Simple Harmonic Motion, spring periods

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SUMMARY

The discussion focuses on the mathematical relationship between the period of a spring and the mass attached to it, specifically using the formula T = 2π√(m/k). Participants confirm that the slope of the graph plotting T² against mass (m) equals (4π²)/k. By determining the slope from the best fit line in Excel, which is 1.5129, users can calculate the spring constant (k) using the rearranged formula k = (4π²)/slope.

PREREQUISITES
  • Understanding of Simple Harmonic Motion principles
  • Familiarity with the formula T = 2π√(m/k)
  • Proficiency in using Excel for data analysis and graphing
  • Basic knowledge of linear relationships and slope calculations
NEXT STEPS
  • Calculate the spring constant (k) using k = (4π²)/slope
  • Explore advanced data analysis techniques in Excel for curve fitting
  • Investigate the effects of varying mass on the period of different spring types
  • Learn about the implications of spring constants in real-world applications
USEFUL FOR

Students studying physics, educators teaching mechanics, and anyone interested in experimental physics and data analysis using Excel.

Navras
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Hi guys! I'd really like help with this as I'm stuck.

Homework Statement



Investigate the mathmatical relationship between period of a spring and mass.
Finding the spring constant (k) from measuring periods with a spring and different hanging masses.

Homework Equations



T = 2pi SQRT(m/k)

or rearranged

T^2 = ((4pi^2)/k) m

The Attempt at a Solution



Used Excel to plot T^2 vs. hanging mass. T^2 in seconds squared (y-axis) and hanging mass in kg (x-axis). Used Excel to find the best fit line.

Since I have plotted T^2 vs. m, does the slope equal (4pi^2)/k ?

I don't know where to go from here to get k though.



---

I've attached the graph I did also
 

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Welcome to PF!

Navras said:
Since I have plotted T^2 vs. m, does the slope equal (4pi^2)/k ?

Yeah. The relationship is linear in m. All the stuff (the coefficient) that multiplies m is therefore the slope, just like in any other linear relationship.

Navras said:
I don't know where to go from here to get k though.

What do you mean? You have the slope, which is 4pi^2 /k, from the best fit line. Therefore, you have k.
 
cepheid said:
What do you mean? You have the slope, which is 4pi^2 /k, from the best fit line. Therefore, you have k.

So, the slope is 1.5129, does that mean k is 1.5129 or

is it (4pi^2)/1.5129 = k?

thanks :)
 
Navras said:
So, the slope is 1.5129, does that mean k is 1.5129 or

is it (4pi^2)/1.5129 = k?

thanks :)

I'm not telling you the answer to that, you should be able to arrive at it using the information you have (and it should be really clear) . Consider these two statements:

1. The slope is 4pi^2/k

2. The slope is 1.5129 (according to the Excel best fit curve).

What do you conclude?
 

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