Phys 223 - Lab Report 7 (Transverse Wave Motion)
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Feb 20, 2024
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PHYS 223
Transverse Wave Motion
Megan Cousins
Tuesday-Thursday
Objective
Our goal is to find the relationship between wavelength and tension in a string undergoing oscillatory motion while the ends are in a fixed position.
Introduction
When a string or rope is set into oscillatory motion, waves are generated and they will travel up and down the string. Each particle of the string will be displaced perpendicular to the general direction of the string. Hence the displacement of the string is perpendicular to the direction of propagation of the wave motion. This type of wave motion is known as transverse wave motion. When a wave traveling down a string comes to the end of the string, it will be reflected back up the string. If the end of the string is fixed, as for instance if it is tied to a support, the wave will also undergo a 180° phase change
Questions
1.
Plot the tension in the string against λ From this straight line calculate f. See Figure 1.
0.3496 = 1/ μ f
2
0.3496 = 1/ 0.001685 (f
2
)
0.3496(f
2
) = 1/ 0.001685
0.000589076(f
2
) = 1
f = √
❑
f = 41.20 hz
2.
Calculate the percentage error for f. 0% error
3.
What general effect does increasing the tension have upon a vibrating string?
The general effect of increasing the tension generally decreases the number of loops, and increases both the velocity and wavelength.
4.
Is there a true node at the vibrator end of the string? Explain.
No, there is not a true node at the vibrating end of the string as the string is still moving and not permanently at rest like the rest of the nodes. 5.
All the strings on a violin, cello, and guitar are the same length. What differences do they have which give them different frequencies? What other way can the frequency be changed? The biggest difference between musical instruments that all have the same length of string is that they get their different frequencies due to the thickness of the strings. Another way that we can change the frequency is by vibrating the strings at a different rate. 6.
A copper wire one meter long weighing 0.02 grams per cm. vibrates in three segments when under a tension of 280 grams. What is the frequency of this vibration?
L = nλ/2
1 = 3λ/2
2 = 3λ
0.667 = λ
λ
=
1
f
√
❑
0.667
=
1
f
√
❑
0.667
=
1
f
(
0.751
)
0.888
=
1
f
0.888
f
=
1
f
=
1.126
hz
Discussion
This experiment was rather short and simple. With a fixed frequency and a rope just a little bit longer than 2 meters also fixed in place, we allowed the string to undergo an oscillatory motion and document the number of visible nodes. Nodes being a point on the string that is not undergoing any
motion, with varying the mass and thus the tension on the string, change in value and position. Doing three trials, where we allow 11 or 10 nodes to be present, one less than the last (10 or 9 nodes) and the last
trial of 6 or 5 nodes, we documented the number of visible nodes and the added mass to achieve that number. Going through calculations to find the corresponding wavelength, this value was squared and plotted against the tension. From this slope, we are able to recalculate the frequency to see if any error was achieved, there was no error in this experiment.
Appendix
Figure 1.
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