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?

f_beat = 210 Hz

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.

what is the actual question though?

get the total number of pieces of fruit and divide it by 4 to get the average. the result is the answer to the problem. 14 total number of fruits and there are 4 members. 14 divided by 4 is equals to 3.5.