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)?










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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)?










    share|improve this question







    New contributor




    nanotek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      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)?










      share|improve this question







      New contributor




      nanotek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      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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      nanotek is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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".






          share|improve this answer

















          • 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


















          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.






          share|improve this answer




























            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
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              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".






              share|improve this answer

















              • 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















              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".






              share|improve this answer

















              • 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













              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".






              share|improve this answer












              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".







              share|improve this answer












              share|improve this answer



              share|improve this answer










              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














              • 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










              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.






              share|improve this answer

























                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.






                share|improve this answer























                  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.






                  share|improve this answer












                  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.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 1 hour ago









                  Carl Witthoft

                  8,03811229




                  8,03811229






















                      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.






                      share|improve this answer

























                        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.






                        share|improve this answer























                          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.






                          share|improve this answer












                          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.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered 54 mins ago









                          coconochao

                          1,10616




                          1,10616






















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