Physics 101 Learning Object 8, Standing Waves on Strings

1. Standing Waves on
Strings
A Physics Learning Object 8 by Nico Utama Alimin

2. Table of Contents
 Definition
 Formulas
 Fundamental Frequency
 Other Harmonics
 Helpful Tips
 Sample Problem
 Clicker Questions

3. 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

4. Formulas
 Sin(
2π
𝜆
L) = 0 
2π
𝜆
L = mπ 
2
m L = 𝜆
 m is an integer 1,2,3,4…
 fm =
𝑣
ߣm
=
𝑚
2𝐿
v =
𝑚
2𝐿
𝑇
𝜇
 fm = m x (
1
2𝐿
𝑇
𝜇
)

5. 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

6. 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

7. 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.

8. 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
2𝐿
𝑣)
ߣ =
2𝐿
𝑚
Thus,
f =
4 𝑥 589 𝑚/𝑠
2 𝑥 15.0 𝑚
f =
2356 𝑚/𝑠
30.0 𝑚
f = 78.533 Hz
ߣ =
2𝐿
𝑚
ߣ =
2 𝑥 15.0 𝑚
4
ߣ = 7.5 m

9. 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

10. 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
2𝐿
v = f
m =
2fL
𝑣
Then,
m =
2 𝑥 192.3 𝐻𝑧 𝑥 13.0 𝑚
1000 𝑚/𝑠
m = 4.9998 ≈ 5

11. 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

12. 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
13.0 𝑚
T = 1100 kgm/s2
T = 1100 N