Respuesta :
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
The tension on the given wire with linear density is 472.7 N.
The given parameters;
- length of the wire, L = 26 cm
- mass density, 20 g/m = 0.02 kg/m
- length of the pipe, = 86 cm = 0.86 m
The wavelength of open-closed pipe for second vibrational mode is calculated as follows;
[tex]L = \frac{3\lambda }{4} \\\\\lambda = \frac{4L}{3} \\\\\lambda = \frac{4\times 0.86}{3} = 1.15 \ m[/tex]
The fundamental frequency of the wave in the pipe is calculated as;
[tex]v = f_o \lambda\\\\f_o = \frac{v}{\lambda} \\\\f_o = \frac{340}{1.15} \\\\f_o = 295.65 \ Hz[/tex]
The speed of the along the wire is calculated as;
[tex]l = \frac{\lambda }{2} \\\\\lambda = 2l\\\\v = f\lambda\\\\v = 295.65 \times (2 \times 0.26)\\\\v= 153.74 \ m/s[/tex]
The tension in the wire is calculated as follows;
[tex]v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = \mu v^2\\\\T = (0.02 \ kg/m) \times (153.74 \ m/s)^2\\\\ T = 472.7 \ N[/tex]
Thus, the tension on the given wire with linear density is 472.7 N.
Learn more here:https://brainly.com/question/14789390
