1. G482 Electricity, Waves & Photons
2.4.1 Wave Motion
EM Waves
2.4.3 Interference
2.4.4 Stationary Waves
Ks5 OCR Physics H158/H558
Index
Mr Powell 2012

2. Practical Skills are assessed using OCR set tasks.
The practical work suggested below may be carried out as part of skill development. Centres are not required to carry out all of these experiments.
Students should gain a qualitative understanding of superposition effects together with confidence in handling experimental data.
Students should be able to discuss superposition effects and perform experiments leading to measurements of wavelength and wave velocity.
Study hearing superposition using a signal generator and two loudspeakers.
Demonstrate stationary waves using a slinky spring, tubes and microwaves.
Determine the speed of sound in air by formation of stationary waves in a resonance tube.

3. 2.4.4 Stationary Waves 3 – String Waves
Assessable learning outcomes....
(d) describe experiments to demonstrate stationary waves using, stretched strings (p160)
(e) determine the standing wave patterns for a stretched string; (p162)
(f) use the equation: separation between adjacent nodes (or antinodes) = /2; (p160) (or equ see right)
(g) define and use the terms: fundamental mode of vibration and harmonics; (p158)
Students can carry out experiments that demonstrate stationary waves for stretched strings, air columns and microwaves.
Invite students to bring along their musical instruments and discuss the formation of stationary waves.
In experiments, students should be aware of the end correction at the open end.
m = number of antinodes
c = speed of wave
L = length of string

4. Why are nodes formed in fixed positions?
We are in effect looking at interference points. Of which there must be a fixed point at either end then a whole number in between.
The first or “fundamental pattern” can be found as 0.
Then it goes up by one antinode at a time for each overtone.
You then simply sub in the fN for f in the usual formula

5. String Theory! (Not that one)
A wave on a string must take a time t to travel along and return the length of a string…
t = 2L/c
Where L is the length of the string and c is the speed of the wave.
The time taken for the mechanical oscillator to pass through a whole number of cycles is….
t = m/f
“m” is the number and “f” is cycles per second.
So we can express the situation to find L in terms of lambda and m.
NB: “m” also represents the number of antinodes

6. String Theory! (Not that one)
A wave on a string must take a time t to travel along and return the length of a string…
t = 2L/c
Where L is the length of the string and c is the speed of the wave.
The time taken for the mechanical oscillator to pass through a whole number of cycles is….
t = m/f
“m” is the number and “f” is cycles per second.
NB: “m” also represents the number of antinodes

7. Summary
Example….
m = 1 fundamental frequency
m = 2 frequency 2 x f0
m = 3 frequency 3 x f0
Now we can use the formulas we have created from first principals to conclude that due to…
Stationary waves fitting these conditions (travel back and forth along string) are formed in multiple frequencies….
f0, 2f0, 3f0, 4f0
Also that the length of the vibrating section of the string is a whole number of half wavelengths….
Example….
m = 1 *1/2
m = 2 *2/2 = 
m = 3 *3/2 = 1.5
NB: “m” also represents the number of antinodes

8. Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Working it all out…
Harder
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Anti- Nodes (m)
2L/m = 
Freq equation
f = c/
Freq value
c = f Speed (ms-1) fo 1
2L/1 = 2L
f0 = c/2L
256
410
2fo
3fo
4fo
5fo
L = length of string
fundamental
1st overtone
2nd overtone
3rd overtone
4th overtone

9. Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Working it all out…
Medium
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Anti- Nodes (m)
2L/m = 
Freq equation
f = c/
Freq value
c = f Speed (ms-1) fo 1
2L/1 = 2L
f0 = c/2L
256
410
2fo 2
f1 = c/L
3fo
2L/3
f2 = 3c/2L
768
4fo
f3 = 2c/L
5fo
L = length of string
fundamental
1st overtone
2nd overtone
3rd overtone
4th overtone

10. Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Working it all out…
Complete
Example if L = 0.8m, fo = 256Hz, c = 410ms-1…….
Anti- Nodes (m)
2L/m = 
Freq equation
f = c/
Freq value
c = f Speed (ms-1) fo 1
2L/1 = 2L
f0 = c/2L
256
410
2fo 2
2L/2 = L
f1 = c/L
512
3fo 3
2L/3
f2 = 3c/2L
768
4fo 4
2L/4 = L/2
f3 = 2c/L
1025
5fo 5
2L/5
f4 = 5c/2L
1281
L = length of string
fundamental
1st overtone
2nd overtone
3rd overtone
4th overtone

11. Data Table?
NB: m is number of antinodes!
Frequency
String Length
Number of anti-nodes
Lambda
Wave Speed
Ave

12. Data Table?
NB: m is number of antinodes!
Frequency
String Length
Number of nodes
Lambda
Wave Speed 8
2.62 2 21 16 3
1.75 28
24.3 4
1.31 32
32.5 5
1.05 34
38.5 6
0.87
Ave 30

13. Example Results?
Frequency
String Length
Nodes
Lambda
Wave Speed 18
2.42 1
43.6 28 2
1.61
45.2
43.4 3
1.21
52.5
57.9 4
0.97
56.0
71.1 5
0.81
57.4
83.1 6
0.69
57.5
95.4 7
0.61
57.7
Ave
52.8

14. b) c = f
 = 1.6m
f = 256Hz
c = f
= 256Hz x 1.6m
= 256s-1 x 1.6m
= 409.6ms-1
= 410ms-1 (3 sf)
c = f
l = 0.8m
f = 256Hz
The fundamental mode must be node to node so is only ½ 
Thus  = (2 x 0.8m) / 1
 = 1.6m (2 sf)
fundamental
1st overtone
2nd overtone

16. b) c = 410ms-1 , f = 384Hz, m = 1
f = mc / 2L
L = mc/ 2f
L = 410ms-1 / (2 x 384Hz)
= 410ms-1 / (2 x 384s-1)
= m
= 0.53m (2sf)
a) c = 410ms-1 , f = 512Hz, m = 1
(assume that this is a fundamental so m = 1)
f = mc / 2L
L = mc/ 2f
L = 1* 410ms-1 / (2 x 512Hz)
= 1* 410ms-1 / (2 x 512s-1)
= m
= 0.40m (2sf)
fundamental
1st overtone
2nd overtone

18. Exam Question
The image shows a side view of a string on a guitar. The string cannot move at either of the two bridges when it is vibrating. When vibrating in its fundamental mode the frequency of the sound produced is 108 Hz.
Sketch the stationary wave produced when the string is vibrating in its fundamental mode. (1 mark)
(b) Calculate the frequency of the overtone. (1 mark)
c) Calculate the speed of a progressive wave on this string. (1mark)

19. Exam Question
Answers
a) one ‘loop’ nodes at A & B
b) f1= 2 x f0 = 2 x 108Hz
= 216Hz
c) λ0 = 2L or λ = 0.64 × 2
= 1.3m or 1.28m
c = f λ = 108Hz × 1.3m
= 138 to 140ms-1
ecf
The image shows a side view of a string on a guitar. The string cannot move at either of the two bridges when it is vibrating. When vibrating in its fundamental mode the frequency of the sound produced is 108 Hz.
Sketch the stationary wave produced when the string is vibrating in its fundamental mode. (1 mark)
(b) Calculate the frequency of the overtone. (1 mark)
c) Calculate the speed of a progressive wave on this string. (1mark)

20. Tension of Wire By increasing the frequency of the vibrator
Different stationary wave (s.w.) patterns
are seen.
Boundary condition for s.w. on a string is that
there must be a node at each end.
Velocity of a transverse wave in a wire or string:
We find that....
T = Tension (N)
 = mass/ unit length kg/m

21. TASK: Annotate this text with some Maths!
More on Strings…
TASK: Annotate this text with some Maths!
How the fundamental frequency of a vibrating string depends on the string's length, tension, and mass per unit length is described by three laws:
The fundamental frequency of a vibrating string is inversely proportional to its length. Reducing the length of a vibrating string by one-half will double its frequency, raising the pitch by one octave, if the tension remains the same.
The fundamental frequency of a vibrating string is directly proportional to the square root of the tension. Increasing the tension of a vibrating string raises the frequency; if the tension is made four times as great, the frequency is doubled, and the pitch is raised by one octave.
The fundamental frequency of a vibrating string is inversely proportional to the square root of the mass per unit length.
This means that of two strings of the same material and with the same length and tension, the thicker string has the lower fundamental frequency. If the mass per unit length of one string is four times that of the other, the thicker string has a fundamental frequency one-half that of the thinner string and produces a tone one octave lower.

22. the frequency of a vibrating string is directly proportional to the square root of the tension.
f  T or T  f2
Thus if the tension is increased the frequency (pitch) goes up as a square relation.
A shorter string or length would yield a higher frequency as the frequency is inversely proportional to its length.
f = mc / 2L or f  1/L
If you half the length you will double its frequency.
This raises the pitch by one octave on a musical scale.
Tension would be the same.
T = Tension (N)
m = string mass kg
L = string length (m)
= mass/ unit length kgm-1
HARDER MATHS!

23. HARDER MATHS! T = Tension (N) m = string mass kg L = string length (m)
= mass/ unit length kgm-1
HARDER MATHS!

24. HARDER MATHS! T = Tension (N) m = string mass kg L = string length (m)
= mass/ unit length kgm-1
HARDER MATHS!
It would make sense that if L, T, diameter are fixed the steel wire being more dense and would have a larger mass.
This would mean that as c2 1/m as the mass went up the wave would travel slower through the wire.
Hence steel is slower than nylon (as it is more dense – makes sense).
As c = f as c goes down so does f as  is fixed from the length of the wire.
So steel strings would be a lower pitch for same conditions as nylon.

25. HARDER MATHS! T = Tension (N) m = string mass kg L = string length (m)
= mass/ unit length kgm-1
HARDER MATHS!

26. Further Reading on Tension & Frequency….
Hanging Mass
Tension
Frequency
3 kg
29.4 N
98 Hz
4 kg
39.2 N
112 Hz
5 kg
49 N
122.5 Hz
6 kg
58.8 N
131 Hz
A brief data set with a steel string is plotted to check the consistency of the data with the equation shown. Hanging masses provide the tension in the string, which was adjusted to a vibrating length of 50 cm. The average slope of the line through zero is taken, the slope is 17.6 HzN-0.5.
Using the frequency relationship above, this corresponds to a mass per unit length m/L = 57 grams/meter.
Key Summary
the shorter the string, the higher the frequency of the fundamental
the higher the tension, the higher the frequency of the fundamental
the lighter the string, the higher the frequency of the fundamental
HSW: Experimental Ideas
Source: Hyper Physics

27. Calculate the velocity of any transverse wave in the wire
Extension Question
A wire of mass per unit length 0.5 x 10-3 kgm-1 has a tension of 60 N and is 50 cm long.
Calculate the velocity of any transverse wave in the wire
Calculate the frequency of the fundamental note.
If nodes are at 17cm and 34cm find the frequency of the vibrating wire.
What must the tension be if a note an octave above the fundamental is required? ( 2x fundamental frequency)
Answers....

28. Revision/ Extension…
Visit the sites show below and then write a paragraph and numerical example for each.
Inverse square law calculation:
Geological example of sound reflection:
Pitch:
Loudness of wave:
String properties:
Wave properties and more :

29. Connection
Connect your learning to the content of the lesson
Share the process by which the learning will actually take place
Explore the outcomes of the learning, emphasising why this will be beneficial for the learner
Demonstration
Use formative feedback – Assessment for Learning
Vary the groupings within the classroom for the purpose of learning – individual; pair; group/team; friendship; teacher selected; single sex; mixed sex
Offer different ways for the students to demonstrate their understanding
Allow the students to “show off” their learning
Consolidation
Structure active reflection on the lesson content and the process of learning
Seek transfer between “subjects”
Review the learning from this lesson and preview the learning for the next
Promote ways in which the students will remember
A “news broadcast” approach to learning
Activation
Construct problem-solving challenges for the students
Use a multi-sensory approach – VAK
Promote a language of learning to enable the students to talk about their progress or obstacles to it
Learning as an active process, so the students aren’t passive receptors

33. Q3
In a stationary wave all the particles in a between a node are acting in phase.. i.e. they travel up at the same time. In a progressive wave each part of the wave is out of phase as you move along the wave through 360…….