In: Physics
A standard violin string is 32.8 cm long from the tuning peg to the bridge. When played open(without pressing a finger against the fingerboard), this string has a fundamental (lowest)frequency of 440 Hz.
(a) What is the wavelength of the wave at this frequency? (Hint: the lowest frequency is alsothe longest wavelength that can produce a standing wave.) b.
(b) What is the speed of the wave on the string? (Hint: this is not the speed of sound!)c.
(c) How far from the end of the string should a finger be placed in order to produce afundamental frequency of 523 Hz.
ANS
It is a physics question, a music related physics
question.
The speed of propagation of a wave in a string (v) is proportional
to the square root of the tension of the string (T) and inversely
proportional to the square root of the linear density (?) of the
string.
V = sqrt(T / ?)
as your question shows there is nothing changed on the string;
then the V must be a constant on both cases. Next equation, V = f *
?, Velocity equals to frequency times the wavelength.
on a violin string, its fundamental frequency wavelength is equal
to two times of the string length. So that, on the case one, we can
find the velocity_1 = 440 * 2 * 0.328 = 288.64 m/s
Now, we can use this result to find the length of the C, its given
frequency is 523 Hz
288.64 = 523 *2 * Length_C
Length_C = 288.64 x 1 / 523 x 1 / 2 = 0.275m
So that, the string length for playing the C is 27.5 cm
It is 32.8-27.5 cm from the end of the string.
Your finger should be place at 5.3 cm from the end of the
string.