A standard violin string is 32.8 cm long from the tuning peg to the bridge. When

Question

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.

Explanation / Answer

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.

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