To show that Frequency is directly proportional to the √Tension
__________________________________________________________________________________
Method:
Set up a sonometer so that a movable bridge shortens the length of the wire to approx one third of its total length.

Place a small paper rider in the middle of l, the length from the fixed to the movable bridge

Strike a tuning fork of the highest frequency and place its stem on the fixed bridge of the sonometer

Adjust the length of the movable bridge slightly until the paper rider begins to resonate i.e. jumps off the wire
in an agitated manner. (Ensure that this is the maximum displacement). This length will now remain fixed for
the rest of the experiment.)

Record the values for tension and length and the frequency of the tuning fork

Repeat this experiment with other tuning forks (i.e. different frequencies) and different values for tension

Plot a graph of F against √Tension

A graph that is straight, best fit and through the origin will tell us Frequency is directly proportional to the
√Tension
Precautions:

To make this experiment a fair test, the length must stay the same for all values of f and T. This ensures there
is always a fixed variable. We are comparing like with like.

When drawing the diagram of the sonometer, the tuning fork, length l, and Newton balance must be included.

You must begin with a tuning fork of the highest frequency and shortest length to ensure you locate the 1st
antinode

The thickness of the wire should be uniform

The tuning fork must always be held at its stem only

Keep the rider as light as possible

The experiment is repeated so that the averages can be taken thus results are more accurate

After drawing the graph, only take data from the graph

Avoid parallax errors when reading lengths and tensions.

The paper rider is placed mid-way between the bridges
Question 1994
The relationship between the natural frequency of a stretched string and its tension was investigated by applying a
force to the end of a wire of fixed length l. The applied force was varied and the resulting values of the natural
frequency f, and the tension T, of the wire were determined. The following results were obtained:
f/Hz
T/N
256
288
320
341
384
427
480
512
5.0
6.0
9.0
10.0
12.0
14.0
18.0
22.0
(a) Draw a suitable graph on graph paper to illustrate the relation between the natural frequency of the wire and its
tension. State the relationship and explain how the graph verifies it.
(18 marks)
(b) Given that the length of the wire was 64 cm, use the graph to calculate a value for the mass per unit length of
the wire. (Assume that the wire was vibrating at its lowest natural frequency, i.e., its fundamental frequency.)
(12
marks)
(c) Explain how the natural frequency of the wire might have been
determined in this experiment. (9 marks)
(ii)
f proportional to √T
(iii)
The graph, being a straight line, verifies this relationship.
Graph: On graph paper, draw a graph of frequency (y-axis) against
the square root of the tension.
Note: Start both axes at zero. A straight line graph through the origin
shows that the frequency of vibration of a stretched string is
proportional to the square root of the tension on the string

Conclusion: The graph is best fit, straight and through the origin. Therefore we can say that Frequency
is directly proportional to the √Tension
(b)
Maths Question:
f
=
1 1 √T
2 L √μ
√μ
=
1 1 √T
√μ
=
1
2
√μ
=
0.00703
1
0.64
2 L f
4
445
√μ
=
(1/2)(1.5625)(0.009)
μ
=
4.94 X 10-5 Kg m-1
Note:


Getting the slope of the graph, rather than taking a point on the graph leads to a better result because some
students take values close to the origin and this increases % error in measurement readings.
I mistakenly took √T as √4 ie 2 (Beware)……………….. √T = 4 which means that T = 16N
(c)
By resonance.
Place a small paper rider on the middle of the string. Adjust the tension of the wire until resonance occurs and the
rider vibrates vigorously or hops off. (May be accompanied by a loud sound)
2009 Papers
f/Hz
T/N
√T/N
256
288
320
341
384
480
2.4
3 .3
3.9
4.3
5.7
8.5
1.5
1.8
2
2.1
2.4
2.9
Conclusion: The graph is best fit, straight and through the origin. Therefore
we can say that Frequency
is directly proportional to
the √Tension
Use the graph to
1. estimate the fundamental frequency of the string when its
tension is 11N
2. calculate the mass per unit length of the string
1.
When tension is 11N then
√Tension
=
3.31, which when you look at the graph tells
√μ
=
us that f = 550Hz
1 1 √T
2 L √μ
2.
f
=
√μ
=
1 1 √T
2 L f
√μ
=
1
2
√μ
=
0.0075
1
0.4
3.31
550
μ
=
(1/2)(2.5)(0.006)
5.625 X 10-5 Kg m-1
(http://www.pbc-cork.ie/UserFiles/File/Physics_PDF_Files/SoundExperiments_000.pdf)
Diagram needs to have the length shown, the tuning fork and the Newton
balance.