Anylon guitar string is fixed between two lab posts 2.00 m apart. the string has a linear mass density of μ=7.20 g/m\mu=7.20~\text{g/m}μ=7.20 g/m and is placed under a tension of 160.00 n. the string is placed next to a tube, open at both ends, of length l. the string is plucked and the tube resonates at the n=1 mode. the speed of sound is 343 m/s. what is the length of the tube?

4.6 m

Explanation:

First of all, we can find the frequency of the wave in the string with the formula:

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

is the mass linear density

Solving the equation,

The frequency of the wave in the string is transmitted into the tube, which oscillates resonating at same frequency.

The n=1 mode (fundamental frequency) of an open-open tube is given by

where

v = 343 m/s is the speed of sound

Using f = 37.3 Hz and re-arranging the equation, we find L, the length of the tube:

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