A tube, open at only one end, is cut into two shorter (nonequal) lengths. The piece that is open at both ends has a fundamental frequency of 416 Hz, while the piece open only at one end has a fundamental frequency of 678 Hz. What is the fundamental frequency of the original tube?

The fundamental frequency of the original tube is 159.169 Hz.

Explanation:

From the orientation of a wave in a tube open at both ends,  we have at the fundamental frequency there is an anti-node at both ends.

Therefore the length of the tube = λ/2

and we have v = fλ where v = speed of sound in air = 343 m / s

f = frequency, therefore

Therefore λ = v/f = 343/416 = 0.824 m, therefore the tube opened at both ends is

λ/2 or 0.824/2 m = 0.4123 m long

The other tube with one end open we have a node at the closed end and an anti-node at the opened end which gives a fundamental frequency with the wavelength = λ/4

Therefore from v = fλ we have λ = v/f  = 343/678 = 0.5059 m

Therefore the length of the tube =  λ/4 = 0.5059÷4 = 0.1265 m

The length of the original tube is then given by the sum of the lengths of the two split tubes, that is =0.1265 + 0.4123 = 0.5387 m

Note that when combined the original tube still has only one end opened

Therefore the length of the tube = λ/4 or  the wavelength = 4×0.5387

λ = 2.1549 m

Therefore we have f =v/λ = 343/2.1549

Frequency, f = 159.169 Hz

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