Wave group question

  • Thread starter asdf1
  • Start date
  • #1
asdf1
734
0
Why is the narrorer the wave group, the greater the range of wavelengths involved?
 

Answers and Replies

  • #2
Galileo
Science Advisor
Homework Helper
1,994
6
Have you tried to look at the expression for a wave-packet and considered its Fourier transform?
 
  • #3
jtbell
Mentor
15,946
4,616
I think it's easier to answer this question qualitatively if you turn it around: why does a small range of wavelengths produce a broad wave packet, and vice versa?

Set up the waves so that they're all in phase at one point (say, the origin). If their wavelengths are nearly the same, they are still nearly in phase and interfere mostly constructively at a large distance from the origin. On the other hand, if they have a wide range of wavelengths, the destructive interference is significant at a short distance from the origin.
 
  • #4
asdf1
734
0
so you mean that the smaller the amount of waves, the less they interfere with each other, so the bigger the range of wavelengths?
@@a
still a little confused~
 
  • #5
asdf1
734
0
@@a
i haven't studied the Fourier transformation yet, so i don't know how to use it~
but i believe that most complicated things can be explained in simple ways~
 
  • #6
jtbell
Mentor
15,946
4,616
asdf1 said:
so you mean that the smaller the amount of waves, the less they interfere with each other, so the bigger the range of wavelengths?

No, I'm talking about the range of wavelengths that the waves have. You're integrating an infinite number of them together via a Fourier integral:

[tex]\psi(x,t) = \int_{\infty}^{-\infty} {A(k) e^{i(kx - \omega t)} dk} [/tex]

[itex]A(k)[/itex] gives the amplitude of the wave with wavelength [itex]2 \pi / k[/itex]. Although we normally write the integral using in infinite range (limits) in [itex]k[/itex], what counts is the range where [itex]k[/itex] is significantly different from zero.

You might say that a large range in [itex]k[/itex] has a larger number of waves in it (i.e. a larger number of values of [itex]k[/itex]), but that's not really correct because any continuous range has an infinite number of values in it!
 
  • #7
asdf1
734
0
@@a
sorry, but I'm still a little confused...
 
  • #8
jtbell
Mentor
15,946
4,616
Well, I guess I'm still a little confused about what you're confused about... :confused:
 
  • #9
reilly
Science Advisor
1,077
1
In reality, most physicists would say that Fourier transforms are simple.

However try building a wave packet graphically -- take several waves with different frequencies, and add them, by hand or whatever, and graph the result. Add some more -- big freqs and small ones. As you do this, you will see the interference patterns that ultimately produce a delta function -- make sure that all waves are of the form exp(kx), where k is the wave number, so that the exp functions are all = one at x=0.

Regards,
Reilly Atkinson
 
  • #10
quasar987
Science Advisor
Homework Helper
Gold Member
4,793
21
It's not easy to explain it from scratch, and let alone, to understand it. But if you really want to know, pick up "Vibrations and Waves in Physics" by Iain G. Main (). It is explained as simply as it can be.
 
Last edited by a moderator:
  • #11
asdf1
734
0
thanks! :)
 

Suggested for: Wave group question

Replies
5
Views
414
  • Last Post
Replies
4
Views
398
  • Last Post
Replies
24
Views
421
  • Last Post
Replies
13
Views
510
  • Last Post
Replies
4
Views
390
Replies
18
Views
630
Replies
21
Views
196
Replies
27
Views
798
Replies
38
Views
915
  • Last Post
Replies
2
Views
265
Top