Solve Physics Lab: Hooke's Law & Simple Harmonic Motion

  • Thread starter kashmirekat
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In summary, Christina Eh says that in order to convert grams to kg, you need to multiply by g and then convert the results to Newtons. She provides a table of masses and their corresponding Newtons. She also says that if you want to find the equation that relates the change in y-coordinate to the change in x-coordinate, you should try to find it.
  • #1
I had a physics lab for Hooke's Law and Simple (ha!) Harmonic Motion. I am not pleased with my grade, but my professor did not indicate what I did wrong and I am banging my head against the wall trying to figure it out (btw, it's not helpful). So if any of you all can take a gander at it and tell me hey! You did this wrong! That would be great. Thank you.

First Section: Spring Elongation

weight(grams)*9.18m/s^2 | Scale reading (mm)
m1g = 114.26                  | y1 4.7
m2g = 124.26                  | y2 = 5.2
m3g = 134.26                  | y3 = 5.9
m4g = 144.26                  | y4 = 6.3
m5g = 154.26                  | y5 = 6.7
m6g = 164.26                  | y6 = 7.3
m7g = 174.26                  | y7 = 7.8
m8g = 184.26                  | y8 = 8.2

Okay, the first thing I did was convert the grams to kg and multiply by g (9.8m/s^2).
For these values I got:
mg1=1.12N, mg2=1.22N, mg3=1.3157N, mg4=1.414N, mg5=1.512N, mg6=1.61N, mg7=1.708N, mg8=1.806N

Then I converted mm to m...0047, .0052, .0059, .0063, etc.

It's asking for k and in parentheses next to it has slope of graph. [ie k(slope of graph)] So k=slope of graph. Slope equals rise/run, (delta y)/(delta x). So I choose two points 1 & 4.
(1.12N - 1.414N) / (.0047m - .0063m) = -0.294N / -.0016m = 183.75 N/m.

His answer is k = 20. Where did go wrong to get a value so far from the answer?
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  • #2
uh I'm not math head but I would recon that you didn't label something correct on your graph...20 what? perhaps he wants you to label everything, and I thought weight was in Newtons therefor you don't have to say weight * 9.8 m/s² because 9.8 m/s² is already included in the Newton measurement...yes? no?
  • #3
According to your table:
weight(grams)*9.18m/s^2 | Scale reading (mm)
m1g = 114.26 | y1 4.7
m2g = 124.26 | y2 = 5.2
m3g = 134.26 | y3 = 5.9
m4g = 144.26 | y4 = 6.3
m5g = 154.26 | y5 = 6.7
m6g = 164.26 | y6 = 7.3
m7g = 174.26 | y7 = 7.8
m8g = 184.26 | y8 = 8.2

"m1g" is ALREADY multiplied by 9.81 (not "9.18") m/s^2. In order to convert that to Newtons, you only need to multiply by 0.001 to convert grams to kg. If that is correct, then m1g is 0.11426 Newtons and m8g is 0.18426. For those two values, y1= 0.0047 m and y8= 0.0082 m so the "slope" of your graph is (0.18426-0.11426)/(0.0082- 0.0047)= 20.
  • #4
Coughlan, k is suppose to be 20N/m.

HallsofIvy, oh if that were only true, but those values did not include the multiplication of g. My masses were 114.26, 124.26, but were not already multiplied by g. Sorry, I should have written g next to them, but that could also be mistaken for grams (which they are in). I would convert to kg before multiplying with 9.8 anyhow, right? The m1g tells us to multiply m by g, ergo .11426kg*9.8m/s^2=1.12N.

Just in case I didn't clear it up, 9.8m/s^2 is not included in the measurement of 114.26, 124.26, etc...those are all masses in grams.

Any more suggestions?
Thank you.
  • #5
Eh sorry about that...I'm fresh out of ideas...

1. What is Hooke's Law?

Hooke's Law is a basic principle in physics that states the force required to stretch or compress a spring is directly proportional to the distance the spring is stretched or compressed.

2. How is Hooke's Law used in a physics lab?

In a physics lab, Hooke's Law is used to study the relationship between the force applied to a spring and the resulting displacement of the spring. This can be done by measuring the force applied to the spring at different lengths and plotting a graph to observe the linear relationship.

3. What is simple harmonic motion?

Simple harmonic motion is a type of periodic motion in which the restoring force is directly proportional to the displacement from equilibrium. This results in a repetitive back and forth motion around the equilibrium point.

4. How is simple harmonic motion related to Hooke's Law?

Simple harmonic motion is related to Hooke's Law because it occurs when an object is subject to a restoring force that is proportional to its displacement. This is the case for a spring obeying Hooke's Law.

5. How can Hooke's Law and simple harmonic motion be applied in real-world situations?

Hooke's Law and simple harmonic motion have many applications in the real world, such as in the design of springs in mechanical systems, pendulum clocks, and musical instruments. They are also used in studying the behavior of materials under stress and in understanding the motion of celestial bodies in orbit.

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