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Speed of Sound Lab (using slope to find speed of sound)

  • Thread starter goman519
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  • #1
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1. Homework Statement

If you had data from a lot of different frequencies, how could you use a slope to find the speed of sound? Explain in detail.

Given/Known: So basically we did a lab where we used 3 different tuning forks and hit them over a tube filled with water. We recorded where we heard each node and calculated what half lambda and lamda were in order to find the speed of sound.


2. Homework Equations

Speed of Sound = frequency * wavelength (lambda)
or
Speed of Sound = 331 + 0.61(T)
T = temperature in celsius (which was 20.2 in our class room)


3. The Attempt at a Solution

Tried taking the slope of my 3 frequencies (523.2, 1024, 2048 (Hz)) and 3 wavelengths (634.67, 316.00, 152.00 (mm) respectively) in excel. First converted the wavelengths to meters. The slope I got was -2.91. Couldn't figure out any correlation and frankly don't think that's the way to approach the problem.


Solution or any kind of tip very much appreciated. Thanks guys.
 

Answers and Replies

  • #2
fzero
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The expected relationship between the frequency and wavelength is one of inverse proportion

[tex] \lambda = \frac{v_s}{f}. [/tex]

The speed of sound [tex]v_s[/tex] could be computed from the slope of a fit to the data where you plot [tex]\lambda[/tex] vs. [tex]1/f[/tex], instead of [tex]\lambda[/tex] vs. [tex]f[/tex].
 
  • #3
2
0
The expected relationship between the frequency and wavelength is one of inverse proportion

[tex] \lambda = \frac{v_s}{f}. [/tex]

The speed of sound [tex]v_s[/tex] could be computed from the slope of a fit to the data where you plot [tex]\lambda[/tex] vs. [tex]1/f[/tex], instead of [tex]\lambda[/tex] vs. [tex]f[/tex].
Ok cool. I see how that works and when I got the slope I had a very small percent error so thanks.

Only thing is, my teacher emailed me about the question and told me to differentiate the equation for velocity of sound. To look at dy/dx. I'm not really sure how to do that correctly.

Wouldn't differentiating V=lambda F give you 1 = lambda + F ? Any idea what she wants me to do?
 

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