Question

A 26-cm-long wire with a linear density of 20 g/m passes across the open end of an 86-cm-long open-closed tube of air. If the wire, which is fixed at both ends, vibrates at its fundamental frequency, the sound wave it generates excites the second vibrational mode of the tube of air.
What is the tension in the wire? Assume the speed of sound is 340 m/s .

Answer 1

Answer: T = 472.71 N

Explanation: The wire vibrates thus making sound waves in the tube.

The frequency of sound wave on the string equals frequency of sound wave in the tube.

L= Length of wire = 26cm = 0.26m

u=linear density of wire = 20g/m = 0.02kg/m

Length of open close tube = 86cm = 0.86m

Sound waves in the tube are generated at the second vibrational mode, hence the relationship between the length of air and and wavelength is given as

L = 3λ/4

0.86 = 3λ/4

3λ = 4 * 0.86

3λ = 3.44

λ = 3.44/3 = 1.15m.

Speed of sound in the tube = 340 m/s

Hence to get frequency of sound, we use the formulae below.

v = fλ

340 = f * 1.15

f = 340/ 1.15

f = 295.65Hz.

f = 295.65 = frequency of sound wave in pipe = frequency of sound wave in string.

The string vibrated at it fundamental frequency hence the relationship the length of string and wavelength is given as

L = λ/2

0.26 = λ/2

λ = 0.52m

The speed of sound in string is given as v = fλ

Where λ = 0.52m f = 295.65 Hz

v = 295.65 * 0.52

v = 153.738 m/s.

The velocity of sound in the string is related to tension, linear density and tension is given below as

v = √(T/u)

153.738 = √T/ 0.02

By squaring both sides

153.738² = T / 0.02

T = 153.738² * 0.02

T = 23,635.372 * 0.02

T= 472.71 N