Question

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass of 9.50 g. It is fixed at both ends and oscillates in its fundamental mode. By resonance, it sets the air column in the tube into oscillation at that column's fundamental frequency. Assume that the speed of sound in air is 343 m/s, find (a) that frequency and (b) the tension in the wire.

Answer 1

71.4583 Hz

67.9064 N

Step-by-step explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

`f=(v)/(4L)\\\Rightarrow f=(343)/(4* 1.2)\\\Rightarrow f=71.4583\ Hz`

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

`\mu=(m)/(l)\\\Rightarrow \mu=(9.5* 10^(-3))/(0.35)\\\Rightarrow \mu=0.02714\ kg/m`

The fundamental frequency of the wire is given by

`f=(1)/(2l)\sqrt{(T)/(\mu)}\\\Rightarrow f^2=(1)/(4l^2)(T)/(\mu)\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2* 0.02714* 4* 0.35^2\\\Rightarrow T=67.9064\ N`

The tension in the wire is 67.9064 N