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course Phy 202
We created standing waves in two different chains, by stretching the chain out and oscillating one end with the necessary frequency. The chain then oscillated at a matching frequency, with nodes at its ends (the end that was used to drive the oscillation wasn't strictly a node, but the oscillations along the chain were of much greater amplitude than at this end, this end was very nearly a node).
The fundamental mode of oscillation is the one of the lowest possible frequency, and it occurs when the only nodes are at the ends. The fundamental mode is also called the first harmonic. This was the mode you used in your first trial, and in most of your trials.
We also achieved an oscillation where the antinodes near the two ends were moving in different directions, with a node in the middle. This mode has the second-lowest frequency and is called the second harmonic (it can also be called the first overtone, with the fundamental being regarded as the basic tone). Including the nodes at the ends, this mode has three nodes.
Higher frequencies can produce modes of oscillation with additional nodes. The third harmonic would have four nodes, one at each end and two in between. It's easy to see why the number of nodes is one greater than the number of the harmonic.
Antinodes (i.e., points along the chain where the amplitude of the oscillation is greatest) occur halfway between adjacent nodes, so that the configuration of the oscillation can be described by its series of alternating nodes and antinodes. The fundamental oscillation of these chains has configuration
•N A N,
i.e., a node at each end with an antinode between.
The second harmonic has configuration
•N A N A N,
i.e., nodes at each end, as well as a node in the middle, with antinodes halfway between the ends and the middle. As you observed when you oscillated the chain, in this mode the antinodes were always moving in opposite directions.
Paperclip chain data:
When the 10-rubber-band chain had respective lengths of 85, 114 and 155 cm, observed frequencies were 55, 71 and 80 cycles / minute.
Rubber band chain data:
When 1 meter of the chain corresponded to the lengths of 15.2, 12.85, 11.5 and 9.8 rubber bands, observed frequencies were 52, 71, 80 and 82 cycles / minute.
Analysis of data:
You have data for the lengths of the long rubber band chain vs. the lengths of the short section of the chain, and also for the length of the short chain vs. the number of 50-gram hooks hanging from it. This is your calibration data, from which we will be able to determine the tensions of the rubber bands in your various trials.
To figure out the tension in your chain for each trial:
I've suggested that you start by plotting the length of a 10-rubber band chain vs. the weight of the hooks.
If a group actually measured their 10-rubber-band chain with different numbers of hooks, it will be possible to construct that graph very easily. Just remember that you will want the tension in Newtons; grams are not units of force.
If data consists of the number of rubber bands in a meter, it is possible to easily infer the length of a 10-rubber-band chain. For example if at some tension 8 rubber bands span a meter, then a 10-rubber-band chain would span 10/8 of a meter, or 125 cm.
From your calibration data, then, you can obtain a table of tension (i.e., the weight of the hooks) vs. length of a 10-rubber-band chain.
You should use this information to construct a graph of tension vs. length of 10-rubber-band chain:
•Graph the points from your table.
•Sketch the straight line you think best fits your points.
•Find the slope and vertical-axis intercept of your line.
Give your table, the slope of your line and its vertical-axis intercept.
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Points are (70cm, 0.49N), (74cm, 0.98N), (78cm, 1.47N), and (82cm, 1.96N)
Slope is 0.1225N/cm
Equation of line is Y=0.1225x-8.085, so vertical -axis intercept is -8.085 newtons.
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Then using your data from the trials where you counted the number of cycles in a minute, find the tension for each trial:
•Based on the lengths of the short rubber band chain (for the paperclip chain) or the short segment of the long rubber band chain for each of the trials, use the length of a 10-rubber-band chain, along with your graph, to figure out the tension. (one group actually measured a 10-rubber-band chain; the group that used the number of rubber bands required to span a meter will need to figure out for each trial how long a 10-rubber-band chain would have been)
Give the length of the 10-rubber-band chain, or the 10-rubber-band segment of the longer rubber-band chain, for each of the trials, for each chain, and give the tension according to your graph:
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For 85cm chain length, tension would be 2.33N
115cm length tension is 6.00N
155cm length tension is 10.9N
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Using the counts for each trial, find the time required for a pulse to make a round trip along the chain:
The time required for a pulse to travel the length of the chain and back is equal to the time required for one complete cycle (that is, one complete cycle of the oscillation mode in which the only nodes are at the ends; for oscillations with one or more nodes between the ends this does not apply). We will see why this is so, but for now we will just assume that it's so.
For each trial you know how many oscillations occurred in a minute, so you can use your results to figure out, for each trial, the time required for a cycle.
Give the time required for a cycle, for each of the tensions, for each of the chains:
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1.09 seconds for a cycle at 2.33N tension
0.85 seconds for a cycle at 6.00N tension
0.75 seconds for a cycle at 10.9N tension
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Calculate the speed of the pulse:
You also know the length of the chain for each trial (for the paperclip chain this will be the same for every trial; the rubber band chain stretched so its length will have varied).
For each trial, then, you can calculate the speed of the pulse.
Give your results for the pulse speed for each trial, and include a detailed calculation of one of the speeds.
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Speed of pulse=2*length/period
Speed of pulse=2*247.5cm/1.09seconds
Speed of pulse=495cm/1.09sec= 454.13cm/second at tension of 2.33N
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Good, but your chain was several meters long. I estimate that the round-trip distance was more like 14 meters, so the speed would be at least double and perhaps that reported.
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582.35cm/sec at tension 6.00N
660cm/sec at tension 10.9N
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Find the mass of the chain:
Using the equation
•c = sqrt(F_tension / (m / L) )
where c is the speed of propagation, F_tension the tension force, m the mass and L the length of the chain:
•for each trial you now know c, F_tension and L, so for each trial you can solve for m.
Give your solutions, including an explanation of how you solved for m and a complete calculation for one of your trials:
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If propagation velocity was 15 m/s, I get a mass per unit length of about 10 grams per meter.
I think it's a little higher than that, since each paperclip is a bit over a gram and its length is less than 10 cm, but 10 grams/meter is in the right ballpark.
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You also observed the second harmonic at one or more tensions. For each tension at which you made this observation, what was the frequency of the first harmonic (in cycles / minute) and the second?
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At close to the same tension, a two node pulse had a frequency of 0.96 cycles/second and a three node wave had a frequency of 2.05 cylces/second.
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Good, but check my notes and see if you can revise some of your results accordingly. We'll be discussing this in class.
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