Waves difference with oscillation
• Waves that travel within or with a medium. As a
wave propagates, it carries energy. The energy in
light waves from the sun warms the surface of our
planet; the energy in seismic waves can crack our
planet's crust.
• Vibration is toing move about the equilibrium
position.
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Reflection
Refraction
Wave’s contidition
Interference
difraction
Polarization
Mechanical Waves,
waves that travel with
some material called
a medium.
Mechanical Waves
• Sound waves in gas,
• Sound waves in liquid
Waves category
• Sound waves in solid
• String waves
Electromagnetic waves
can propagate even in
empty space, where there
is no medium.
Electromagnetic waves
• Radio waves
• light
The displacements of the medium
are perpendicular or transverse to
the direction of travel of the wave
along the medium, this is called a
transverse wave.
Transverse wave
• Electromagnetic waves
waves
The motions of the particles of the
medium are back and forth along the
same direction that the wave travels.
We call this a longitudinal wave.
• Repples on a pond
• String waves
Longitudinal wave
• Sounds Wave
• Springs wave
The wavelength () is the
minimum distance between any
two identical points (such as the
crests) on adjacent waves
The period (T) is the time required
for two identical points (such as the
crests) of adjacent waves to pass by
a point.
The frequency of a periodic wave (f) is the number of crests (or
troughs, or any other point on the wave) that pass a given point in
a unit time interval.
The maximum displacement of a particle of the medium is called
the amplitude (A) of the wave.




Wave number (rad/m),
k = 2/
Angular frequency
(rad/dt),
 = 2  /
= 2  v/ 
=2f
v=f
v = /k

1/2 
1/4
3/4 
That is, the traveling sinusoidal wave moves to the right a
distance vt in the time t, with amplitude A (m), frequency f (Hz.),
wavelength  (m), and wave speed v (m/s) as shown in Figure)
a.
t=0
X
 2 
x,0  A sin  x 
  
b.
t=
X
v
c.
t=2
 2

x,   A sin x  v
 

X
2v
 2

x,2  A sin  x  2v
 

If the wave moves to the right with a speed v, then the
wave function at some later time t is :
 2

x, t   A sin x  vt
 

If the wave moves to the left with a speed v, then the
wave function at some later time t is :
 2

x, t   A sin x  vt
 

generally express the wave function in the form,
 2

x, t   A sin x  vt
 

The wave function can be express
x, t   A sin kx  vt
x, t   A sin kx  t 
x t 
x, t   A sin 2  
 T
x 
x, t   A sin 2f   t 
v 
When the vertical displacement y is’n zero at x=0 and t=0
Wave phase,
 x, t   A sinkx  t   
  ( kx  t  )
 is phase constant.
The combination of separate waves in the same region
of space to produce a resultant wave is called
interference
Y1(x,t) = A sin (kx-t)
Y2(x,t) = A sin (kx+ t)
Y(x,t) = A sin (kx-t) + A sin (kx+ t)
Remember,
sin α + sin β = 2 sin ½(α+ β) cos ½(α-β)
Y(x,t) = 2A sin ½ (kx-t+ kx+ t) cos ½(kx-t- kx- t)
Y(x,t) = 2A sin kx cos t = Asw sin kx cos t
standing wave on a string, fixed end at x = 0
The standing wave amplitude Asw is twice the amplitude
A of either of the original traveling waves: Asw = 2A
EXAMPLE 1
A sinusoidal wave traveling in the positive x direction
has an amplitude of 15.0 cm, a wavelength of 40.0
cm, and a frequency of 8.00 Hz. The vertical
displacement of the medium at x=0 and t=0 is also 15.0
cm, as shown in Figure. (a) Find the angular wave
number k, period T, angular frequency, and speed v of
the wave. (b) Determine the phase constant , and write
a general expression for the wave function.
Because A = 15 cm and because y = 15 cm at x=0 and t=0
substitution into Equation
By inspection, we can see that the wave function
must have this form, noting that the cosine function has
the same shape as the sine function displaced by 90°.
Substituting the values for A, k, and  into this
expression, we obtain
The speed of a wave (v) traveling on a taut
string of mass per unit length  and tension
T is
The intensity I at any
distance r is therefore
inversely proportional to r2
average power, sinusoidal
wave on a string
The Doppler Effect
d=
a
Stationary Source
b Source moves to
the right
vs

ds = vs T
 =Distance between two adjacent wave front.
vs = speed of source,
T = time = period,
First wave front was move as far as
d = v T, or  = v T
v = Speed of wave sound on the air
At the same time, source have traveled with distance
ds = vs T.
Hence, the new wave length :
 = d - ds =  - vs T =  - vs  / v =  (1- vs / v)
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When a source move toward a stationary
observer, frequency heart by observer is:
v
v
1
v
f ' 

f 
f

'
v  vs
 vs   vs 
 1   1  
 v  v
For a source moving with a speed Vs away a stationary
observer,
v
v
1
v
f   

f 
f


v  vs
 vs   vs 
 1   1  
 v  v
 Doppler effect also happen, when a source is
motionless and an observer is moving.
 Wave speed relative to the observer is change.
 If the observer move toward a stationary source,
Wave speed relative to observer is:
v = v + vp ,
v = Sound speed in air
Vp= Speed of observer.
So:
v v  v p
f  
λ
λ
because  = v/f, so:
 v  Vp 
 
f   f 
 v 
 Vp 
f 1  
v 

Generally rule,
 v  Vp 
 f
f   
 v  Vs 
Contoh Soal 2
As an ambulance travels east down a highway at a speed
of 33.5 m/s, its siren emits sound at a frequency of 400 Hz.
What frequency is heard by a person in a car traveling west at
24.6 m/s (a) as the car approaches the ambulance and (b) as the
car moves away from the ambulance?
Solution a)
Solution b)
what happens when the speed vS of a source or speed
of an observer equal with the wave speed v? what
happens when the speed vS of a source exceeds the
wave speed v?
If an observer toward a source, frequency heart by
observer is zero.
If a source toward an observer, frequency heart by
observer is unlimited. (shock waves)
Human’s ear only can heart sound with frequency
between 20-20.000 Hz.
If the speed vS of a source exceeds the wave speed v,
happen shock waves and produce sonic boom.
The shock wave carries a great deal of energy
concentrated on the surface of the cone, with
correspondingly great pressure variations. Such shock
waves are unpleasant to hear and can cause damage to
buildings when aircraft fly supersonically at low altitudes.
If the medium is not stationary
 v  Vm  V p 
 f
f   
 v  Vm  Vs 
Contoh Soal 3
Sebuah kereta api yang mendekati sebuah bukit
dengan kelajuan 40 km/jam membunyikan peluit
dengan frekuensi 580 Hz ketika kereta berjarak 1 km
dari bukit. Angin dengan kelajuan 40 km/jam bertiup
searah dengan gerak kereta.
Tentukan frekuensi yang didengar oleh seorang
pengamat di atas bukit. Cepat rambat bunyi di udara
adalah 1200 km/jam
Tentukan jarak dari bukit di mana gema dari bukit
didengar oleh masinis kereta. Berapa frekuensi bunyi
yang didengar oleh masinis tersebut?
Jawab:
Frekuensi yang didengar oleh pengamat di bukit
 v  Vm  V p 
1200 40  0 

 f  
f   
580  599Hz
 1200 40  40 
 v  Vm  Vs 
C
A
1-x
B
x
1 km
Misalkan masinis mendengar gema peluit kereta oleh
dinding bukit ketika berjarak x km dari bukit.
Waktu tempuh kereta dari A ke B adalah
AB
1 x
t 

Vker eta
40
Waktu bunyi merambat dari A ke C kemudian
dipantulkan ke B adalah
AC  CB 1  x
t

Vbunyi
1200
Waktu keduanya sama, maka
1  x 1 x

40 1200
30(1  x) 1  x
30  30x 1  x
31x  29
x  935m
Frekuensi pantul yang didengar oleh
masinis
 v  Vm  V p 
 f 
f "  
 v  Vm  Vs 
 1200 40  40 

599
 1200 40  0 
 620Hz