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Waves Motion

  • Thread starter ubiquinone
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Hi there, I'm trying to teach myself some concepts involving waves. I tried a problem from an old physics exercise book, but I am having difficulty in getting the correct answer. I was wondering if someone can please point me into the right direction or show me a link where I can read more about this kind of stuff. Thank You!

Question: A string, fixed at both ends, oscillates at its fundamental frequency. The length of the string is [tex]60.0cm[/tex], the speed of the wave is [tex]140m/s[/tex], and the maximum displacement of a point at the middle of the string is [tex]1.40mm[/tex] and occurs at [tex]t=0.00s[/tex]. Calculate the displacement of the string at [tex]x=20.0cm[/tex] and [tex]t=0.0380s[/tex].

I think the general equation of a wave is [tex]y=A\cos(kx+\omega t-\phi)[/tex], where [tex]A[/tex] is the amplitude, [tex]k=\frac{2\pi}{\lambda}[/tex], [tex]\omega = 2\pi f[/tex] and [tex]\phi = \frac{2\pi\times\text{phaseshift}}{\lambda}[/tex]
 

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Does anyone happen to know how to do this?
 
jtbell
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I think the general equation of a wave is [tex]y=A\cos(kx+\omega t-\phi)[/tex]
This is for a traveling wave. Your question is about a standing wave. If your book doesn't discuss standing waves, first try a Google search for "standing wave" (with the quotes, to keep the words together).
 
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Hi jtbell! Thank you very much for replying to my question. I read up some more information regarding standing waves from a few resource sites. Particularly I learned alot more about their properties from here:
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/standw.html and
http://hyperphysics.phy-astr.gsu.edu/hbase/waves/standw.html#c4
However, I've still not figured out how to solve this problem. May you please offer me with another hint on how to solve this problem? I'm willing to try. Thank you again!
 
BobG
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Try this link instead: http://www.cord.edu/dept/physics/p128/lecture99_35.html [Broken]
 
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Well, a standing wave is just a traveling wave + its reflection. When two waves encounter one another, the resultant wave is just the sum of the two waves. For instance, the equation for a standing wave that occurs between a sound source and a wall is the result of adding the first wave to the reflected wave.

The equation for a standing wave on a string is given by doing the same thing with a wave and its reflection. Since both ends of the string are fixed, the displacement is always zero at those end points. If you can figure out how a wave is reflected from a hard boundary (it is phase-shifted 180 degrees (pi radians)) you'll know which two wave equations to add together.
 
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Hi I tried the problem again and I hope someone could please check over my work. Thank You!

The equation for a standing wave: [tex]\displaystyle y(x,t)=[2y_m\sin(kx)]\cos(\omega t)[/tex]
Since [tex]\frac{1}{2}\lambda=L\Leftrightarrow \lambda =2L[/tex]
Thus, [tex]k=\frac{2\pi}{\lambda}=\frac{2\pi}{2 L}=\frac{\pi}{L}[/tex]
Also, [tex]v=f\lambda\Leftrightarrow f=\frac{v}{\lambda}[/tex]
Therefore, [tex]\omega= 2\pi f=\frac{2\pi v}{\lambda}=\frac{2\pi v}{2L}=\frac{\pi v}{L}[/tex]
The standing wave equation could then be expressed as:
[tex]\displaystyle y=(x,t)=\left[2y_m\sin\left[\left(\frac{\pi}{L}x\right)\right]\cos\left(\frac{\pi v}{L}t\right)[/tex]
When [tex]x=30 cm[/tex] and at time, [tex]t=0[/tex],
[tex]\displaystyle y(30,0)=\left [ 2y_m\sin\left(\frac{30\pi}{60}\right )\right ]\cos(0)=1.40mm[/tex]
Solving for [tex]y_m[/tex]: [tex]\displaystyle y_m\sin\left(\frac{\pi}{2}\right)=0.7\Leftrightarrow y_m=0.7mm[/tex]
When [tex]x=20 cm[/tex] and at time [tex]t=0.0380s[/tex] and substituting [tex]y_m=0.7mm[/tex], we get
[tex]\displaystyle y(20,0.0380)=\left [ 2(0.7mm)\sin\left(\frac{20\pi}{60}\right )\right ]\cos\left(\frac{\pi\times 140m/s}{0.60m}(0.0380s)\right )[/tex]
We have [tex]\displaystyle y(20,0.0380) = -1.107mm\approx -1.11mm[/tex]
 

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