Topic 12.6
More about stationary
waves on strings
Understanding stationary waves
Aims
This activity will enable you to show your knowledge of fundamental and overtone
oscillations as well as providing practice in handling the wave equation. You are also asked
to describe an experiment, an important skill in logical communication. You will compare
units on two sides of an equation and will need to think about the meaning of a newton, from
Chapter 9. Finally you will evaluate an incorrect argument and, hopefully, show your mastery
of standing waves on a string.
Waves on a vibrating string
The string is vibrating at 42 Hz; this is the 6th overtone also known as the 7th harmonic. It is
not the fundamental frequency.
1 (a) Explain what is meant by the fundamental frequency and draw a diagram to show
what the string looks like when it is vibrating at its fundamental frequency.
(b) How you can tell that the string is vibrating at its 6th overtone (7th harmonic)?
(c) What is the fundamental frequency of vibration of the string?
(d) Calculate the wavelength of the wave.
(e) Calculate the speed of the wave along the string.
2 Suggest how the apparatus can be used to measure the speed of the wave along the string.
Decide what measurements should be made and then state clearly:
(a) how each actual measurement is made
(b) any extra apparatus that is used in each measurement
(c) how the speed is calculated from the actual measurements.
3 The formula for the speed c of a wave on a string is c =
T

where T is the tension in the
string and is the mass per unit length of the string.
The units on both sides of an equation must be equal.
(a) Show that the units are equal on both sides of the wave equation c = f
(b) Show that the units are equal on both sides of the equation c =
T

(Hint: F = ma enables you to express the N in more basic units.)
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
1
Topic 12.6
More about stationary
waves on strings
(c) The tension in the string is increased by a factor of four while the oscillator vibrates at
the same frequency as in the diagram.
What happens to the speed and the wavelength of the wave?
Explain why the string does not now show a standing wave pattern.
4 Someone says to you, ‘The wave on the string creates a sound. The speed of sound is the
speed of the wave on the string.’
Why are they wrong? What wave property is the same for the string and the sound that is
created?
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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Topic 12.6
More about stationary
waves on strings
Answer Sheet
Aims

To show knowledge of fundamental and overtone oscillations.

To practise handling the wave equation.

To describe a simple experiment, developing skill in logical communication.

To compare units on two sides of an equation.

To evaluate an incorrect argument, developing skills of synthesis.
Teaching notes

and the wave equation c = f.
2
They need to know a definition of the fundamental frequency and how to find a particular
overtone. They may need to be shown how to take various units and express them in terms of
basic units.
Students need to know that the node-to-node distance is
Answers
5 (a) The fundamental frequency is the simplest mode of operation with only one antinode
on the string and the string vibrates at the lowest frequency.
7
2
The wavelength of the standing wave is one-seventh the wavelength of the
fundamental and thus seven times the frequency. This makes the vibration the 6th
overtone and the 7th harmonic.
(d) There are seven antinodes, so the string has a length of
(e) 6 Hz
(f) Four sections of the string cover 20 cm so

= 5 and  = 10 cm.
2
(g) v = f = 42  0.10 = 4.2 m s−1
6 Frequency is measured by either attaching a frequency meter to the ac signal from the
signal generator or by attaching an oscilloscope to the signal generator. Using the
oscilloscope, the time for one wave is measured by multiplying the distance along the
x-axis of one oscillation by the number of seconds per division.
1
time period
The wavelength is measured by measuring the distance between adjacent nodes with a
metre rule. For accuracy more than one node-to-node distance can be measured. The
wavelength  = 2  node-to-node distance and v = f.
The frequency =
1
7 (a) Units of f = Hz  m =    m = m s−1
s
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
3
Topic 12.6
More about stationary
waves on strings
(b) Units of T = N = kg m s−2
2
Units of T = kg m s1 = m2 s−2.

kg m
Taking the square root gives m s−1, which is the same as for the velocity.
(c) The speed doubles; the wavelength doubles (the frequency is constant).
7

The string (of length 20 
= 35 cm) is not now a multiple of
(which is now
4
2
10  2 = 20 cm).
8 The string vibrates up and down making the air molecules vibrate with the same
frequency as the string. The sound has the same frequency as the wave on the string but
since the speed of sound (340 m s−1) is different, the wave on the string and the sound
wave have different wavelengths.
AQA Physics A AS Level Extension Activity © Nelson Thornes Ltd 2008
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