- #1

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The graph makes sense, the higher the mass the less time it takes to do an oscillation. However, you may not have correctly calculated the slope of the graph.

- #1

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- #2

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Y: Dependent variable.

So, did you change mass to find different periods or did you change the periods of find different masses?

- #3

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i added mass to see the variation in oscillation periods

- #4

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Remember:

X is what you change

Y is what you measure

- #5

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k = (4pi²)/slope

but the slope of the graph has to be t² and kilograms correct? i have the values in t² and the mass in grams

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- #7

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My assignment says i have to do 2 tables where one measures force against displacement and than this one above

for force vs displacement i get k as 34.6 while when i do it with the formula i got above i get k as-8.77, is this normal?

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- #9

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k = (4pi²)/slope

which gave me -8.77 N/m

- #10

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If you plot displacement against weight, yes. If you plot it against mass, the k = m * 9.8

- #11

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y axis - Force (N)

x axis - displacement (m)

Where i got the slope of 34.6 which the instructions say its the k constant

the second graph

y axis - t²

x axis - mass (kg)

where i got slope of -4.5

which i imputed here

k = (4pi²)/slope

which gave me -8.77 N/m

Now i was supposed to calculate the difference between the two

with these values i get a difference of -125% which i think is a huge difference

- #12

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I'm not quite sure how you ended up with a negative slope, there should be a strong positive correlation between mass and period squared. Besides, the spring constant can't be negative. Can you list some data points?

- #13

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First graph

x f(N)

0.98

1.96

2.94

3.92

y displacement (m)

0.03

0.06

0.088

0.115

Second graph

x mass(kg)

.3

.4

.5

.6

y t²(s)

2.72

2.13

1.66

1.41

i used this website to help input data to get info

http://www.shodor.com/unchem/math/lls/leastsq.html

x f(N)

0.98

1.96

2.94

3.92

y displacement (m)

0.03

0.06

0.088

0.115

Second graph

x mass(kg)

.3

.4

.5

.6

y t²(s)

2.72

2.13

1.66

1.41

i used this website to help input data to get info

http://www.shodor.com/unchem/math/lls/leastsq.html

Last edited:

- #14

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- #15

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we had to count how many secs it took to do 25 oscillations

.3kg - took 15.19 sec

.4kg - 17.16 sec

.5kg - 19.38 sec

.6kg- 21.09 sec

than i took the average period T (sec)

divided 25 (oscillations) by the secs it took and squared them

25/15.19 = 1.65 = t² = 2.72

25/17.16 = 1.46 = t² = 2.13

25/19.38 = 1.29 = t² = 1.66

25/21.09 = 1.19 = t² = 1.41

and that's how i got my values.

in a way the graph makes sense, the higher the mass the less time it takes to do an oscillation but i know o have something wrong on this graph

.3kg - took 15.19 sec

.4kg - 17.16 sec

.5kg - 19.38 sec

.6kg- 21.09 sec

than i took the average period T (sec)

divided 25 (oscillations) by the secs it took and squared them

25/15.19 = 1.65 = t² = 2.72

25/17.16 = 1.46 = t² = 2.13

25/19.38 = 1.29 = t² = 1.66

25/21.09 = 1.19 = t² = 1.41

and that's how i got my values.

in a way the graph makes sense, the higher the mass the less time it takes to do an oscillation but i know o have something wrong on this graph

Last edited:

- #16

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- #17

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So i am supposed to divide these values by one and than square them, is that what i did wrong?

25/15.19 = 1.65 = t² = 2.72

25/17.16 = 1.46 = t² = 2.13

25/19.38 = 1.29 = t² = 1.66

25/21.09 = 1.19 = t² = 1.41

25/15.19 = 1.65 = t² = 2.72

25/17.16 = 1.46 = t² = 2.13

25/19.38 = 1.29 = t² = 1.66

25/21.09 = 1.19 = t² = 1.41

Last edited:

- #18

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The frequency of the Earth is 1/24 hours = 0.04 per hour (meaning that it makes 1/24 of a rotation per hour) and its period is 24 hours (meaning that it takes 24 hours to make one revolution.

- #19

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- #20

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Yes, so long as the spring constant is N/m and the displacement is m.

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