 m/s Ans.
Given, 𝑘 =
π
3
rad/cm=
π
0.03
rad/m; 𝜔 = 40π rad/s
(c) ∆𝑥 =
𝜆
2
=
2π
k
2
=
π
𝑘
=
π
π
0.03
= 0.03 m Ans.
Solution:

30. 15.8 Normal Modes of a String:
• If a string with length L is fixed in both ends,
a standing wave can exist only if its wavelengths
satisfies
𝐿 = 𝑛
𝜆
2
, (n=1,2,3,…….)
• The stable wavelengths:
𝜆𝑛 =
2𝐿
𝑛
, (n=1,2,3,…….)
• The stable frequencies:
𝑓𝑛 =
𝑣
𝜆𝑛
=
𝑛
2𝐿
𝜏
𝜇
➢These stable wavelengths/frequencies are
called harmonics or resonant wavelengths/
frequencies, and series is called harmonic
series.
L

31. Harmonic Series:
➢ Figure shows the first four normal modes of a
string fixed at both ends. n=1: fundamental mode
(2 nodes, 1 antinodes), n=2: second harmonic (3
nodes, 2 antinodes), n=3: third harmonic (4 nodes,
3 antinodes), so on.
𝜆𝑛 =
2𝐿
𝑛
, (n=1,2,3,…….)
𝑓𝑛 =
𝑣
𝜆𝑛
=
𝑛
2𝐿
𝜏
𝜇

32. ❑ A 120 cm length of string is stretched between fixed supports. What are the (a) longest,
(b) second longest, and (c) third longest wavelength for waves traveling on the string if
standing waves are to be set up? (d) Sketch those standing waves.
Solution:

33. ❑ A 125 cm length of string has a mass 2.00 g and tension 7.00 N between
fixed supports. (a) What is the wave speed for this string? (b) What is the
lowest resonant frequency of this string?
a v = √
τ
μ
= √
7.00
0.0016
= √(4375) = 66.14 m/s Ans.
(b) For the lowest resonant frequency, n = 1: f = 𝑛
2𝐿
𝜏
𝜇
Solution: Here, L = 125 cm =
125
100
= 1.25 m
m = 2.00 gm =
2
1000
kg = 0.002 kg
τ = 7.00 N
μ =
m
L
=
0.002
1.25
kg/m = 0.0016 kg/m
f1 = 1
2L
τ
μ
=
1
2L
v=
1
2 1.25
66.14 =26.46 Hz

34. Problems for
Practice
(University
Physics,
Chapter 15)
➢ Students will solve the problems
with the help of their faculties.
15.3, 15.4, 15.10, 15.15,
15.16, 15.40 and 15.42.