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 422 Hz, while the piece open only at one end has a fundamental frequency of 666 Hz. What is the fundamental frequency of the original tube?

159.1 Hz

Explanation:

The formula for the fundamental frequency of an open pipe is given as,

f' = v/2l' Equation 1

Where f' = fundamental frequency of the open pipe, v = velocity of sound in air, l' = length of the open pipe.

make l' the subject of the equation

l' = v/2f' Equation 2

Given: f' = 422 Hz, v = 343 m/s

Substitute into equation 2

l' = 343/(2×422)

l' = 0.41 m.

Also, for the closed pipe

f = v/4l

Where f = fundamental frequency of the closed pipe, l = length of the closed pipe.

make l the subject of the equation

l = v/4f Equation 4

Given: f = 666 Hz, v = 343 m/s.

Substitute into equation 4

l = 343/(4×666)

l = 0.129 m.

But,

Length of the original tube = l+l' = 0.41+0.129 = 0.539 m.

L = 0.539 m.

Note: The tube is a closed pipe.

F = v/4L Equation 5

Where F = Fundamental frequency of the original tube.

F = 343/(4×0.539)

F = 159.1 Hz.

Hence the fundamental frequency of the original tube = 159.1 Hz

the answer would be c. motion and direction

i hope this : )

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