Data Sheet

 PLEASE VERIFY THESE NUMBERS AND COMPUTATIONS; CHECK THEM FOR ERRORS .

Room temperature = 22.5 (using LC = 1 and error = LC/4 = 0.25 degrees,)
Frequency 1 = __512__HZ_________________

 

Error = LC/4 = ¼ = 0.25 =0.30 =  ΔLinst

 

 

Position of water level

 

L1

 

 

17.0


Not Applicable

16.5

17.0

16.5

Average L1= 16.75

(L1max – L1min)/4 = (17.0 - 16.5)/4 = 0.125

ΔL inst = 0.25 = 0.30

ΔL (larger of previous two) =  0.30

 

L3

50.0

 

Average L3 - L1 =
=  50.50 - 16.75 = 33.75

50.5

50.5

51.0

Average L3= 50.50

(L3max – L3min)/4 = (51.0 - 50.0)/4 = 0.25

ΔL inst = 0.25 = 0.30

ΔL (larger of previous two) =  0.30

 

L5

 

 

 

83.5

 

Average L5 – L3 =

 = 83.50 - 50.50 =33.00
=

84.0

83.0

83.5

Average L5 = 83.50

(L5max – L5min)/4 = (84.0 - 83.0)/4 = 0.25

ΔL inst = 0.25 = 0.30

ΔL (larger of previous two) =  0.30


Average   =  (33.75 + 33.00)/2 = 33.375

 

 

 

 

Average   = 2*(33.375) = 66.75

 

Average  = (512)*(0.6675) = 341.76  m/s

 
accepted v = (331.00 + 0.6*T) = (331.00 + 0.6*22.5) = (331.000 +  14.5) = 345.5 m/s .

|average v – accepted v | =
Compare  |average v – accepted v | with the overall error, which gives the range, as  discussed in class. Does the accepted value fall within the range centered at the average value?
Answers to this are in the class notes; I can give you hints now with gaps you can fill later on using other resources including the internet.
The  overall error or uncertainty in the speed of the sound is the following excluding the following elements: the error in the accepted speed of sound value via errors in the temperature using the Least Count (LC) indicated , 1 degree Celsius    AND errors in the frequency used , 512 HZ. A section on the missing elements will be added to the lab report, so please keep posted by checking back to this internet  page often . Add this  to your Firefox bookmark..
(Overall uncertainty)2 = f2*4ΔL2 excluding above mentioned elements. See class notes where we used Math 3 concepts.  Thus, overall uncertainty   =   f*2*(0.0030) = (512)*0.0060.  =   3.072
Now, we subtract to get the left hand side of the equation: 

|341.76  m/s - 345.5 m/s | = 3.74 m/s.

Since, the absolute value of the difference is larger, this begs the question of whether we should include errors flowing from the frequency and from the  temperature . The percent error computed in the traditional way is indeed about 1 %., so this really begs the question !  What are the other errors !?