Example 13-6 G-String Physics
The third string on an electric guitar is 0.640 m long and has a linear mass density of 1.14 * 10-3 kg>m. This string is
to play a note of frequency 196 Hz (to a musician, this note is G3 or G below middle C). (a) What must be the tension in
the string? (b) What is the wavelength of the fundamental standing wave on the string? (c) What is the speed of transverse
waves on this string? (d) What is the wavelength of the sound wave produced by the wave?
Set Up
We are given the string length L = 0.640 m
and the mass per length of the string
m = 1.14 * 10-3 kg>m. We’ll use Equation
13-19 to find the tension F that gives a fundamental frequency f1 = 196 Hz. We’ll use
Equation 13-18 to find the wavelength l of the
standing wave, and use this along with Equation 13-2 to find the speed of waves on the
string. The sound wave produced by the guitar
has the same frequency as the standing wave
on the string. We’ll find the wavelength of the
sound wave using Equations 13-2 and 13-13.
Frequencies for a standing wave
on a string:
fn =
F
n
where n = 1, 2, 3, c
2L A m
(13-19)
Wavelengths for a standing wave on a string:
nl
L =
where n = 1, 2, 3, c 2
(a) The fundamental frequency is given by
Equation 13-19 with n = 1. Solve this equation
for the string tension F.
(13-18)
Propagation speed, frequency, and wavelength of a wave:
(13-2)
vp = fl
Speed of sound in dry air at 20°C:
vsound = 343 m>s
Solve
G string
(13-13)
The fundamental frequency (n = 1) is
f1 =
F
1
2L A m
To solve for the string tension F, first square both sides to get rid of
the square root:
f 21 =
1 F
4L2 m
Multiply both sides by 4L2m:
F = 4L2mf 21
= 410.640 m2 2 11.14 * 10-3 kg>m2 1196 Hz2 2
= 71.8 kg # m # Hz 2 = 71.8 kg # m>s 2
= 71.8 N
(Recall that 1 Hz = 1 s21.)
(b) For the fundamental mode (the n =
1 standing wave) of a string held at both
ends, the length of the string equals one
half-wavelength.
From Equation 13-18,
l
L = for n = 1
2
l = 2L = 2(0.640 m) = 1.28 m
(wavelength of the standing wave on the string)
(c) Find the wave speed on the string using
Equation 13-2.
Speed of transverse waves on this string:
vp = fl
= (196 Hz)(1.28 m)
= 251 m>s
(d) Use Equations 13-2 and 13-13 to find the
wavelength of the sound wave produced by the
guitar.
For the sound wave,
vsound = fl
The frequency of the sound wave is the same as the frequency of the
standing wave on the spring, but the wavelength is different because
the wave speed is different:
343 m>s
343 m>s
vsound
=
=
f
196 Hz
196 s -1
= 1.75 m
l =
Reflect
There are six strings on the guitar, each of which is under about the same tension. So the total force that acts on the
­guitar at the points where the strings are attached is about 6 * 71.8 N = 431 N (about 97 lb). The guitar must be of
sturdy construction to withstand these forces.
You can check our result for the speed of transverse waves on the string by using Equation 13-10, vp = 2F>m, and
the value of F that we found in part (a). Do you get the same answer this way?
Note that our answer for part (d) is reasonable. The wavelength l of the sound wave is greater than the wavelength
of the standing wave on the string because, although the frequency f is the same for both waves, the propagation speed
vp is greater for the sound wave and vp = fl.