Question

A child's train whistle replicates a classic conductor's whistle from the early 1900s. This whistle has two open-open tubes that produce two different frequencies. When you hear these two different frequencies simultaneously, you may have the perception of also hearing a lower note, called a difference tone, that is at the same frequency as the beat frequency between the two notes. The two tubes of the whistle are 14 cm and 12 cm in length.Assuming a sound speed of 350 m/s, what is the frequency of this difference tone?

Answer 1

f_beat = 210 Hz

Step-by-step explanation:

Solution:

- The frequency of the difference tone - beat frequency is given by:

f_beat = |f_1 - f_2|

Where,

f_1 = Frequency of first component

f_2 = Frequency of second component

- The whistle is a tube with openings at both ends. The wavelength (λ) of the fundamental tone in a pipe of length L which is open at both ends is:

λ = 2*L

- For all waves, the relationship between frequency (f) , wavelength (λ), and propagation speed (v) is:

λ*v = c

- So the fundamental frequency of a pipe of length L, open at both ends is:

f = v/(2*L)

- In this case, we are told that v = 350 m/s and the pipes have lengths 12 cm and 14 cm, so the beat (difference) frequency is:

f_beat = | (350 m/s)/(2*0.12 m) - (350 m/s)/(2*0.14 m)|

= 208.33 Hz

- All the input data for this problem were only given to 2 significant digits, so your result should only be reported to 2 significant digits, which in this case would be 210 Hz.