Physics lab problem. Ambiguous directions. What are they asking?

[tony873004]

We took timings in lab of a pendulum.

Activity 1: release the pendulum from 10 degrees. Time how long it takes to do 10 oscillations.

Activity 3, shorten the length of the string to 2/3 original, then to 1/3 original and time 10 oscillations from a 10 degree release point.

Conclusion:
Using your data from activities 1 and 3, compute ln(T/1s), ln(L/1cm), for each length and their worst-case uncertainties. Describe your reasoning in detail and show all calculations.

Plot a graph of ln(T/(1s)) vs ln(L/(1cm)) and determine its slope. Discuss how you use uncertainty in the data to determine worst-case uncertainty in the slope. Is your slope consistent with the expected value of n=1/2? Explain your reasoning carefully.



What does this mean?

It never tells us what the formulas ln(T/(1s)) and ln(L/(1cm)) mean. So how can I expect that the slopes of their graphs will equal 1/2? How can I draw any conclusions when I don't know why I'm applying that formula?

Also, what's the point of dividing T by 1? It's just going to give me T? And why do they want me to divide L by 1? It's just going to give me L.

 

[dextercioby]

Division is being performed in order to graph numerical functions,i.e.mathematical objects,which should have no physical dimansion.

I don't know why they asked you to logarithmate,the formula is not an exponential,not even in the general case.

Daniel.

 

[tony873004]

*** bump! ***
I still don't get it.

 

[OlderDan]

We took timings in lab of a pendulum.

Activity 1: release the pendulum from 10 degrees. Time how long it takes to do 10 oscillations.

Activity 3, shorten the length of the string to 2/3 original, then to 1/3 original and time 10 oscillations from a 10 degree release point.

Conclusion:
Using your data from activities 1 and 3, compute ln(T/1s), ln(L/1cm), for each length and their worst-case uncertainties. Describe your reasoning in detail and show all calculations.

Plot a graph of ln(T/(1s)) vs ln(L/(1cm)) and determine its slope. Discuss how you use uncertainty in the data to determine worst-case uncertainty in the slope. Is your slope consistent with the expected value of n=1/2? Explain your reasoning carefully.



What does this mean?

It never tells us what the formulas ln(T/(1s)) and ln(L/(1cm)) mean. So how can I expect that the slopes of their graphs will equal 1/2? How can I draw any conclusions when I don't know why I'm applying that formula?

Also, what's the point of dividing T by 1? It's just going to give me T? And why do they want me to divide L by 1? It's just going to give me L.

*** bump! ***
I still don't get it.

dextercioby gave you the reason for the division. You are not dividing by 1 in either case. You are dividing the time by a time (1s) and the length by a length (1cm) in order to achieve dimensionless quantities. If you did not do that, you would wind up taking logarithms of quantities that involved dimensions. The arguments of functions like logs and exponentials and trig functions must always be dimensionless.

Now why plot logarithms? Maybe you have to be old to know this. Before everyone had a whiz-bang electronic calculator in their pocket that could do all sorts of regression analyses, graphs were often drawn on log-log paper to figure out the power in a relationship, and on semi-log paper to figure out the multiplying factor in an exponential relationship.

Suppose you suspect that the period of oscillation is proportional to some power of the length of a pendulum. How would you write such a relationship? You could write

$$T = aL^n$$

where a would have to have dimensions of $$time/length^n$$ If you divide the equation by 1 sec you get

$$\frac{T}{sec} = a/sec L^n$$

If you multiply the right hand side by

$$\frac{cm^n}{cm^n}$$

you get

$$\left[\frac{T}{sec}\right] = \left[a\frac{cm^n}{sec}\right] \left[\frac{L}{1cm}\right]^n$$

You now have a dimensionless equation. Take the log of both sides

$$ln\left[\frac{T}{sec}\right] = ln\left[a\frac{cm^n}{sec}\right] + n \ ln \left[\frac{L}{1cm}\right]$$

You now have a linear equation whose slope is the power and whose intercept is the log of the dimensionless constant. You can get both of those things from your log-log graph.

 

[tony873004]

Thanks, Dan. I think I got it now.

Another question: If I'm asked to plot a graph of A vs. B, does it matter which goes on the x axis. Would it be A, or does it matter?