Gelombang Berjalan

1.

A wave travels through points A and B, in A to B direction. When t = 0 the wave displacement at point A is 0. Known that A and B is 8 cm apart, the wavelength and the amplitude are 12 cm and 4 cm respectively. Determine the wave displacement at point B when the phase of point A is /2!

→Suatu gelombang berjalan melalui titik A dan B yang berjarak 8 cm dalam arah dari A ke B. Pada saat t = 0 simpangan gelombang di A adalah 0. Jika panjang gelombangnya adalah 12 cm dan amplitudonya = 4 cm, tentukan simpangan titik B pada saat fase titik A /2!

Discussion

Traveling wave equation for point B:

YB = A sin 2π ( t/T − x/λ)
2 π t/T = /2
t/T = 3/4
YB = 4 sin 2π ( 3/4 − 8/12)
YB = 4 sin 2π ( 9/12 − 8/12)
YB = 4 sin ( π/6 ) = 4 sin 30° = 4 ( 0.5) = 2 cm

2.

A wave travels from point A to point B with a speed of 5 m/sec. The wave period is 0.4 sec and the phase difference between A and B is /5, then the AB distance is….
A. 0.6 m
B. 0.8 m
C. 1.0 m
D. 1.2 m
E. 1.4 m
→ Sebuah gelombang berjalan dari titik A ke B dengan kelajuan 5 m/s. Periode gelombang tersebut adalah 0,4 s. Jika selisih fase antara titik A dan B adalah /5 maka jarak AB adalah…..
A. 0,6 m
B. 0,8 m
C. 1,0 m
D. 1,2 m
E. 1,4 m

Discussion

The informations that we can get from above problem are:
wave speed ν = 5 m/sec
wave period T = 0.4 sec
AB phase (angle) difference Δθ = /5
or
AB phase difference Δφ = /5 : 2π = 3/5
AB = ….m

Use the wave basic formula to find the wavelength first,
λ = Tν = (0.4 sec)(5 m/sec) = 2 m

as we know that the phase difference between two points (A and B) :
Δφ = x/λ where x is the AB distance. So then
3/5 = x/2

x = (2)(3/5) = 6/5 = 1.2 m

3.

Given a standing wave equation traveling along a string:


with y and x in m and t in sec.
Determine the speed of a particle in the string at x = 1.2 x 10−2 m when t = 11/6 sec!

→ Diberikan sebuah persamaan gelombang stationer pada seutas kawat sebagai berikut:


dengan y dan x dalam m dan t dalam s. Tentukan kecepatan sebuah partikel tali pada x = 1,2 x 10−2 m saat t = 11/6 s!

Discussion

 

Get the speed equation (ν) from the displacement y equation:

Insert the values of x and t :

 

4.

Given a standing wave equation traveling along a string:

y = 0.2 sin (5πx) cos (20πt)

with y and x in m and t in sec. Find the wave speed!

→Berikut persamaan sebuah gelombang stationer pada tali

y = 0.2 sin (5πx) cos (20πt)

y dan x dalam satuan meter dan t dalam sekon. Tentukan kecepatan gelombang!

Discussion

Use the standing wave equation form:
y = 2A sin kx cos ωt
y = 0.2 sin (5πx) cos (20πt)
Find the values of wavelength (λ) and wave frequency (f) to find the wave speed ( velocity):

k = 5π as we knew in the previous, that k = /λ
so then

5π = /λ
λ = 0.4 m

ω = 20π as we knew ω = 2πf

so then

20π = 2πf
f = 10 Hz

The relationship between λ, f and ν :

ν = λ f

so then we get

ν = (0.4)(10) = 4 m.s−1

 

5.

Given two equations of traveling waves on a string:

y = 0.04 sin (2πx + 10πt)
y = 0.04 sin (2πx − 10πt)

with y and x in m and t in sec. Find magnitude of standing wave amplitude that is formed by these two waves for x = 1/12 m!

→Berikut dua buah persamaan gelombang berjalan sinusoidal pada sebuah tali:

y = 0.04 sin (2πx + 10πt)
y = 0.04 sin (2πx − 10πt)

dengan y dan x dalam satuan m dan t dalam satuan s. Tentukan besar amplitudo gelombang stationer yang dihasilkan oleh superposisi dua buah gelombang tersebut untuk nilai x = 1/12 m!

Discussion

These two sinusoidal waves have the same wavelength and amplitude and traveling in opposite directions. The first is to the left, the second is to the rigth direction. By applying the superposition principle we’ll get a standing wave equation:

y = 2A sin kx cos ω t

so we have to get the needed quantities from the two above:

A = 0.04 m
ω = 10π
k = 2π

so then our equation becomes

y = 2(0.04) sin 2πx cos 10π t
y = 0.08 sin 2πx cos 10π t

0.08 sin 2πx that what we called the amplitude of standing wave (As). For x = 1/4 m

As = 0.08 sin 2πx
As = 0.08 sin 2π(1/12)
As = 0.08 sin (π/6)
As = 0.08 (0.5) = 0.04 m

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Tentang yogabudibhaktifisika

physics teacher
Tulisan ini dipublikasikan di Uncategorized. Tandai permalink.

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