Standing Waves on

StringsA Physics Learning Object 8 by Nico Utama Alimin

Table of Contents

Definition

Formulas

Fundamental Frequency

Other Harmonics

Helpful Tips

Sample Problem

Clicker Questions

What are Standing Waves on Strings?

Standing waves are waves propagated in a fixed length of string, with three

possibilities:

Open on both ends

Open on one end, closed on the other

Closed on both ends

Standing waves are stationary, unlike travelling waves.

Images taken from:

http://whs.wsd.wednet.edu/faculty/busse/mathhomepage/busseclasses/apphysics/studyguides/chapter16_2008/chapter16studyguid

e2008.html

Formulas

Sin(2π

𝜆L) = 0

2π

𝜆L = mπ

2

m L = 𝜆

m is an integer 1,2,3,4…

fm = 𝑣

mߣ=

𝑚

2𝐿v =

𝑚

2𝐿

𝑇

𝜇

fm = m x (1

2𝐿

𝑇

𝜇)

Fundamental Frequency

The fundamental frequency, or firs harmonic, has the lowest frequency, and

longest wavelength

The formula for fundamental frequency is fm = m x (1

2𝐿

𝑇

𝜇), where m is 1

The formula would be f1 = 1

2𝐿

𝑇

𝜇

Image from: http://www.antonine-education.co.uk/Pages/Physics_2/Waves/WAV_05/Waves_Page_5.htm

Other Harmonics

There are other harmonics, which corresponds to higher integers in m.

2nd Harmonic: m = 2; f2 = 2

2𝐿

𝑇

𝜇= 2 f1

3rd Harmonic: m = 3; f3 = 3

2𝐿

𝑇

𝜇= 3 f1

4th Harmonic: m = 4; f4 = 4

2𝐿

𝑇

𝜇= 4 f1

And so on…

http://improvisingguitar.blogspot.ca/2006/10/harmonics-pt-0-physics-of_06.html

Helpful Tips

To quickly find the wavelength, divide 2L with the number of waves.

The number of integral corresponds to the number of antinodes when both

ends are fixed.

A string open at one end has half the frequency and number of waves

compared to both closed and both open strings.

The pressure is highest on the node, and lowest on the antinode

The displacement node is at the fixed ends, and displacement antinode at

open ends.

Sample Problem

What is the frequency and wavelength of a 4th harmonic wave that has a speed of 589 m/s on a 15.0m string, and has a linear mass density of 90kg/m?

Known:

m = 4

v = 589 m/s

L = 15.0

𝜇 = 90 kg/m

Formula: fm = m x (1

2𝐿𝑣)

ߣ = 2𝐿

𝑚

Thus,

f = 4 𝑥 589𝑚/𝑠

2 𝑥 15.0𝑚

f = 2356𝑚/𝑠

30.0 𝑚

f = 78.533 Hz

ߣ = 2𝐿

𝑚

ߣ = 2 𝑥 15.0 𝑚

4

ߣ = 7.5 m

Clicker Question 1

What harmonic is it for a string to have the length of 13.0 m, a frequency of

192.3 Hz, and a speed of 1000m/s?

A. 4

B. 5

C. 6

D. 7

E. 8

Clicker Question 1 Answer

What harmonic is it for a string to have the length of 13.0 m, a frequency of

192.3 Hz, and a speed of 1000m/s?

A. 4

B. 5

C. 6

D. 7

E. 8

Since

L = 13.0m

f = 192.3 Hz

v = 1000 m/s

Formula: 𝑚

2𝐿v = f

m = 2fL𝑣

Then,

m = 2 𝑥 192.3 𝐻𝑧 𝑥 13.0 𝑚

1000𝑚/𝑠

m = 4.9998 ≈ 5

Clicker Question 2

For the same string as sample problem 1, what is the tension of the string

that has a mass of 14.3 g?

A. 1000 N

B. 1100 N

C. 1430 N

D. 14300 N

E. 13400 N

Clicker Question 2 Answer

For the same string as sample problem 1, what is the tension of the string

that has a mass of 14.3 g?

A. 1000 N

B. 1100 N

C. 1430 N

D. 14300 N

E. 13400 N

Since

v = 1000 m/s

M = 14.3 g = 0.0143 kg

L = 13.0 m

Formulas:

𝜇 = 𝑀

𝐿

v = 𝑇

𝜇

Then,

T = v2 x 𝜇

T = v2 x 𝑀

𝐿

T = (1000m/s)2 x 1.43 𝑥 10−3 𝑘𝑔

13.0𝑚

T = 1100 kgm/s2

T = 1100 N