|Mar4-07, 02:28 AM||#1|
We have been studying mechanical waves, using the mathematical description of a wave, however, im not quite sure how to solve the following, so if anyone has any pointers or hints that would be great!
You are told that two points x=0 and x=0.0900m are within one wavelength of each other. If the wave is moving in the +x-direction, determine the wavelength. We are also given the Amp = 4mm and the period which = 0.04 s.
Also see attached pic.
I tried simply using the following (however obviously I get the wrong ans, which should be 0.14m)
solve for lambda given
3mm = 4 mm . (cos (2pi (0.09/lambda)-0))
but my answer comes out as lambda = 0.782m???
|Mar4-07, 03:52 AM||#2|
Is this the exact wording of the problem? If the two points are simply "within" one wavelength of each other, then it could really be anything. The information that you stated is not enough to calculate the wavelength. Are you sure there is no more information, such as the wave speed, the mass per unit length (I assume this is a transverse wave on a string), or the tension?
|Mar4-07, 07:37 AM||#3|
no thats the exact Q....comes from University Physics...addison wesley....hence why i had trouble with it.
"A sinusoidal wave is propagating along a stretched stringas a function of time is graphed in the figure (attached to the original post above) for particles at x = 0 and x =0.0900m. a) what is the period of the wave ( my ans = 0.04 sec
b) what is the amplitude of the wave ( my ans = 4mm)
c)you are told that the 2 points x=0 and x=0.0900m are within one wavelength of each other. If the wave is moving in the +x direction, determine the wavelenght and the wave speed."
I just shortende the Q above for ease?...how about now...is the Q answerable?? cheers
|Apr28-08, 03:04 AM||#4|
I was just working on this same problem and couldn't find any help online, but I figured it out myself; so, although this is rather late, perhaps it will help someone else.
Anyway, at first I wanted to solve for wavelength the same way that ku1005 did, but, in fact, this is not necessary.
Inspection of the figure in the text that corresponds to this problem (as shown in ku1005's original post) shows that the wavespeed is 3.6 m/s. Just look at the wave y(x,t) = 0 (this is easiest) for x =0 and t = 0 and then find when the wave is in the same part of the phase (i.e., when y = 0) for x = 0.09m; that happens at t = 0.025s. So, 0.09m/0.025s = 3.6m/s.
Similarly, inspect the figure to determine the period T. I find it easiest looking at y for maxima/minima and y = 0. Here we need only be concerned with either the point on the string at x = 0.0m or at x = 0.090m since the frequency is constant. Choosing to inspect the point at x = 0.0m (the red curve in the figure), we see that it takes 0.040s for the wave to complete one full cycle. So, the period T = 0.040s. Frequency is inversely proportional to period: 1/period = 1/0.040s = 25s^-1.
wavespeed = wavelength * frequency --> wavelength = wavespeed/frequency = (3.6m/s) / (25s^-1) = 0.144m
Part (d) of the question asks you to find the same things if the wave is traveling in the opposite direction.
Inspecting the figure as for part (c) shows that it takes 0.015s to travel 0.090m.
So, the wavespeed = 0.090m / 0.015s = 6.0 m/s.
The frequency is unchanged and is equal to 25 cycles/second.
Using the relation v = lambda*f shows that lambda = 0.24m.
Finally, part (e) of the question asks if it would be possible to determine wavelength if you were not told that x = 0.0m and x = 0.090m were within one wavelength of each other. The answer to this is simply no. You could determine the frequency but not the wavespeed, and therefore you could not determine the wavelength.
|Similar Threads for: Mechanical Waves|
|Mechanical waves||Introductory Physics Homework||7|
|Light waves vs. mechanical waves||Introductory Physics Homework||3|
|mechanical waves||General Physics||5|
|mechanical waves||Classical Physics||1|
|Mechanical Waves||Introductory Physics Homework||12|