1. WAVES 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.

2. Reflection
Refraction
Wave’s contidition
Interference
difraction
Polarization

3. Electromagnetic waves
Mechanical Waves, waves that travel with some material called a medium.
Mechanical Waves
Sound waves in gas,
Sound waves in liquid
Sound waves in solid
String waves
Waves category
Electromagnetic waves can propagate even in empty space, where there is no medium.
Electromagnetic waves
Radio waves
light
gelombang radio
gelombang TV

4. Transverse wave waves Longitudinal wave
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
Repples on a pond
String waves
waves
Longitudinal wave
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.
Sounds Wave
Springs wave

5. 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.

6. Wave number (rad/m), k = 2/
Angular frequency (rad/dt),
 = 2  /
= 2  v/ 
= 2  f
v = f 
v = /k

7. 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 b.
t = 
v  c.
t = 2 
2 v 

8. If the wave moves to the right with a speed v, then the wave function at some later time t is :
If the wave moves to the left with a speed v, then the wave function at some later time t is :
generally express the wave function in the form,

9. The wave function can be express
When the vertical displacement y is’n zero at x=0 and t=0
Wave phase,
 is phase constant.

10. 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

11. EXAMPLE 1
A sinusoidal wave traveling in the positive x direction has an amplitude of cm, a wavelength of 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.

12. Because A = 15 cm and because y = 15 cm at x=0 and t=0 substitution into Equation

13. 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

15. The speed of a wave (v) traveling on a taut string of mass per unit length  and tension T is

16. The intensity I at any distance r is therefore inversely proportional to r2
average power, sinusoidal wave on a string

17. The Doppler Effect

18. Stationary Source
d =  vs
ds = vs T  b a
Source moves to the right

19.  =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)

20. When a source move toward a stationary observer, frequency heart by observer is:
For a source moving with a speed Vs away a stationary observer,

21. 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:

22. because  = v/f, so:
Generally rule,

23. 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)

24. 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 Hz.

25. 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.

26. If the medium is not stationary

27. 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?

28. Frekuensi yang didengar oleh pengamat di bukit
Jawab:
Frekuensi yang didengar oleh pengamat di bukit
1 km A B C
1 - x x
Misalkan masinis mendengar gema peluit kereta oleh dinding bukit ketika berjarak x km dari bukit.
Waktu tempuh kereta dari A ke B adalah

29. Waktu bunyi merambat dari A ke C kemudian dipantulkan ke B adalah
Waktu keduanya sama, maka
Frekuensi pantul yang didengar oleh masinis

31. SEKIAN
TERIMAKASIH