How Do Spring Length Calculations Relate to Frequency and Harmonics?

[lc99]

Homework Statement

Homework Equations

v = sqrt(F/(m/L))

The Attempt at a Solution

Trying to find the length of the spring. But i can find the tension of the wire at all.
I'm not sure what i can do with the frequencies? Maybe I'm forgetting a concept cause i thought the equation v^2/F = L would give me the length and since i can neglect mass of the wire.

I'm am just super stuck!

 

[haruspex]

v^2/F = L

If that F represents tension then you have omitted the mass of the wire.
But that is not the useful equation here.
What is the relationship between frequency and wavelength?
For a given length of wire, what wavelengths of standing wave can there be?

 

[lc99]

If that F represents tension then you have omitted the mass of the wire.
But that is not the useful equation here.
What is the relationship between frequency and wavelength?
For a given length of wire, what wavelengths of standing wave can there be?

Freq and wave length are inversely proportional
Is it frequency of nth tone = n*v/4L??
So for fundamental freq, v/(4L) = 500 then solve for L?

 

[haruspex]

Freq and wave length are inversely proportional
Is it frequency of nth tone = n*v/4L??
So for fundamental freq, v/(4L) = 500 then solve for L?

You do not know that the fundamental frequency is 500Hz. What you do know is the frequencies of two consecutive harmonics.

 

[lc99]

You do not know that the fundamental frequency is 500Hz. What you do know is the frequencies of two consecutive harmonics.

I think the wave lengths are 1/10 apart for the frequencies. If I calculate wave length with v = flambda

 

[haruspex]

I think the wave lengths are 1/10 apart for the frequencies. If I calculate wave length with v = flambda

I don't know what that means. Can you be clearer?

 

[lc99]

I don't know what that means. Can you be clearer?

For frequency 500, v =500lambda , lambda= 3/5
For 600, lambda= 1/2

6/10-5/10= 1/10 m wavelength difference between the two consecutive harmonics. Is this info any helpful?

 

[haruspex]

For frequency 500, v =500lambda , lambda= 3/5
For 600, lambda= 1/2

6/10-5/10= 1/10 m wavelength difference between the two consecutive harmonics. Is this info any helpful?

Ok, I could not understand your previous post because you left out the units.
Yes, the difference in the two wavelengths is indeed 0.1m. But you have not used the fact that these are consecutive harmonics. If a string has length L, what are the possible wavelengths?

 

[Merlin3189]

You could think about harmonics in general and deduce your own formulae.
Say you had a string whose fundamental frequency was 30 Hz. What are the first few harmonic frequencies that are possible?
What do you notice about the differences between consecutive harmonics?

 

[lc99]

You could think about harmonics in general and deduce your own formulae.
Say you had a string whose fundamental frequency was 30 Hz. What are the first few harmonic frequencies that are possible?
What do you notice about the differences between consecutive harmonics?

They will differ by the number of antinodes (n)

 

[lc99]

Ok, I could not understand your previous post because you left out the units.
Yes, the difference in the two wavelengths is indeed 0.1m. But you have not used the fact that these are consecutive harmonics. If a string has length L, what are the possible wavelengths?

All i understand now is that the number of antinodes give a certain length. So for the consecutive frequencies, the antinodes are responsible for making the lengths equal

 

[haruspex]

All i understand now is that the number of antinodes give a certain length. So for the consecutive frequencies, the antinodes are responsible for making the lengths equal

If the string length is L, what is the wavelength of the fundamental? What is the wavelength of the next harmonic?

 

[lc99]

If the string length is L, what is the wavelength of the fundamental? What is the wavelength of the next harmonic?

I think the wavelength is 2L = lambda. for fundamental
next harmonic is L = lambda

 

[haruspex]

I think the wavelength is 2L = lambda. for fundamental
next harmonic is L = lambda

Right, so what about the n^th harmonic?

 

[lc99]

Right, so what about the n^th harmonic?

Lambda = 2L/n?

 

[haruspex]

Lambda = 2L/n?

Right, so if two consecutive harmonics have frequencies 500Hz and 600Hz, which two are they?

 

[lc99]

Right, so if two consecutive harmonics have frequencies 500Hz and 600Hz, which two are they?

5th and 6th harmonic?

 

[haruspex]

5th and 6th harmonic?

Right. So how do their wavelengths relate to L?

 

[lc99]

Right. So how do their wavelengths relate to L?

for 5th harmonic, wavelength = 2L/5
for 6th harmonic, wavelength = 2L/6 = 1L/3

 

[haruspex]

for 5th harmonic, wavelength = 2L/5
for 6th harmonic, wavelength = 2L/6 = 1L/3

Right. So what is the length L?

 

[lc99]

Right. So what is the length L?

Okay, 1.5 m? right. And it's because we know that consequtive harmonics help us find the n

 

[haruspex]

Okay, 1.5 m? right. And it's because we know that consequtive harmonics help us find the n

Right.

 

[lc99]

Right.

So i have a question. In terms of difficulty, how difficult is this question for an exam?

 

[haruspex]

So i have a question. In terms of difficulty, how difficult is this question for an exam?

Were I doing it I would have started by immediately recognizing that these must be the 5th and 6th harmonics. Next would be your post #7 calculation for the wavelength of, say, the 500Hz tone, then multiply by 5/2 to get the string length.
How long would that have taken?

 

[lc99]

Were I doing it I would have started by immediately recognizing that these must be the 5th and 6th harmonics. Next would be your post #7 calculation for the wavelength of, say, the 500Hz tone, then multiply by 5/2 to get the string length.
How long would that have taken?

not long if you know what you are doing! I do get it now. I didn't recognize that was about harmonics.

 

[Merlin3189]

So i have a question. In terms of difficulty, how difficult is this question for an exam?

I think it's easy, but it's interesting, because we have different perspectives on this topic.

They will differ by the number of antinodes (n)

Honestly, I can never remember which are nodes and which antinodes! The nodes of Ranvier are the unmyelinated bits of nerve fibre, so I guess nodes on the string are where there's no transverse displacement. But if it was important for a question I might well look it up to check before answering.*

Here I'd thought to help you by getting you to say 30, 60, 90, 120, ...Hz, differing by 30 Hz, the fundamental. But even though I asked about frequency, you thought about wavelengths.

When I first saw the question, 500 Hz and 600 Hz as consecutive harmonics meant 100 Hz fundamental (as Haruspex said) and then I'd have had to think a bit about what that meant for the length of the string. I don't actually know the formula you quoted! (Another thing I would have looked up, though I might have been able to work it out, because they're all the same really.) Though, as you found, we don't need it.**

But it reflects a difference, not so much in understanding of physics, as view of the world. I see a vibrating string and think of frequency, like a guitar string. If I press my finger on a fret, the pitch or frequency is higher and that's what I'm thinking, not that the string or wavelength is shorter. You, not unreasonably, look and see nodes and thus wavelengths.
I also use old fashioned radio a lot, where frequencies are the defining characteristic. Wavelength is a bit of a moveable feast that we talk about in round numbers, but only pin down when we need it in specific situations. Most people have no need to think of wavelength or frequency if they use radio, because it's all channels and the rest is hidden away by the computers that control the equipment. Harmonics there can pop up all across the band and they don't tell you whether they are the 5th, 6th or 1739th. But the spacing tells you the fundamental frequency, so, maybe, where it's coming from. This is the sort of experience that creates the intuition I benefited from here.

So all questions are going to be easier for some people and harder for others, because we all have different experiences. Either way, I'd rate this as easy physics, because waves are a core topic and the maths is very basic.

* Nodes: it just occurs to me that you might be better off thinking of internodes, antinodes or bumps, rather than nodes. The fundamental has one bump, the 2nd harmonic two bumps, the 3rd has three bumps etc. With nodes, you always have one extra (or one less if you don't count the ends.)

** I'm never sure what PF means by "relevant equations". Here you seem to have picked a formula you thought was relevant before you started, then found it not useful (albeit still relevant?) Maybe you are supposed to fill this in when you've finished, with just the formulae you used?