Asked by genevieve
A particularly beutiful note reaching your ear from a stradivarius violin has a wavelength of 39.1cm. The room is slightly warm, so the speed of sound is 344m/s. If the string's linear density is 0.670g/m and the tension is 140N, how long is the vibrating section of the violin string?
Answers
Answered by
drwls
First calculate the frequency of vibration of the sound wave.
f = Vsound/(wavelength) = 880 Hz
(which would be an A note one octave above the standard A)
Next calculate the wavelength of the waves traveling on the violin string (if waves were traveling; actually there is a standing wave there). You need to know the wave speed on the string, which is sqrt(T/density) = sqrt(140/.000670) = 457 m/s
The wavelength on the string is
Vwave/f = 0.519 m = 51.9 cm.
Assuming this is a fundamental mode of the string, the length of the vibrating string is half a wavelength, or 26 cm.
f = Vsound/(wavelength) = 880 Hz
(which would be an A note one octave above the standard A)
Next calculate the wavelength of the waves traveling on the violin string (if waves were traveling; actually there is a standing wave there). You need to know the wave speed on the string, which is sqrt(T/density) = sqrt(140/.000670) = 457 m/s
The wavelength on the string is
Vwave/f = 0.519 m = 51.9 cm.
Assuming this is a fundamental mode of the string, the length of the vibrating string is half a wavelength, or 26 cm.
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