Calculating Frequency of Standing Wave on a Fixed String | Physics Homework Help

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A student is struggling to calculate the frequency of a standing wave on a fixed string, given a length of 2.0 meters and a wave speed of 78 m/s. They initially used the formula f = v/(2L) but obtained an incorrect frequency of 19.5 Hz. After realizing the need to account for the number of wavelengths, they attempted to adjust their calculations but still arrived at an incorrect answer. The student is confused about the correct number of wavelengths in the provided figure and questions the validity of their formula. The discussion highlights the importance of accurately identifying the number of wavelengths and correctly applying the wave frequency formula.
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I've never taken physics before and this is my first physics class in university.

Homework Statement


A string, 2.0 meters in length, is fixed at both ends and tightened until the wave
speed is 78 m/s. What is the frequency of the standing wave shown in the figure?


Homework Equations


v=nv/2L because it is fixed at both ends


The Attempt at a Solution


so f = (78m/s)/(2)(2)
and I get 19.5 Hz but that's not the right answer. It's too small. Help please?
 
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It seems like there's some information missing in your outline of the problem. Perhaps try multiplying by an integer, n, that is the number of wavelengths the standing wave is comprised of.
 
Okay I actually just tried that. Here's the picture to the problem:

ScreenShot2012-09-16at80313PM.png


I thought the number of wavelengths was 6, but that doesn't seem to be right? Am I reading the picture wrong?
 
Theres only 3 wavelengths in that picture;)
 
haha okay I see that now. So then I'd do

f = (78m/s)(3m)/(2m)(2)

and I'd get 58.5 which doesn't make sense. my answer either has to be A) 350Hz B) 230Hz C) 120Hz D) 470Hz and it's neither. so is my formula wrong maybe? because isn't frequency supposed to have units s? because that gives me units m/s.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

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