Having trouble understanding the relationship between these expressions

  • Thread starter coldjeanz
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  • #1
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I understand everything up until:

√MgL/m = L/t

Why is that equal to L/t?

I also don't understand this either:

U8IiJ.png


This is the problem itself however I don't feel it's that important since I am trying to figure out how they are arriving at their conclusions.

An astronaut on a small planet wishes to measure the local value of the free-fall acceleration by timing pulses traveling down a wire that has an object of large mass suspended from it. Assume a wire has a mass of 4.10 g and a length of 1.60 m and that a 3.00 kg object is suspended from it. A pulse requires 43.6 ms to traverse the length of the wire. Calculate g of planet from these data. (You may ignore the mass of the wire when calculating the tension in it.)
 

Answers and Replies

  • #2
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They are not saying that the last equality follows from something about [itex]\sqrt{MgL/m}[/itex], but rather that the wave speed [itex]v[/itex] is equal to both [itex]\sqrt{T/u}[/itex] AND [itex]L / t[/itex] (since speed is just distance divided by time).

The second section is just an algebraic rearrangement of the second equation in the first section.
 
  • #3
22
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Yeah okay that went through my head for a split second but I couldn't find that in my book so I didn't know where they pulled it from. Thank you!
 

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