Why do lower pitched string instruments have a larger body?
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I know that increasing the length of the string will decrease the pitch, but that doesn't explain why the body of a cello is larger than a violin, only longer.
What would be the "problem" with having a violin with a really long neck so it's the length of cello, and putting cello strings on it.
Is there some physical issue with trying to achieve a low pitch in a small resonating body (or visa versa)?
theory learning strings instruments string-instruments
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up vote
3
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favorite
I know that increasing the length of the string will decrease the pitch, but that doesn't explain why the body of a cello is larger than a violin, only longer.
What would be the "problem" with having a violin with a really long neck so it's the length of cello, and putting cello strings on it.
Is there some physical issue with trying to achieve a low pitch in a small resonating body (or visa versa)?
theory learning strings instruments string-instruments
New contributor
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
I know that increasing the length of the string will decrease the pitch, but that doesn't explain why the body of a cello is larger than a violin, only longer.
What would be the "problem" with having a violin with a really long neck so it's the length of cello, and putting cello strings on it.
Is there some physical issue with trying to achieve a low pitch in a small resonating body (or visa versa)?
theory learning strings instruments string-instruments
New contributor
I know that increasing the length of the string will decrease the pitch, but that doesn't explain why the body of a cello is larger than a violin, only longer.
What would be the "problem" with having a violin with a really long neck so it's the length of cello, and putting cello strings on it.
Is there some physical issue with trying to achieve a low pitch in a small resonating body (or visa versa)?
theory learning strings instruments string-instruments
theory learning strings instruments string-instruments
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New contributor
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asked 7 hours ago
nanotek
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3 Answers
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The lower the sound the instrument is producing, the larger the vibrating plate needs to be to produce that sound. The vibrations from the strings are being transferred to to the face of the instrument by the bridge, and the face of the instrument vibrates in response, making the air move, producing the waves that we hear as sound.
The larger face plate also increases the volume of the lower sound. If you had a cello with a violin sized body, you wouldn't hear the instrument very well, and the low notes would sound very "weak" and "thin".
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
add a comment |
up vote
1
down vote
You are only partly correct. For a given tension a string's fundamental resonant frequency scales with length. However, to within mechanical and material limits, you can increase or decrease tension on a fixed-length string to achieve any resonant frequency desired.
The string on a musical instrument is analogous to the oscillator in an electronic audio generator: it produces the source frequency(ies). The bridge and body of the instrument are analogous to the amplifier, transferring the energy from the string to the atmosphere with (one hopes!) minimal impedance mismatch (and yes, there is such a thing as acoustic impedance in air) so as to maximize the sound volume.
add a comment |
up vote
1
down vote
Just to complement Alphonso's answer: the lower the pitch, the lower the frequency. The lower the frequency, the larger the wavelength. Surfaces respond better to waves, when their wavelength is close to the size of the surface.
Imagine you try to shake a sheet of paper, by holding it vertically. There is a minimum frequency you need to apply to make it undulate. Now, if you try to apply this same frequency to a post-it, it won't shake, just move along with your hand. To make it undulate you would need to shake it much faster (higher frequency, smaller wavelength), because the post-it is smaller then a sheet of paper. Also, if you take a meter long sheet of paper and shake it that fast, it won't undulate a lot too. You would probably see the waves fading out through the paper, while it's other end barely moves. There is of course a lower frequency that would make it undulate entirely.
Instruments function in a similar way. The string has to shake the body, to produce a louder sound, and therefore the body size must minimally match the wavelength of the sound waves you want to produce.
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
The lower the sound the instrument is producing, the larger the vibrating plate needs to be to produce that sound. The vibrations from the strings are being transferred to to the face of the instrument by the bridge, and the face of the instrument vibrates in response, making the air move, producing the waves that we hear as sound.
The larger face plate also increases the volume of the lower sound. If you had a cello with a violin sized body, you wouldn't hear the instrument very well, and the low notes would sound very "weak" and "thin".
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
add a comment |
up vote
2
down vote
The lower the sound the instrument is producing, the larger the vibrating plate needs to be to produce that sound. The vibrations from the strings are being transferred to to the face of the instrument by the bridge, and the face of the instrument vibrates in response, making the air move, producing the waves that we hear as sound.
The larger face plate also increases the volume of the lower sound. If you had a cello with a violin sized body, you wouldn't hear the instrument very well, and the low notes would sound very "weak" and "thin".
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
add a comment |
up vote
2
down vote
up vote
2
down vote
The lower the sound the instrument is producing, the larger the vibrating plate needs to be to produce that sound. The vibrations from the strings are being transferred to to the face of the instrument by the bridge, and the face of the instrument vibrates in response, making the air move, producing the waves that we hear as sound.
The larger face plate also increases the volume of the lower sound. If you had a cello with a violin sized body, you wouldn't hear the instrument very well, and the low notes would sound very "weak" and "thin".
The lower the sound the instrument is producing, the larger the vibrating plate needs to be to produce that sound. The vibrations from the strings are being transferred to to the face of the instrument by the bridge, and the face of the instrument vibrates in response, making the air move, producing the waves that we hear as sound.
The larger face plate also increases the volume of the lower sound. If you had a cello with a violin sized body, you wouldn't hear the instrument very well, and the low notes would sound very "weak" and "thin".
answered 6 hours ago
Alphonso Balvenie
4,236516
4,236516
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
add a comment |
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
1
1
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
Yep. I'd just add that as a rule of thumb, to most efficiently get air moving for a particular frequency, the vibrating plate should be at least 1/4 of the wavelength of sound to be produced long. That makes a cello plate about right for viola c = 128 Hz, but only half the desired length for the bottom C at 64 Hz. A violin is way too short to get air moving efficiently at 64 Hz.
– Scott Wallace
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
You should incorporate @ScottWallace comment into your answer. As it stands, you have pretty much written a tautology. The whole point is that you can't produce a note of X wavelength with a body less than (at best) X/2 dimension.
– Carl Witthoft
1 hour ago
add a comment |
up vote
1
down vote
You are only partly correct. For a given tension a string's fundamental resonant frequency scales with length. However, to within mechanical and material limits, you can increase or decrease tension on a fixed-length string to achieve any resonant frequency desired.
The string on a musical instrument is analogous to the oscillator in an electronic audio generator: it produces the source frequency(ies). The bridge and body of the instrument are analogous to the amplifier, transferring the energy from the string to the atmosphere with (one hopes!) minimal impedance mismatch (and yes, there is such a thing as acoustic impedance in air) so as to maximize the sound volume.
add a comment |
up vote
1
down vote
You are only partly correct. For a given tension a string's fundamental resonant frequency scales with length. However, to within mechanical and material limits, you can increase or decrease tension on a fixed-length string to achieve any resonant frequency desired.
The string on a musical instrument is analogous to the oscillator in an electronic audio generator: it produces the source frequency(ies). The bridge and body of the instrument are analogous to the amplifier, transferring the energy from the string to the atmosphere with (one hopes!) minimal impedance mismatch (and yes, there is such a thing as acoustic impedance in air) so as to maximize the sound volume.
add a comment |
up vote
1
down vote
up vote
1
down vote
You are only partly correct. For a given tension a string's fundamental resonant frequency scales with length. However, to within mechanical and material limits, you can increase or decrease tension on a fixed-length string to achieve any resonant frequency desired.
The string on a musical instrument is analogous to the oscillator in an electronic audio generator: it produces the source frequency(ies). The bridge and body of the instrument are analogous to the amplifier, transferring the energy from the string to the atmosphere with (one hopes!) minimal impedance mismatch (and yes, there is such a thing as acoustic impedance in air) so as to maximize the sound volume.
You are only partly correct. For a given tension a string's fundamental resonant frequency scales with length. However, to within mechanical and material limits, you can increase or decrease tension on a fixed-length string to achieve any resonant frequency desired.
The string on a musical instrument is analogous to the oscillator in an electronic audio generator: it produces the source frequency(ies). The bridge and body of the instrument are analogous to the amplifier, transferring the energy from the string to the atmosphere with (one hopes!) minimal impedance mismatch (and yes, there is such a thing as acoustic impedance in air) so as to maximize the sound volume.
answered 1 hour ago
Carl Witthoft
8,03811229
8,03811229
add a comment |
add a comment |
up vote
1
down vote
Just to complement Alphonso's answer: the lower the pitch, the lower the frequency. The lower the frequency, the larger the wavelength. Surfaces respond better to waves, when their wavelength is close to the size of the surface.
Imagine you try to shake a sheet of paper, by holding it vertically. There is a minimum frequency you need to apply to make it undulate. Now, if you try to apply this same frequency to a post-it, it won't shake, just move along with your hand. To make it undulate you would need to shake it much faster (higher frequency, smaller wavelength), because the post-it is smaller then a sheet of paper. Also, if you take a meter long sheet of paper and shake it that fast, it won't undulate a lot too. You would probably see the waves fading out through the paper, while it's other end barely moves. There is of course a lower frequency that would make it undulate entirely.
Instruments function in a similar way. The string has to shake the body, to produce a louder sound, and therefore the body size must minimally match the wavelength of the sound waves you want to produce.
add a comment |
up vote
1
down vote
Just to complement Alphonso's answer: the lower the pitch, the lower the frequency. The lower the frequency, the larger the wavelength. Surfaces respond better to waves, when their wavelength is close to the size of the surface.
Imagine you try to shake a sheet of paper, by holding it vertically. There is a minimum frequency you need to apply to make it undulate. Now, if you try to apply this same frequency to a post-it, it won't shake, just move along with your hand. To make it undulate you would need to shake it much faster (higher frequency, smaller wavelength), because the post-it is smaller then a sheet of paper. Also, if you take a meter long sheet of paper and shake it that fast, it won't undulate a lot too. You would probably see the waves fading out through the paper, while it's other end barely moves. There is of course a lower frequency that would make it undulate entirely.
Instruments function in a similar way. The string has to shake the body, to produce a louder sound, and therefore the body size must minimally match the wavelength of the sound waves you want to produce.
add a comment |
up vote
1
down vote
up vote
1
down vote
Just to complement Alphonso's answer: the lower the pitch, the lower the frequency. The lower the frequency, the larger the wavelength. Surfaces respond better to waves, when their wavelength is close to the size of the surface.
Imagine you try to shake a sheet of paper, by holding it vertically. There is a minimum frequency you need to apply to make it undulate. Now, if you try to apply this same frequency to a post-it, it won't shake, just move along with your hand. To make it undulate you would need to shake it much faster (higher frequency, smaller wavelength), because the post-it is smaller then a sheet of paper. Also, if you take a meter long sheet of paper and shake it that fast, it won't undulate a lot too. You would probably see the waves fading out through the paper, while it's other end barely moves. There is of course a lower frequency that would make it undulate entirely.
Instruments function in a similar way. The string has to shake the body, to produce a louder sound, and therefore the body size must minimally match the wavelength of the sound waves you want to produce.
Just to complement Alphonso's answer: the lower the pitch, the lower the frequency. The lower the frequency, the larger the wavelength. Surfaces respond better to waves, when their wavelength is close to the size of the surface.
Imagine you try to shake a sheet of paper, by holding it vertically. There is a minimum frequency you need to apply to make it undulate. Now, if you try to apply this same frequency to a post-it, it won't shake, just move along with your hand. To make it undulate you would need to shake it much faster (higher frequency, smaller wavelength), because the post-it is smaller then a sheet of paper. Also, if you take a meter long sheet of paper and shake it that fast, it won't undulate a lot too. You would probably see the waves fading out through the paper, while it's other end barely moves. There is of course a lower frequency that would make it undulate entirely.
Instruments function in a similar way. The string has to shake the body, to produce a louder sound, and therefore the body size must minimally match the wavelength of the sound waves you want to produce.
answered 54 mins ago
coconochao
1,10616
1,10616
add a comment |
add a comment |
nanotek is a new contributor. Be nice, and check out our Code of Conduct.
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