Y-intercept of a lambda square VS tension of standing wave

  • #1

Main Question or Discussion Point

Hi all!

I am doing an experiment where we create a standing wave by attaching a string to a hanging mass at one end and to a string vibrator at the other (the string passes through a pulley). When plotting the graph, the slope is inevitably 1/(u*f^2) where u is the linear density and f the frequency.

In this context, what would be the Y-intercept of the graph. Theoretically, there shouldn't be any, I know, but what does it represents in the reality of this experiment?

My guess is that it represents the friction by the air. There wouldn't be any y-intercept if this experiment was to be done in vaccum.
 

Answers and Replies

  • #2
anorlunda
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That sounds like a good question. Can you post a copy of the graph? You can use the UPLOAD button down there next to POST REPLY and PREVIEW to take a JPG or BMP or PNG file from your computer and put it in a post.
 
  • #3
That sounds like a good question. Can you post a copy of the graph? You can use the UPLOAD button down there next to POST REPLY and PREVIEW to take a JPG or BMP or PNG file from your computer and put it in a post.
 

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