Solving a Wave Problem: A Step-by-Step Guide

[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 $$\displaystyle y(20,0.0380) = -1.107mm\approx -1.11mm$$