Question

A nylon 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?

Answer 1

Answer:

4.6 m

Explanation:

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

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where we have

L = 2.00 m is the length of the string

T = 160.00 N is the tension

$\mu =7.20 g/m = 0.0072 kg/m$ is the mass linear density

Solving the equation,

$f=\frac{1}{2(2.00 m)}\sqrt{\frac{160.00 N}{0.0072 kg/m}}=37.3 Hz$

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

$f=\frac{v}{2L}$

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:

$L=\frac{v}{2f}=\frac{343 m/s}{2(37.3 Hz)}=4.6 m$