Standing waves are created by two harmonic waves of equal amplitude, wavelength and frequency moving in opposite directions. Nodes occur where the amplitude is always zero, while antinodes occur where the amplitude is at its maximum between -2A and 2A. The distance between nodes and antinodes is λ/2, while the distance between a node and the nearest antinode is λ/4. The frequency of the mth harmonic is given by fm = m*f1, where f1 is the fundamental frequency.

1. Standing Waves on
Strings
Physics 101

2. What is a Standing Wave?
 A wave that does not appear to travel.
 Created by two harmonic waves of equal amplitude, wavelength and
frequency, moving in opposite directions.

3. Nodes and Antinodes
 Nodes occur where the amplitude is always zero.
Node Node NodeNode

4. Nodes and Antinodes
 Antinodes occur where the amplitude moves between -2A and 2A, the
maximum amplitude. (A – amplitude of constituent harmonic waves)
Antinode
Antinode

5. λ
4
λ
2
3λ
4
λ 5λ
4
λ
2
λ
2
Node to Node
Antinode to Antinode
Nodes and Antinodes
 Consecutive nodes and consecutive antinodes are
𝜆
2
(half a wavelength) apart

6.  The distance between a node and the nearest antinode is
𝜆
4
.
Nodes and Antinodes
λ
4
λ
2
3λ
4
λ 5λ
4
λ
4
Antinode
to Node
λ
4
Node to
Antinode

7. Frequency of Standing Waves on Strings
 𝒇 𝒎 =
𝒗
𝝀 𝒎
=
𝒎
𝟐𝑳
𝒗 =
𝒎
𝟐𝑳
𝑻
𝝁
=
𝒎
𝟐𝑳
𝑻
𝑴
𝑳
 m = mode number (1, 2, 3, …) (see next slide)
 v = wave speed
 𝜆 = wavelength
 L = length of string
 T = tension of the string
 M = mass of string
 Recall that 𝜇 is the linear mass density of the string
𝑀
𝐿
.

8. Normal Modes
 Represented as n in the diagram
 The mode number corresponds to the
number of antinodes on the vibrating string
 A mode of 2 is double the frequency of the
fundamental frequency (n = 1)
 The relationship between the first harmonic
and other harmonics is represented by this
equation:
 𝒇 𝒎 = 𝒎𝒇 𝟏
 m is a positive integer > 0

9. Harmonics/Resonant Frequencies
 The allowed frequencies represented by the equation:
 𝒇 𝒎 = 𝒎𝒇 𝟏
 In lab 5, you found the fundamental frequency, the lowest frequency that
results in a single antinode at its maximum frequency.
 Knowing the fundamental frequency, we can find the harmonics/resonant
frequencies.

10. Questions

11. #1 - The frequency of the fourth
harmonic is…
 A) Same as the frequency of the 2nd harmonic.
 B)
4
3
times greater than the frequency of the 3rd harmonic.
 C) Triple the frequency of the fundamental frequency.
 D) Double the frequency of the 3rd harmonic.

12. The frequency of the fourth harmonic
is…
 A) Same as the frequency of the 2nd harmonic.
 B)
𝟒
𝟑
times greater than the frequency of the 3rd harmonic.
 C) Triple the frequency of the fundamental frequency.
 D) Double the frequency of the 3rd harmonic.
 𝒂⦁𝒇 𝟑 = 𝒇 𝟒
 𝒂⦁𝟑𝒇 𝟏 = 𝟒𝒇 𝟏
 𝒂 =
𝟒𝒇 𝟏
𝟑𝒇 𝟏
=
𝟒
𝟑

13. #2 - A guitar string that is plucked produces a
standing wave. Its angular velocity is 5 rad/m.
The amplitude is 8mm. The length the string
vibrating is 40cm. The mass of the string is 0.2g.
The tension of the string is equal to the weight
of a 36kg mass on the moon.
 A) Imagine that the standing wave is produced by two harmonic waves of equal
amplitude, wavelength, and frequency travelling opposite directions. What would
be the amplitude of the two constituent waves?
 B) Write an equation to represent the amplitude of the standing wave.
 C) What is the distance between a node and the nearest antinode?
 D) What is the wave speed in the string?
 E) What is the fifth harmonic?

14. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
 A) Imagine that the standing wave is produced by two harmonic waves of
equal amplitude, wavelength, and frequency travelling opposite directions.
What would be the amplitude of the two constituent waves?
 The amplitude of the standing wave is double the amplitude of the
constituent wave.
 2A = 8mm
 A = 4mm

15. A guitar string that is plucked produces a standing wave. Its angular velocity is
5.0 rad/m. The amplitude is 8.0mm. The length the string vibrating is 40cm. The
mass of the string is 0.2g. The tension of the string is equal to the weight of a
36kg mass on the moon.
 B) Write an equation to represent the amplitude of the standing wave.
 A(x) = A sin(ωx)
 A(x) = (8.0 mm)sin(5.0x)

16. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
 C) What is the distance between a node and the nearest antinode?
 Let us find the distance between nodes.
 Recall from the text (eq 14-44): 𝑥 = 𝑚
𝜆
2
, where m = 0, ±1, ±2, ±3, …
 We know that ω = 5rad/m and ω =
2𝜋
𝜆
 5 =
2𝜋
𝜆
𝜆 =
2𝜋
5
 m = 0  x = 0
 m = 1  x =
𝜆
2
=
2𝜋
2(5)
=
𝜋
5
 m = 2  x = 𝜆 =
2𝜋
5
 (continued on next slide)

17. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass
of the string is 0.2g. The tension of the string is equal to the weight of a 36kg
mass on the moon.
 C) What is the distance between a node and the nearest antinode?
 The location of the nodes are at: x = 0,
𝜋
5
,
2𝜋
5
, …
 The location of the first antinode is at:
𝜆
4
=
2𝜋
(4)5
=
𝜋
10
 We can find the distance between any node the nearest antinode by finding
the distance between the first node and antinode.
 The first node is at x = 0
 The first antinode is at x =
𝜋
10

𝜋
10
- 0 =
𝝅
𝟏𝟎

18. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass of
the string is 0.2g. The tension of the string is equal to the weight of a 36kg mass on
the moon.
 D) What is the wave speed in the string?
 𝑣 =
𝑇
𝜇
 𝜇 =
𝑚 𝑠𝑡𝑟𝑖𝑛𝑔
𝐿
=
0.2 𝑥 10−3 𝑘𝑔
0.4𝑚
=
5 𝑥 10−4 𝑘𝑔
𝑚
 𝑇 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 36𝑘𝑔 𝑚𝑎𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑚𝑜𝑜𝑛
 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑔 𝑚𝑜𝑜𝑛
 𝑔 𝑚𝑜𝑜𝑛 =
1
6
𝑔 𝑒𝑎𝑟𝑡ℎ =
1
6
(9.81
𝑚
𝑠2) = 1.635
𝑚
𝑠2
 𝑇 = 𝑚𝑔 𝑚𝑜𝑜𝑛 = 36𝑘𝑔 1.635
𝑚
𝑠2 = 58.86𝑁
 𝑣 =
58.86
5 𝑥 10−4 = 𝟑𝟒𝟐
𝒎
𝒔

19. A guitar string that is plucked produces a standing wave. Its angular velocity is 5
rad/m. The amplitude is 8mm. The length the string vibrating is 40cm. The mass of
the string is 0.2g. The tension of the string is equal to the weight of a 36kg mass on
the moon.
 E) What is the fifth harmonic?
 Find the fundamental frequency first.
 𝑣 = 𝜆𝑓
 𝑓1 =
𝑣
𝜆
 𝑣 = 342.34
𝑚
𝑠
 𝜆 = 2𝐿 = 2 0.4𝑚 = 0.8𝑚 𝑎𝑡 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦
 𝑓1 =
342.34
𝑚
𝑠
0.8𝑚
= 427.925Hz
 𝑁𝑜𝑤 𝑤𝑒 𝑐𝑎𝑛 𝑓𝑖𝑛𝑑 𝑡ℎ𝑒 𝑓𝑖𝑓𝑡ℎ ℎ𝑎𝑟𝑚𝑜𝑛𝑖𝑐 𝑢𝑠𝑖𝑛𝑔 𝑡ℎ𝑒 𝑓𝑢𝑛𝑑𝑎𝑚𝑒𝑛𝑡𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦.
 𝑓5 = 5𝑓1 = 5 427.925Hz = 𝟐𝟏𝟒𝟎𝐇𝐳

20. Works Cited
 Wave on a String PhET simulation (wave diagrams)
 https://phet.colorado.edu/en/simulation/wave-on-a-string
 Physics 101 Textbook (definitions and equations)
 Physics for Scientists and Engineers: An Interactive Approach, 1st
Edition
 Robert Hawkes, Javed Iqbal, Firas Mansour, Marina Milner-Bolotin,
and Peter Williams