A string with a mass per length of 2.00 g/m is stretched with a force of 150 N b

Question

A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends. The velocity of sound in air at 0 A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends. The velocity of sound in air at 0 A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends. The velocity of sound in air at 0 A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends. The velocity of sound in air at 0 A string with a mass per length of 2.00 g/m is stretched with a force of 150 N between two points that are 0.500 m apart. The fundamental frequency of the stretched string is in tune with the fundamental frequency of an organ pipe filled with air and open at both ends. The velocity of sound in air at 0

Explanation / Answer

v = sqrt(T/u) = sqrt(150/.002) = 273.86m/s

fundamental frequancy = 273.86/2*.5 = 273.86

it is equal to fundamental frequancy of organ pipe

V/2l = 273.86

331/2l = 273.86

l = .604m

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