Fundamental Frequency (String Resonance) - stumped

[Pruzhinkin]

Fundamental Frequency (String Resonance) - velocity problem!

Homework Statement

Hey guys, i just started working on Fundamental Frequency but am getting confused!

A string that is 6.0m long is vibrating with three loops in it. The frequency of the source is 16.5Hz.
A)What is the fundamental frequency of the string?
B)What is the speed of the wave in the string?
C)Will 28.5Hz produce a standing wave in the string or not?

Here is what i was doing, so please correct me and explain!

Homework Equations

3λ/2 = L or λ = 2/3 * L
v = f3λ = f3 * 2L/3 or f3 = 3v/2L = 3 * v/2L = 3f1, fn = nf1

The Attempt at a Solution

A)f3 = 3f1
f1 = 16.5Hz ÷ 3 = 5.5 Hz
The fundamental frequency is 5.5 Hz
I think this is correct so far.

B) - Now here is where i get confused, what i was thinking at first was
v = f3λ = f3 * 2L/3 so plugging in the numbers this would give me
v = (16.5 Hz) * (4m) = 16.5 Hz * (2*6) ÷ 3 = 66m/s
After this i went to yahoo answers, and someone suggested that each loop is λ/2 long.
So am i supposed to do this instead?
v = f3 * λ/2 = f3 * L/3 so plugging in the numbers this would give me
v = (16.5 Hz) * (4/2) = 16.5 Hz * 6/3 = 33m/s
Is this the right answer? If so, or if not please tell me! Explain the reason behind this, please.

C) 28.5 Hz = n*(5.5 Hz)
n = 28.5/5.5 = 5.2
28.5 Hz will not produce a standing wave as it does not satisfy fn = nf1, where n = 1, 2, 3,..
I think that this one is correct as well.

P.S. I am doing this course online, so i can not always get the feedback from the teacher right away, especially regarding this question as it is supposed to be sent in for evaluation.
I really want to know the correct way of solving this, as it has been bugging me for the whole day and not letting me focus on anything else.
Thank-you in advance!

 

[phyzguy]

The speed of the wave is always equal to the frequency times the wavelength, so your first answer in (B) is correct. The statement in Yahoo that each loop is lambda/2 long is also correct and is consistent with what you did - i.e. if there are three loops in 6 m, then lambda/2 = 2m, so lambda = 4m. (A) and (C) look correct as well.

 

[Pruzhinkin]

The speed of the wave is always equal to the frequency times the wavelength, so your first answer in (B) is correct. The statement in Yahoo that each loop is lambda/2 long is also correct and is consistent with what you did - i.e. if there are three loops in 6 m, then lambda/2 = 2m, so lambda = 4m. (A) and (C) look correct as well.

So what answer for B am i supposed to write then? The first or the second?
There is a big difference between 33-66! Phyzguy said that speed is always equal to the frequency times the wavelength, therefore the answer that am i supposed to use is the first one i wrote? How can the yahoo answer be correct as well if it gives me 33 instead of 66? Maybe 33 because lambda/2 and 33 is the answer for the loop, where as the whole wavelength is lambda/2 therefore 33*2 = 66 m/s speed of the whole wavelength? Either way, which answer do i use:)

 

[phyzguy]

Did you read what I wrote?

 

[Pruzhinkin]

Sorry i was sleepy when i read your post the first time, totally got it now! I was just overcomplicating things! Thank you