Area under a frequency vs wavelength graph?

In summary: V sound = 223.03 meters/second. Can you please help me understand why this is inaccurate?In summary, the student attempted to find the speed of sound using the formula given, but the data was inaccurate because only one tuning fork was used.
  • #1
JosephTraverso2
8
0

Homework Statement


Hi Everyone,

So I'm doing writing up my weekly physics lab report and I had an idea to better present my findings. I have a chart displaying the frequencies of numerous tuning forks as well as their experimentally determined wavelengths and I have to find the speed of sound within our classroom. If I plot a frequency vs. time graph, can I obtain the average speed of sound by finding the area under the graph or am I mistaken? Thank you for your time.

Homework Equations


V = F * ƛ
 
Physics news on Phys.org
  • #2
This may be a good time to think about the dimensions or units of measurement of the quantities in your equations (i.e. dimensional analysis).
For example, in your equation V = F * λ. The dimensions of λ are meters (feet, miles, etc.). The dimensions of F are 1/seconds (hertz, "per hour", etc). When you multiply them together you get meters/seconds, which are the dimensions of velocity. So this all works out ok, you expected a velocity out of that equation.
Now look at the area under a frequency versus time curve. For dimensional analysis, finding the area under a curve will have the units of the two axes multiplied together. For example, if you graph the velocity of a car versus time the area would be the distance traveled; eg. miles/hour * hours = miles.
So for your proposal of the area under the graph of frequency versus time what would the resulting dimensions be? Are those the dimensions of the answer you expect (or want)?
 
  • Like
Likes berkeman
  • #3
JosephTraverso2 said:
If I plot a frequency vs. time graph
My mistake I meant frequency vs. Wavelength graph.
 
  • #4
I went ahead and did a definite integral on my calculator and got V sound = 223.03 meters/second. Why is this so inaccurate?
 
  • #5
Yes, those units are correct. This doesn't necessary mean your equation is right, of course. This method is good for spotting poorly constructed equations, not so good at proving if the equation is correct. For example, it is useless in finding if you should or shouldn't multiply something by π, which has the same dimensions of 1, 2, e, etc.
 
  • #6
DaveE said:
Yes, those units are correct. This doesn't necessary mean your equation is right, of course. This method is good for spotting poorly constructed equations, not so good at proving if the equation is correct. For example, it is useless in finding if you should or shouldn't multiply something by π, which has the same dimensions of 1, 2, e, etc.
Ok great, thank you for answering all my questions. Have a nice day!
 
  • #7
I am assuming that you have a pair of measurements for each tuning fork (F, λ), so your graph has a few distinct points, one for each tuning fork. What you are trying to do is find the speed of sound using the formula you were given V = F * λ? What if you only had one tuning fork to measure, could you do it with those data?
 
  • #8
DaveE said:
What you are trying to do is find the speed of sound using the formula you were given V = F * λ?
I'm sorry can you retype this? I cannot understand what you are trying to say.
 
  • #9
I was just trying to understand the question you were asked to solve and what your data and graph are (and why you have a graph of the data).
 
  • #10
DaveE said:
I was just trying to understand the question you were asked to solve and what your data and graph are (and why you have a graph of the data).
the question I have is "What is the speed of sound within our classroom" and the data I got was this (Frequency in hertz on the Y axis and Wavelength on the X-axis in meters)

upload_2019-3-4_16-19-16.png
 

Attachments

  • upload_2019-3-4_16-19-16.png
    upload_2019-3-4_16-19-16.png
    8 KB · Views: 768

1. What is the purpose of an area under a frequency vs wavelength graph?

The area under a frequency vs wavelength graph represents the total amount of energy emitted by a source, such as a light bulb, over a specific range of wavelengths. It is used to understand the intensity and distribution of energy emitted by a source, and can also provide information about the composition and temperature of the source.

2. How is the area under a frequency vs wavelength graph calculated?

The area under the graph is calculated by finding the integral of the function that represents the relationship between frequency and wavelength. This can be done using mathematical techniques such as integration by parts or the trapezoidal rule. Alternatively, software such as Microsoft Excel can be used to calculate the area under the graph.

3. What does the shape of the area under a frequency vs wavelength graph indicate?

The shape of the area under the graph can indicate the type of emission or absorption spectrum of the source. For example, a continuous spectrum will have a smooth and broad area under the graph, while a line spectrum will have distinct peaks and valleys. The shape can also reveal information about the temperature and composition of the source.

4. How is the area under a frequency vs wavelength graph used in scientific research?

The area under the graph is used in various fields of science, such as astronomy, chemistry, and physics. It can help scientists understand the properties of different types of radiation, such as visible light, radio waves, and X-rays. It is also used to analyze the emissions from different sources, such as stars, galaxies, and chemical compounds.

5. Are there any limitations to using the area under a frequency vs wavelength graph?

One limitation is that the accuracy of the graph depends on the precision of the data used to create it. Inaccuracies in measurements or assumptions about the shape of the graph can affect the reliability of the calculated area. Additionally, the area under the graph may not accurately represent the total amount of energy emitted by a source if the emission is not within the measured range of frequencies and wavelengths.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
412
  • Introductory Physics Homework Help
Replies
15
Views
7K
  • Introductory Physics Homework Help
Replies
6
Views
951
  • Introductory Physics Homework Help
Replies
10
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
7K
  • Introductory Physics Homework Help
Replies
1
Views
912
  • Introductory Physics Homework Help
Replies
7
Views
979
Back
Top