Question

A violin string which is 30 cm long is tuned using a 440 Hz reference tone.

(a) What is the wavelength fundamental mode of the string?

(b) When the string has a tension of 34 N a violinist hears 5.7 beats per second. What are the frequencies at which the string might be vibrating?

(c) The tension is increased and the violinist hears 3.8 beats per second. What is the original frequency of the string?

(d) What should the tension of the string be so that the fundamental frequency is 440 Hz.

Answer 1

A) the fundamental wavelength is twice the length of the string. 2 x 30 cm = 60 cm = .6m

B) the difference between the 440 Hz reference tone and the frequency of the violin is 5.7 so the string is either vibrating at 434.3 hz or 445.7 hz

C) the velocity of a wave is v=sqrt(FT/u) where FT is the tension and u is the linear density.

    also v = lambda * f where lambda is the wavelength and f is the frequencey.

    therefore lambda * f = sqrt(FT/u)

    f = sqrt(FT/u)/lambda

therfore if tension increased then frequency should be increased.

the beat frequency is 3.8 and the reference tone is 440 so the frequency is now either 443.8 or 436.2
so the frequency will be = 443.8hz

    

D) at 34 N of tension:

f*lambda = sqrt(FT/u)

434.3 * .6 = sqrt(34/u) 260.58

            = sqrt(34/u) 67902 = 34/u

                u= 34/67902

                  = 5x10^-4 where u is the linear density if we want f = 440 FT

                  = u (lambda * f)^2 = (5x10^-4) (.6 * 440)^2 = 34.8N