Pendulum lab
inital height of spring:11.5cm
m=2000g d=8cm
m=3000g d=6cm
m=2500g d=7cm
I need to calculate C....I'm not exactly sure how to do that...
Maybe there's a mindreader cruising by who can shed light on what the symbol C means. Barring that, perhaps you could provide more information - enough in fact for someone not familiar with your pendulem lab or your experiment - to understand the question.
When posting a problem, you've got to ask yourself a question (and it's not "do I feel lucky?"):
"Have I posted enough information so that someone else with no prior familiarity with the problem will understand it well enough to provide an answer?"
I've done some channeling and have seen that you need to find the spring constant.
This is probably best done graphically, but I'm too lazy to use Excel right now. The other way is to calculate a k for each data point and then get the average and standard error.
The equation you are trying to fit is
$\displaystyle F = k(x - x_0)$
Since you have a hanging mass F = w = mg. So
$\displaystyle mg = k(x - x_0)$
Now, $\displaystyle x - x_0$ is the amount the spring has stretched from its equilibrium length. Given what the data look like (more clairvoyance) I would guess that the 11.5 cm is the end of the spring with no weight on it and the d values are the position the mass hangs at. Thus the spring will have stretched a distance of $\displaystyle 11.5~cm - d$, so this is the $\displaystyle x - x_0$.
So finally:
$\displaystyle mg = k(0.115 - d)$
(To keep the units consistent I am using d in meters and m in kilograms.)
I get k = 560.000, 534.545, 544.444 N/m (respectively). So the experimental value for k will be $\displaystyle k = 550 \pm 7.4 ~ N/m$.
-Dan
Not really maths but ..... I think there's scope for valuable comment on significant figures and rounding in this sort of problem .....
The least significant data has 1 sig fig .... you can't give an answer more accurate than the least accurate data ... so an answer correct to 1 sig fig would be the technically correct answer to give .....
Personally, I'd be inclined to state the answer as $\displaystyle 550 \pm 10$ N/m ......
All comment welcome.
I never know how many digits to keep when I'm doing a standard error problem, so I just keep the number of sig digs the problem has.
We actually probably have two sig digs in the "d" measurement since any reasonable meter stick is measured out in mm. Likewise I would imagine the mass measurements are correct to 4 places.
However, the student did not report the numbers that way. So I must have done a bit more ESP than I mentioned earlier.
-Dan