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How can overtones exist when plucking open string?
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I read the following from wikipedia:
When a string is plucked normally, the ear tends to hear the
fundamental frequency most prominently, but the overall sound is also
colored by the presence of various overtones (frequencies greater than
the fundamental frequency).
If an open string is plucked, I just notice a node at each end and an anti node at the middle. How can we have overtones in addition to the fundamental frequency? It seems to be counter intuitive for me.
waves acoustics string harmonics
edited 42 mins ago
Qmechanic♦
97k121631031
asked 1 hour ago
Artificial Stupidity
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up vote
2
down vote
favorite
I read the following from wikipedia:
When a string is plucked normally, the ear tends to hear the
fundamental frequency most prominently, but the overall sound is also
colored by the presence of various overtones (frequencies greater than
the fundamental frequency).
If an open string is plucked, I just notice a node at each end and an anti node at the middle. How can we have overtones in addition to the fundamental frequency? It seems to be counter intuitive for me.
waves acoustics string harmonics
edited 42 mins ago
Qmechanic♦
97k121631031
asked 1 hour ago
Artificial Stupidity
1165
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I read the following from wikipedia:
When a string is plucked normally, the ear tends to hear the
fundamental frequency most prominently, but the overall sound is also
colored by the presence of various overtones (frequencies greater than
the fundamental frequency).
If an open string is plucked, I just notice a node at each end and an anti node at the middle. How can we have overtones in addition to the fundamental frequency? It seems to be counter intuitive for me.
waves acoustics string harmonics
edited 42 mins ago
Qmechanic♦
97k121631031
asked 1 hour ago
Artificial Stupidity
1165
I read the following from wikipedia:
When a string is plucked normally, the ear tends to hear the
fundamental frequency most prominently, but the overall sound is also
colored by the presence of various overtones (frequencies greater than
the fundamental frequency).
If an open string is plucked, I just notice a node at each end and an anti node at the middle. How can we have overtones in addition to the fundamental frequency? It seems to be counter intuitive for me.
waves acoustics string harmonics
waves acoustics string harmonics
edited 42 mins ago
Qmechanic♦
97k121631031
asked 1 hour ago
Artificial Stupidity
1165
edited 42 mins ago
Qmechanic♦
97k121631031
asked 1 hour ago
Artificial Stupidity
1165
edited 42 mins ago
Qmechanic♦
97k121631031
edited 42 mins ago
Qmechanic♦
97k121631031
edited 42 mins ago
Qmechanic♦
97k121631031
97k121631031
asked 1 hour ago
Artificial Stupidity
1165
asked 1 hour ago
Artificial Stupidity
1165
asked 1 hour ago
Artificial Stupidity
1165
1165
add a comment |Â
add a comment |Â
2 Answers
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5
down vote
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
answered 50 mins ago
S. McGrew
3,9432620
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up vote
1
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Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m sin(pi m x/L)$, where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location $x_0 $means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
answered 47 mins ago
Maxim Umansky
2,80831025
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
answered 50 mins ago
S. McGrew
3,9432620
add a comment |Â
up vote
5
down vote
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
answered 50 mins ago
S. McGrew
3,9432620
add a comment |Â
up vote
5
down vote
up vote
5
down vote
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
answered 50 mins ago
S. McGrew
3,9432620
The only way to avoid overtones would be to pluck the string in such a way that its initial shape is sinusoidal. However, that would be nearly impossible. In practice, the initial shape is almost always triangular.
If you are familiar with Fourier transforms, consider how you would do a discrete Fourier decomposition of the string's initial shape. The Fourier components correspond to the overtones.
answered 50 mins ago
S. McGrew
3,9432620
answered 50 mins ago
S. McGrew
3,9432620
answered 50 mins ago
S. McGrew
3,9432620
answered 50 mins ago
S. McGrew
3,9432620
3,9432620
add a comment |Â
add a comment |Â
up vote
1
down vote
Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m sin(pi m x/L)$, where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location $x_0 $means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
answered 47 mins ago
Maxim Umansky
2,80831025
add a comment |Â
up vote
1
down vote
Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m sin(pi m x/L)$, where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location $x_0 $means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
answered 47 mins ago
Maxim Umansky
2,80831025
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m sin(pi m x/L)$, where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location $x_0 $means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
answered 47 mins ago
Maxim Umansky
2,80831025
Eigenmodes of a string have sinusoidal spatial form $f_m(x) = C_m sin(pi m x/L)$, where x is the parallel coordinate and L is the length of the string. Plucking a string at a fixed location $x_0 $means giving it a non-sinusoidal initial perturbation, e.g., something like a piece-wise linear function, $f(x) = A x/x_0$ for $x le x_0$ and $f(x) = A (L-x)/(L-x_0)$ for $x ge x_0$. Expanding the initial perturbation $f(x)$ in eigenmodes $f_m(x)$ shows how much each harmonic is excited initially, in general it would be a full spectrum of eigenmodes.
answered 47 mins ago
Maxim Umansky
2,80831025
answered 47 mins ago
Maxim Umansky
2,80831025
answered 47 mins ago
Maxim Umansky
2,80831025
answered 47 mins ago
Maxim Umansky
2,80831025
2,80831025
add a comment |Â
add a comment |Â
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