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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Wave’s contidition Reflection Refraction Interference difraction Polarization

Mechanical Waves, waves that travel with some material called a medium.

Waves category Electromagnetic waves can propagate even in empty space, where there is no medium.

Mechanical Waves

•

Sound waves in gas,

•

Sound waves in liquid

•

Sound waves in solid

•

String waves 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 waves

•

Electromagnetic waves

•

Repples on a pond

•

String 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.

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  v =

f

 v =  /k

f

1/4  1/2  3/4 

That is, the traveling sinusoidal wave moves to the right a distance vt in the time t, with amplitude A (m), frequency wavelength 

f

(Hz.), (m), and wave speed v (m/s) as shown in Figure ) a .

t = 0 b.

c.

t =  t = 2  v      A sin X 2   x     A sin 2  X   x  v        x , 2    A sin X 2    x  2 v      2 v 

If the wave moves to the right with a speed v, then the wave function at some later time t is :     A sin 2    x  vt     If the wave moves to the left with a speed v, then the wave function at some later time t is :     A sin 2    x  vt     generally express the wave function in the form,     A sin 2    x  vt    

The wave function can be express

    A sin k  x  vt      A sin  kx   t        A sin A sin 2   2  f x t  x v T t When the vertical displacement y is ’n zero at x=0 and t=0 Wave phase,     ,

t

  (

A

kx sin    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 Y 1 (x,t) = A sin (kx  t) Y 2 (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 Y(x,t) = 2A sin kx cos  t = A sw sin kx cos  t  t) standing wave on a string, fixed end at x = 0 The standing wave amplitude A sw is twice the amplitude A of either of the original traveling waves: A sw = 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 r 2 average power, sinusoidal wave on a string

The Doppler Effect

a Stationary Source d =  b Source moves to the right v s d s = v s T 



=Distance between two adjacent wave front. v s = 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 d s = v s T.

Hence, the new wave length



:

 = d - d s =  - v s T =  - v s  / v =  (1- v s / v) www.themegallery.com

When a source move toward a stationary observer, frequency heart by observer is:

f

' 

v

 ' 

v

    1 

v s v

    1    1 

v s v

  

f

 

v



v v s



f

For a source moving with a speed V s observer, away a stationary

f

 

v

  

v

    1 

v s v

       1  1

v s v

  

f

 

v



v v s



f

 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, v  Wave speed relative to observer is: = v + v p , v = Sound speed in air V p = Speed of observer. So: f   v  λ  v  v p λ

because



= v/f, so:

f

 

f

 

v v V p

   

f

   1 

V p v

   Generally rule,

f

   

v v V



V s p

  

f

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 v S of a source or speed of an observer equal with the wave speed v? what happens when the speed v S wave speed v?

of a source exceeds the 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 v S 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

f

   

v v



V m



V m



V p



V s

 

f

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:

f

Frekuensi yang didengar oleh pengamat di bukit    

v v

 

V V m m

 

V V p s

 

f

 1200 1200   40 40   0 40   580  599

Hz

C A B 1 km x Misalkan masinis mendengar gema peluit kereta oleh dinding bukit ketika berjarak x km dari bukit.

Waktu tempuh kereta dari A ke B adalah

t



AB V

ker

eta

 1  40

x

Waktu bunyi merambat dari A ke C kemudian dipantulkan ke B adalah

t



AC



CB V bunyi

 1 

x

1200 Waktu keduanya sama, maka 1  40 30 ( 1

x

 

x

) 1  1200  1

x



x

Frekuensi pantul yang didengar oleh masinis 30 31

x

  30

x

29  1 

x f

"   

v v

 

V m V m



V p



V s

 

f



x

 935

m

 1200 1200   40 40   40 0   599  620

Hz