Question

The E string on an electric bass produces the instrument’s lowest possible frequency, 41 Hz. The string has a length of 0.86 m and a mass per unit length of 0.015 kg/m. (a) Calculate the tension in this string. (b) Calculate the frequency and wavelength of the second harmonic. (c) Calculatethe frequency and wavelength of the third harmonic. So a) is 72N and b) is 0.86m and c) 123Hz

How would i solve this problem? With the same values of ? and FT, what length of string would be needed for a fundamental frequency of 21 Hz ?

Answer 1

a)
The lowest possible frequency corresponds to the highest possible wavelength.
= 2L = 2 x 0.86 = 1.72 m
Velocity of waves, v = f
= 41 x 1.72 = 70.52 m/s

Velocity, v = SQRT[T/]
Where T is the tension and is the mass per unit length
T = v²
= 0.015 x (70.52)²
= 74.596 N

b)
Frequency of the firs harmonic, f = 41 Hz
Frequency of the second harmonic = 2f = 82 Hz
Wavelength of the second harmonic = v/2f = 70.52/82 = 0.86 m

c)
Frequency of the third harmonic = 3f = 3 x 41 = 123 Hz
Wavelength of the third harmonic = v/3f = 70.52/123 = 0.573 m

d)
Velocity remains the same, v = 70.52 m/s
Wavelength corresponds to fundamental frequency, = 2L
v = f
70.52 = 21 x 2L
L = 70.52/42
= 1.68 m