How Do You Calculate Displacement in a Standing Wave Problem?

[ubiquinone]

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 60.0cm, the speed of the wave is 140m/s, and the maximum displacement of a point at the middle of the string is 1.40mm and occurs at t=0.00s. Calculate the displacement of the string at x=20.0cm and t=0.0380s.

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

 

[ubiquinone]

Does anyone happen to know how to do this?

 

[jtbell]

I think the general equation of a wave is y=A\cos(kx+\omega t-\phi)

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).

 

[ubiquinone]

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 a lot 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]

Try this link instead: http://www.cord.edu/dept/physics/p128/lecture99_35.html

 

[gabee]

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.

 

[ubiquinone]

Hi I tried the problem again and I hope someone could please check over my work. Thank You!

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