Solving Simple Wave Problem: Math and Conversion

  • Thread starter Ramses The Pharaoh
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In summary, the conversation is about solving a simple wave problem involving a wave function in the form y(x,t)=A*sin(kx-wt). The problem is to convert the given data y(x,t)=4.2*cos(0.2x)*sin(300t) into the given form. Some confusion arises about the number of sin() terms and how to extract the wavelength and frequency from the equation. The experts suggest thinking about the physical meaning of these terms and working out the frequency and wavelength from first principles rather than relying on formulae. The conversation also touches on the idea of double frequency and doubly periodic functions.
  • #1
Ramses The Pharaoh
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Simple wave problem

Hi guys!
I'm trying to solve a wave problem for homework and I hope that I can extract all data, once I have my wave function in the form: y(x,t)=A*sin(kx-wt) :biggrin: . The simple problem is to convert the given data: y(x,t)=4.2*cos(0.2x)*sin(300t) into the above form :uhh: . It seems that I've forgotten mathematics at all! Can you please show me how such a conversion works, because when I tried the f-la for sin(a)*cos(b) I obtain two sin()s when I need only one. :confused:
It is later asked for the wavelength, frequency, speed, etc. But I hope I can solve this with the "proper stated" wave function. :wink:
Thanks in advance!
 
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  • #2
Just a guess, but maybe your wave is the superposition of two sinusoids?
 
  • #3
Hurkyl said:
Just a guess, but maybe your wave is the superposition of two sinusoids?
OooooK, but then my problem becomes more difficult :biggrin:. How to extract the wavelength and frequency of such an equation? Any hints? :uhh:
 
  • #4
Think about what they mean, physically, rather than how to read them off of a sinusoid.

(Though, I suspect you could get the answer by reading them off of the individual sinusoids, but that's merely educated guessing, and that's not good enough!)
 
  • #5
The problem is that I don't know how to deal with such separated expressions - one for sin (containing "t") and one for cos (containing "x")... If it is a superposition, then how to separate the two functions?!
 
  • #6
I obtain two sin()s when I need only one.

I thought this meant that you got an expression for your wave as a sum of two sinusoids!



Anyways, I'm pretty sure that you don't need to do any algebraic manipulations on this problem -- you should be able to read the wavelength and frequency straight from the expression as given... if you remember what those terms mean.
 
  • #7
Hurkyl said:
I thought this meant that you got an expression for your wave as a sum of two sinusoids!
That's right and exactly this confuses me :rolleyes:. But leaving the confusion aside, I obtain from [tex]y(x,t)=4.2\cos(0.2t)\sin(300t)[/tex] the function [tex]y(x,t)=2.1(\sin(300t + 0.2t) + \sin(300t - 0.2t))[/tex]. So if [tex]\lambda=\frac{2\pi}{k}[/tex] and I leave the sign, I get [tex]\lambda=10\pi[/tex]. For [tex]f=\frac{\omega}{2\pi}=\frac{150}{\pi}[/tex]. Correct? Because I'm not sure how to deal with the two different signs...
Moreover, I'm asked to find the speeds. Is it possible to write the equation in this form: [tex]y(x,t)=2.1(\sin(0.2(x+\frac{300}{0,2}t)+\sin(0.2(x-\frac{300}{0,2}t))[/tex] and obtain the speed -> 300/0.2 = 1500?
Can you please check if the logic is correct and provide some hints if I'm going completely the wrong way :redface:
 
  • #8
This is what I meant by "educated guessing", and it sounds like you've figured it out. (Though, I think you're reading the parameters wrong off of your sinusoids)


I still advise you figure out how to work out the frequency and the wavelength from first principles, rather than from the canned formulae that work only for sine waves. If your professor gave you this problem, I wouldn't put it past him to put similar problems on tests! And those might not be so easily worked out by educated guessing!

(Not to mention you could use it to promote your work on this from educated guessing to rigorous proof!)
 
  • #9
Hurkyl said:
I still advise you figure out how to work out the frequency and the wavelength from first principles, rather than from the canned formulae that work only for sine waves.
I suppose first principles means straight from the given equation, without transforming it in any way. :shy:
However, being new to this field, makes it a little more difficult for me to understand such hints. Probably if you provide a little more specific tips or advices, I will be able to grasp the idea and solve my problem. :wink:
Thanks for your time :rolleyes:
 
  • #10
I have -- can you quote the definition of wavelength and frequency, or at least give a reasonable approximation?
 
  • #11
OK, If I take in mind that wavelength is the distance between successive points of equal amplitude and phase on a wave, and frequency is the number of cycles or events per unit time, I take the given equation [tex]y(x,t)=4.2\cos(0.2x)\sin(300t)[/tex] and I see that it has a double frequency (due to cos()*sin()... or at least my teacher presented it so) :). This means that the frequency should be half of the value I formerly gave and the wavelength - double. Am I reading the parameters correct off my sinusoids? :)
 
  • #12
Ramses The Pharaoh said:
OK, If I take in mind that wavelength is the distance between successive points of equal amplitude and phase on a wave, and frequency is the number of cycles or events per unit time, I take the given equation [tex]y(x,t)=4.2\cos(0.2x)\sin(300t)[/tex] and I see that it has a double frequency (due to cos()*sin()... or at least my teacher presented it so) :). This means that the frequency should be half of the value I formerly gave and the wavelength - double. Am I reading the parameters correct off my sinusoids? :)

The sin(300t) term has all the time dependence. For any constant value of x, y varies in time in proportion to that sin(300t) term. You get the frequency directly from that. There is no "double frequency". The spatial variation of the waveform at some moment in time is completely described by the cos((0.2x) term. Look at a value of x for which the cosine term is zero. Can y ever be anything other than zero at that x by changing t? What about the two separate sinusoids you found. Do they behave the same way as their combination? How do those two sinusiods differ from one another?
 
  • #13
I see that it has a double frequency

If your professor used those exact words, he's wrong. :tongue2: He probably said it's doubly periodic, meaning it has two different periods. (A period is, minor details aside, something you can add to the inputs that doesn't change the outputs)


If I take in mind that wavelength is the distance between successive points of equal amplitude and phase on a wave

Right -- a slightly different way of saying it is that, no matter where you are, if you move a distance of one wavelength, you get the same amplitude!


and frequency is the number of cycles or events per unit time

Right. IMHO it's conceptually easier to look for the time per cycle, and then invert to get the frequency.


(You should also note there's a duality here -- shouldn't everything said about the wavelength also apply in a similar, but not quite the same, way to the frequency?)
 
  • #14
Following the advices of you both, I calculated the values for x when the cos is 0: [tex]x=\frac{\pi}{0.4}[/tex] and [tex]x=\frac{5\pi}{0.4}[/tex], so the distance between two successive knots is [tex]10\pi[/tex]. This gives me the wavelength [tex]\lambda[/tex]. :tongue2:
The period for one full cycle I obtained in the same way, using the sin this time -> [tex]t=\frac{\pi}{150}[/tex], so the frequency is the reciprocal: [tex]f=\frac{150}{\pi}[/tex].
The speed I find when I multiply them both - [tex]v=f\lambda=1500[/tex].
Did I get the point now? :biggrin:
 
  • #15
Ramses The Pharaoh said:
Following the advices of you both, I calculated the values for x when the cos is 0: [tex]x=\frac{\pi}{0.4}[/tex] and [tex]x=\frac{5\pi}{0.4}[/tex], so the distance between two successive knots is [tex]10\pi[/tex]. This gives me the wavelength [tex]\lambda[/tex]. :tongue2:
The period for one full cycle I obtained in the same way, using the sin this time -> [tex]t=\frac{\pi}{150}[/tex], so the frequency is the reciprocal: [tex]f=\frac{150}{\pi}[/tex].
The speed I find when I multiply them both - [tex]v=f\lambda=1500[/tex].
Did I get the point now? :biggrin:
Fortunately, you calculated a full period of the cosine function, not the interval between successive zeros. The full spatial period is the wavelength. You should recognize that the function you are working with describes a standing wave, which is the superposition of two traveling waves, one moving in the positive x direction and the other in the negative x direction. I had not noticed the typo error you made earlier.
[tex]y(x,t)=4.2\cos(0.2t)\sin(300t)[/tex]

[tex]y(x,t)=2.1(\sin(300t + 0.2t) + \sin(300t - 0.2t))[/tex]
should have said (when combined)

[tex]y(x,t)=4.2\cos(0.2x)\sin(300t) = 2.1\left[\sin(300t + 0.2x) + \sin(300t - 0.2x)\right][/tex]

One of the last two terms is a wave traveling in the +x direction, and the other is a wave traveling in the -x direction with angular frequency [tex] \omega = 300 [/tex] and wave number [tex] k = 0.2 [/tex]
 
  • #16
Oh, yeah, sorry for the mistake :tongue2: ...
As far as I've understood you correctly, you say that everything is OK now, right? :bugeye:
 
  • #17
Ramses The Pharaoh said:
Oh, yeah, sorry for the mistake :tongue2: ...
As far as I've understood you correctly, you say that everything is OK now, right? :bugeye:

It appears to me that you have correctly identified the wavelength and frequency, so the velocity is correct. I just wanted to make sure you understand the connection between that standing wave and the decomposition into the two traveling waves.
 
  • #18
Your work looks perfect.
 
  • #19
Thank you guys, you really helped me! :smile: It seems that I've got the point now...
However, I have one or two more questions about this problem (I've not stated them yet), but I think I will post them in a new thread, because they are not "simple math problems" any more :rolleyes:
So for now thank you and see you in the next thread :biggrin:
 

1. What is a simple wave problem?

A simple wave problem is a mathematical exercise that involves solving for various properties of a wave, such as its frequency, wavelength, and speed.

2. How do you convert between frequency, wavelength, and speed?

To convert between frequency (f), wavelength (λ), and speed (v), you can use the formula v = fλ. This means that speed is equal to the product of frequency and wavelength. You can rearrange this formula to solve for any of the three variables.

3. What units are used for frequency, wavelength, and speed?

The most common units for frequency are hertz (Hz) or cycles per second. Wavelength is typically measured in meters (m). Speed can be measured in meters per second (m/s) or any other unit of distance over time.

4. How do you solve for the missing variable in a simple wave problem?

To solve for the missing variable in a simple wave problem, you will need to use the formula v = fλ, where v is the speed, f is the frequency, and λ is the wavelength. Plug in the known values and solve for the unknown variable.

5. Can you provide an example of solving a simple wave problem?

Sure! Let's say we have a wave with a frequency of 10 Hz and a wavelength of 2 meters. To find the speed of the wave, we would use the formula v = fλ. Plugging in the values, we get v = (10 Hz)(2 m) = 20 m/s. Therefore, the speed of the wave is 20 meters per second.

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