Fundamental Frequency of nylon string

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Fundamental Frequency!!! [SOLVED]

A nylon string is stretched between fixed supports 0.75m apart. Experimental plucking of the string shows that several standing waves can exist on the string. Two such standing waves have frequencies of 225Hz and 300Hz with no other frequencies in between.

Q1. What is the lowest frequency (fundamental) frequency that can exist on the string?

I tried to use f=v/2L = Root(t/m/L)/2L
But the tension and mass was not given. I'm pretty sure that would have something to do with those frequencies provided. But I have no clue of how to do it.

Thanks ^^
 
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  • #2
Hootenanny
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A nylon string is stretched between fixed supports 0.75m apart. Experimental plucking of the string shows that several standing waves can exist on the string. Two such standing waves have frequencies of 225Hz and 330Hz with no other frequencies in between.

Q1. What is the lowest frequency (fundamental) frequency that can exist on the string?

I tried to use f=v/2L = Root(t/m/L)/2L
But the tension and mass was not given. I'm pretty sure that would have something to do with those frequencies provided. But I have no clue of how to do it.

Thanks ^^
Welcome to Physics Forums.

HINT: How is the frequency of the nth harmonic related to the fundamental mode?
 
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Welcome to Physics Forums.

HINT: How is the frequency of the nth harmonic related to the fundamental mode?
fn=nv/2L So is it when n=1 its the first harmonic? And its the fundamental frequency?
I'm not sure very sure...
 
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Hootenanny
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fn=nv/2L So is it when n=1 its the fundamental frequency?
I was thinking more of something of the form fn = n*f1, where fn is the nth harmonic and f1 is the fundamental frequency.

Can you use this expression together with the information provided to create a system of two equations?
 
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I was thinking more of something of the form fn = n*f1, where fn is the nth harmonic and f1 is the fundamental frequency.

Can you use this expression together with the information provided to create a system of two equations?
So is it 225Hz=nf1 and 300Hz=(n+1)f1 ?
Throught my calculations....
So f1=300-225?
 
  • #6
Hootenanny
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So is it 225Hz=nf1 and 300Hz=(n+1)f1 ?
Throught my calculations....
So f1=300-225?
Looks good to me :approve:
 
  • #7
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Ohh ! I got it!!!
Because the Nth frequency is the n*fundamental frequency!
Is that right?
 
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Thank u !!! :D
 
  • #9
HallsofIvy
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Okay, exactly what answer did you get? It does NOT follow that because [itex]225= n*f_1[/itex], that 330 must be the next harmonic. [itex]330= m*f_1[/itex], certainly, but m is not necessarily n+1.

330- 225= 105. But 105 is NOT a divisor of 330: that is, 330 is not equal to n*105 for any integer n so 105 is NOT the "fundamental frequency". What is true is that [itex]f_n= n*f_1[/itex] so the fundamental frequence must be a factor of both 330 and 225 (and, so, 105). 330= 3*110= 3*5*22= 2*3*5*11. 225= 5*45= 5*5*9= 32*52. What is the largest number that divides both of those?
(Even then there is no guarentee that a fundamental frequency is the greatest common factor. There four possible fundamental frequencies for this.
 
  • #10
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Okay, exactly what answer did you get? It does NOT follow that because [itex]225= n*f_1[/itex], that 330 must be the next harmonic. [itex]330= m*f_1[/itex], certainly, but m is not necessarily n+1.

330- 225= 105. But 105 is NOT a divisor of 330: that is, 330 is not equal to n*105 for any integer n so 105 is NOT the "fundamental frequency". What is true is that [itex]f_n= n*f_1[/itex] so the fundamental frequence must be a factor of both 330 and 225 (and, so, 105). 330= 3*110= 3*5*22= 2*3*5*11. 225= 5*45= 5*5*9= 32*52. What is the largest number that divides both of those?
(Even then there is no guarentee that a fundamental frequency is the greatest common factor. There four possible fundamental frequencies for this.
Sorry to cause confusing there. The two frequencies provided was supposed to be 225 Hz and 300 Hz. Not 330 Hz for the second one.
 

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