Question

Help!!!

Answer 1

sup

Answer 2

Its Activity 3 is the only thing I need help with

Answer 3

The experimental period is the average of the 3 periods you found above (110s, 100s, 69s) since these were the periods you got in the experiment. This is for the longer pendulum. You need to repeat the experiment for the shorter pendulum.

Answer 4

I have them too just a second

Answer 5

ok

Answer 6

I haven't changed this one into seconds yet

Answer 7

this is for the shorter pendulum right?

Answer 8

yes

Answer 9

ok so the average of those times will be the average period for the shorter pendulum

Answer 10

and that will be an experimental period

the theoretical period will come from the formula given

Answer 11

ok

Answer 12

you will need to solve for t first

afterwards, you would plug in the given length of the pendulum into \(\Large \ell\) then evaluate

Answer 13

ok

Answer 14

If you're curious (or stuck) on how to solve for t, here's one way to do it:

\[\Large \ell = \frac{980t^2}{4\pi^2}\]


\[\Large \ell*4\pi^2 = 980t^2\]


\[\Large 980t^2 = \ell*4\pi^2\]


\[\Large t^2 = \frac{\ell*4\pi^2}{980}\]


\[\Large t = \sqrt{\frac{\ell*4\pi^2}{980}}\]

Answer 15

you're welcome

Answer 16

ok so im still a little confused sorry

Answer 17

where at?

Answer 18

this last part still the activity 3

Answer 19

im confused on how to solve the equation and how to average the times

Answer 20

to average the 3 times, you add up the times, then divide that sum by 3

Answer 21

ok

Answer 22

like all of them for each expirement

Answer 23

for 1 coin, you average the 3 trials to get the average time for 1 coin (ignore 2 coins or 3 coins)

then for 2 coins, you find the average (ignore 1 coin or 3 coins)

etc etc

Answer 24

ok thanks but how do i do the next step

Answer 25

to find the theoretical period?

Answer 26

yeah

Answer 27

what is the length of the longer pendulum?

Answer 28

25.6cm

Answer 29

10 inches

Answer 30

For the longer pendulum, \(\Large \ell = 25.6\)

so plug it in and evaluate to find t

Answer 31

ok and then do the same thing with the smaller pendulum?

Answer 32

yes

Answer 33

ok thank you again lol

Answer 34

you're welcome

Answer 35

ok for the first on would the answer be 1.01552 if so how do i like down size this? And is the second one 0.715268

Answer 36

or is it different cause i just did it online and gat 501.376 for the second one

Answer 37

1010.65 for the first one

Answer 38

for \(\Large \ell = 25.6 \ \text{cm}\) I'm getting \(\Large t = 1.015516100 \ \text{seconds}\)

for \(\Large \ell = 10 \ \text{cm}\) I'm getting \(\Large t = 0.6346975625 \ \text{seconds}\)

Answer 39

the second one isnt 10cm its 12.7cm

Answer 40

oh ok, one sec

Answer 41

alright

Answer 42

for \(\Large \ell = 25.6 \ \text{cm}\) I'm getting \(\Large t = 1.015516100 \ \text{seconds}\)

for \(\Large \ell = 12.7 \ \text{cm}\) I'm getting \(\Large t = 0.7152678272 \ \text{seconds}\)

both times are approximate

Answer 43

so in other words they are pretty close and do i just put the whole thing in my graph

Answer 44

yeah the experimental periods should be close to the theoretical periods (but will never line up exactly)

Answer 45

ok thanks so much you helped a lot it makes more sense.

Answer 46

I'm glad I was able to help out.

Answer 47

you are like a lifesaver lol

Answer 48

lol glad I could be