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If all harmonics are generated by plucking, how does a guitar string produce a pure frequency sound?

How are overtones produced by plucking a string?Where are the harmonics in the radio spectrum?Why do harmonics occur when you pluck a string?How to find the first-harmonic frequency from the frequency spectrum of a recording of this harmonic being struck on a guitar?Why are the harmonics of a piano tone not multiples of the base frequency?Harmonic frequencies of a guitar stringtrying to understanding resonance, harmonics wrt standing wavesWhich frequency in a Fourier spectrum determines the frequency of the superposition / pitch of the result?Does plucking a guitar string create a standing wave?At what frequency does a string vibrate?How does a guitar work?

7

2

$begingroup$

A guitar is a plucked instrument and it is played by plucking a string at an off-centre point fixed at two ends. In general, Fourier analysis tells that all harmonics (the resonant frequencies of the string) will be excited and the string will vibrate in a superposition of different harmonics. It is true that for $n^{th}$ harmonic the amplitude goes like $1/n$ suggesting that the fundamental contributes the most. In this situation, I am confused. How does a guitar string produce a pure tone/pure frequency sound instead of a noise?

waves acoustics string vibrations harmonics

share|cite|improve this question

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: physics.stackexchange.com/q/429495
  $endgroup$
  – PM 2Ring
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Note: It's also a myth that all harmonics are generated by plucking. Pluck a string exactly in the middle, and you get a tone that only includes the even harmonics. This tone sounds very different from the tone that you get by plucking near the end of the string where you get a near complete set of harmonics. Also, the relative strengths of the harmonics varies with plucking location, which also contributes to slight changes of sound. Try it out if you have a guitar handy! E-guitarists use this effect a lot in conjunction with pick-up location to enable sounds otherwise impossible.
  $endgroup$
  – cmaster
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can reduce the number of harmonics by playing a string harmonic. It sounds quite different from the same pitch played "normally" which contains the full set of harmonics.
  $endgroup$
  – Peter A. Schneider
  25 mins ago
  

  
  
  
  

add a comment | 

7

2

$begingroup$

A guitar is a plucked instrument and it is played by plucking a string at an off-centre point fixed at two ends. In general, Fourier analysis tells that all harmonics (the resonant frequencies of the string) will be excited and the string will vibrate in a superposition of different harmonics. It is true that for $n^{th}$ harmonic the amplitude goes like $1/n$ suggesting that the fundamental contributes the most. In this situation, I am confused. How does a guitar string produce a pure tone/pure frequency sound instead of a noise?

waves acoustics string vibrations harmonics

share|cite|improve this question

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Related: physics.stackexchange.com/q/429495
  $endgroup$
  – PM 2Ring
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Note: It's also a myth that all harmonics are generated by plucking. Pluck a string exactly in the middle, and you get a tone that only includes the even harmonics. This tone sounds very different from the tone that you get by plucking near the end of the string where you get a near complete set of harmonics. Also, the relative strengths of the harmonics varies with plucking location, which also contributes to slight changes of sound. Try it out if you have a guitar handy! E-guitarists use this effect a lot in conjunction with pick-up location to enable sounds otherwise impossible.
  $endgroup$
  – cmaster
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can reduce the number of harmonics by playing a string harmonic. It sounds quite different from the same pitch played "normally" which contains the full set of harmonics.
  $endgroup$
  – Peter A. Schneider
  25 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

A guitar is a plucked instrument and it is played by plucking a string at an off-centre point fixed at two ends. In general, Fourier analysis tells that all harmonics (the resonant frequencies of the string) will be excited and the string will vibrate in a superposition of different harmonics. It is true that for $n^{th}$ harmonic the amplitude goes like $1/n$ suggesting that the fundamental contributes the most. In this situation, I am confused. How does a guitar string produce a pure tone/pure frequency sound instead of a noise?

waves acoustics string vibrations harmonics

share|cite|improve this question

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

$endgroup$

share|cite|improve this question

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

share|cite|improve this question

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

edited 12 hours ago

Qmechanic♦

105k121921207

edited 12 hours ago

Qmechanic♦

105k121921207

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

asked 14 hours ago

mithusengupta123mithusengupta123

1,32411436

3 Answers
3

active

oldest

votes

11

$begingroup$

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like:

$$ A(t,x) = sum_{i=0}^infty A_i sin(nomega_i t - k_i x) $$

i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with frequency $f = 2piomega_0$ so the ear/brain team perceives it as a tone not a noise.

Constructing a noise is actually quite complicated as we need to include all frequencies, not just integer multiples of a fundamental, and there will be a phase term in the equation that is not constant i.e. the sine waves making up the noise are not coherent.

share|cite|improve this answer

answered 14 hours ago

John RennieJohn Rennie

276k44549793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I may ask a related question here. What it means to tune a detuned guitar and why do different strings make different sounds?
  $endgroup$
  – mithusengupta123
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 That's probably better asked on the Music SE. The details of what overtones are produced will be a complicated function of lots of factors like the string density and it's position wrt the body of the guitar. Predicting the frequency spectrum from first principles using physics is impossibly hard. I guess some form of numerical finite element calculation would be needed.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK. I think that the only parameter than can change from one string to another is the tension T because the length L is same for all strings. That can change the fundamental and overtones from one string to another. Not very sure though
  $endgroup$
  – mithusengupta123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 the string linear density also changes. The high strings are thinner than the low strings. Also the position of the string relative to the guitar body will matter because a lot of the sound comes from vibrations induced in the guitar body.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically, for a guitar, the different strings have a different linear density. Then you adjust the tension of the rope to get the right note. The basic formula is $f=frac{1}{2L}sqrt{frac{T}{lambda }}$ In reality, things are of course more complicated. Because of the stiffness of the string, the modes are not strictly harmonic. A physicist alone could not tune a piano!
  $endgroup$
  – Vincent Fraticelli
  12 hours ago
  

  
  
  
  

 | 
show 3 more comments

11

$begingroup$

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern among their weights), where many non-pure tones are still recognized as dominated by a fundamental frequency. The additional frequencies add what we call the tone color or timbre of the sound.

In general, the exact weight of each harmonics can be somewhat varid according to how and where the chord is plucked. You might find interesting this study on the subject.

share|cite|improve this answer

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

$endgroup$

add a comment | 

7

$begingroup$

For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency.

For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$

To complete, the dependence in $n$ is rather $approx 1/{{n}^{2}}$ for a plucked string.

Sorry for my english !

share|cite|improve this answer

edited 13 hours ago

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not clear. All harmonics are present at the same time and the string doesn't vibrate with a particular resonant frequency
  $endgroup$
  – mithusengupta123
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$
  $endgroup$
  – Vincent Fraticelli
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK. I have completed !
  $endgroup$
  – Vincent Fraticelli
  13 hours ago
  

  
  
  
  

add a comment | 

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11

$begingroup$

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like:

$$ A(t,x) = sum_{i=0}^infty A_i sin(nomega_i t - k_i x) $$

i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with frequency $f = 2piomega_0$ so the ear/brain team perceives it as a tone not a noise.

Constructing a noise is actually quite complicated as we need to include all frequencies, not just integer multiples of a fundamental, and there will be a phase term in the equation that is not constant i.e. the sine waves making up the noise are not coherent.

share|cite|improve this answer

answered 14 hours ago

John RennieJohn Rennie

276k44549793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I may ask a related question here. What it means to tune a detuned guitar and why do different strings make different sounds?
  $endgroup$
  – mithusengupta123
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 That's probably better asked on the Music SE. The details of what overtones are produced will be a complicated function of lots of factors like the string density and it's position wrt the body of the guitar. Predicting the frequency spectrum from first principles using physics is impossibly hard. I guess some form of numerical finite element calculation would be needed.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK. I think that the only parameter than can change from one string to another is the tension T because the length L is same for all strings. That can change the fundamental and overtones from one string to another. Not very sure though
  $endgroup$
  – mithusengupta123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 the string linear density also changes. The high strings are thinner than the low strings. Also the position of the string relative to the guitar body will matter because a lot of the sound comes from vibrations induced in the guitar body.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically, for a guitar, the different strings have a different linear density. Then you adjust the tension of the rope to get the right note. The basic formula is $f=frac{1}{2L}sqrt{frac{T}{lambda }}$ In reality, things are of course more complicated. Because of the stiffness of the string, the modes are not strictly harmonic. A physicist alone could not tune a piano!
  $endgroup$
  – Vincent Fraticelli
  12 hours ago
  

  
  
  
  

 | 
show 3 more comments

11

$begingroup$

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like:

$$ A(t,x) = sum_{i=0}^infty A_i sin(nomega_i t - k_i x) $$

i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with frequency $f = 2piomega_0$ so the ear/brain team perceives it as a tone not a noise.

Constructing a noise is actually quite complicated as we need to include all frequencies, not just integer multiples of a fundamental, and there will be a phase term in the equation that is not constant i.e. the sine waves making up the noise are not coherent.

share|cite|improve this answer

answered 14 hours ago

John RennieJohn Rennie

276k44549793

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think I may ask a related question here. What it means to tune a detuned guitar and why do different strings make different sounds?
  $endgroup$
  – mithusengupta123
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 That's probably better asked on the Music SE. The details of what overtones are produced will be a complicated function of lots of factors like the string density and it's position wrt the body of the guitar. Predicting the frequency spectrum from first principles using physics is impossibly hard. I guess some form of numerical finite element calculation would be needed.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK. I think that the only parameter than can change from one string to another is the tension T because the length L is same for all strings. That can change the fundamental and overtones from one string to another. Not very sure though
  $endgroup$
  – mithusengupta123
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @mithusengupta123 the string linear density also changes. The high strings are thinner than the low strings. Also the position of the string relative to the guitar body will matter because a lot of the sound comes from vibrations induced in the guitar body.
  $endgroup$
  – John Rennie
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Basically, for a guitar, the different strings have a different linear density. Then you adjust the tension of the rope to get the right note. The basic formula is $f=frac{1}{2L}sqrt{frac{T}{lambda }}$ In reality, things are of course more complicated. Because of the stiffness of the string, the modes are not strictly harmonic. A physicist alone could not tune a piano!
  $endgroup$
  – Vincent Fraticelli
  12 hours ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Human perception is involved here because when you humans talk about noise this generally means a sound that is aperiodic. However the tone produced by a guitar will be something like:

$$ A(t,x) = sum_{i=0}^infty A_i sin(nomega_i t - k_i x) $$

i.e. a superposition of the frequencies $f$, $2f$, $3f$, etc. The function $A(t,x)$ is periodic in time with frequency $f = 2piomega_0$ so the ear/brain team perceives it as a tone not a noise.

Constructing a noise is actually quite complicated as we need to include all frequencies, not just integer multiples of a fundamental, and there will be a phase term in the equation that is not constant i.e. the sine waves making up the noise are not coherent.

share|cite|improve this answer

answered 14 hours ago

John RennieJohn Rennie

276k44549793

$endgroup$

share|cite|improve this answer

answered 14 hours ago

John RennieJohn Rennie

276k44549793

answered 14 hours ago

John RennieJohn Rennie

276k44549793

answered 14 hours ago

John RennieJohn Rennie

276k44549793

11

$begingroup$

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern among their weights), where many non-pure tones are still recognized as dominated by a fundamental frequency. The additional frequencies add what we call the tone color or timbre of the sound.

In general, the exact weight of each harmonics can be somewhat varid according to how and where the chord is plucked. You might find interesting this study on the subject.

share|cite|improve this answer

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

$endgroup$

add a comment | 

11

$begingroup$

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern among their weights), where many non-pure tones are still recognized as dominated by a fundamental frequency. The additional frequencies add what we call the tone color or timbre of the sound.

In general, the exact weight of each harmonics can be somewhat varid according to how and where the chord is plucked. You might find interesting this study on the subject.

share|cite|improve this answer

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

$endgroup$

add a comment | 

$begingroup$

Usually a guitar does not produce a pure tone/frequency. If so, its sound would be very close to a diapason. The difference between noise and a musical tone is not that a tone is made by a unique frequency, but there is a continuum between a pure tone (one frequency) and noise (all frequencies, not only multiple of a fundamental, without any regular pattern among their weights), where many non-pure tones are still recognized as dominated by a fundamental frequency. The additional frequencies add what we call the tone color or timbre of the sound.

In general, the exact weight of each harmonics can be somewhat varid according to how and where the chord is plucked. You might find interesting this study on the subject.

share|cite|improve this answer

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

$endgroup$

share|cite|improve this answer

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

answered 14 hours ago

GiorgioPGiorgioP

3,4251425

7

$begingroup$

For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency.

For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$

To complete, the dependence in $n$ is rather $approx 1/{{n}^{2}}$ for a plucked string.

Sorry for my english !

share|cite|improve this answer

edited 13 hours ago

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Not clear. All harmonics are present at the same time and the string doesn't vibrate with a particular resonant frequency
  $endgroup$
  – mithusengupta123
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$
  $endgroup$
  – Vincent Fraticelli
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  OK. I have completed !
  $endgroup$
  – Vincent Fraticelli
  13 hours ago
  

  
  
  
  

add a comment | 

7

$begingroup$

For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency.

For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$

To complete, the dependence in $n$ is rather $approx 1/{{n}^{2}}$ for a plucked string.

Sorry for my english !

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edited 13 hours ago

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

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  Not clear. All harmonics are present at the same time and the string doesn't vibrate with a particular resonant frequency
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  – mithusengupta123
  14 hours ago
  

  
  
  
  


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  For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$
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  – Vincent Fraticelli
  14 hours ago
  
  
  

  
  
  
  


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  OK. I have completed !
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  – Vincent Fraticelli
  13 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

For an ideal string, the key point is that all the harmonics are "harmonic" : their frequency is a integer multiple of the frequency of the fundamental. So the movement of the string is periodic and has a well definite frequency.

For an ideal string, the harmonics have frequency ${{f}_{1}}$ , $2{{f}_{1}}$, $3{{f}_{1}}$.....and ${{A}_{1}}cos (2pi {{f}_{1}}t)+{{A}_{2}}cos (2pi 2{{f}_{1}}t)+{{A}_{3}}cos (2pi 3{{f}_{1}}t)....$ is a periodic function of frequency ${{f}_{1}}$

To complete, the dependence in $n$ is rather $approx 1/{{n}^{2}}$ for a plucked string.

Sorry for my english !

share|cite|improve this answer

edited 13 hours ago

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

$endgroup$

share|cite|improve this answer

edited 13 hours ago

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125

answered 14 hours ago

Vincent FraticelliVincent Fraticelli

1,569125