How Is the Spring Constant Calculated from Oscillation and Mass Changes?

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Homework Help Overview

The problem involves calculating the spring constant from the behavior of a spring when a mass is added and the spring is set into oscillation. The context includes the mass of the object, the additional mass, the stretch of the spring, and the amplitude of oscillation.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the application of Hooke's law and question the linearity of the spring's behavior under the given conditions. There is an attempt to calculate the spring constant using the additional mass and stretch, with some uncertainty about the correctness of the approach.

Discussion Status

Some participants have provided feedback on the calculations and emphasized the importance of units. There is an ongoing exploration of the assumptions regarding the linearity of the spring's response to force.

Contextual Notes

Participants note that not all information provided in the problem is necessary for the solution, indicating potential constraints in the approach to the problem.

ConstableZiM
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Homework Statement



A 100g object is suspended from a spring. When 40g are added, the spring stretches an additional 5.0cm. With the total mass of 140g, the spring is set into vertical oscillations with an amplitude of 10 cm. (a) What is the force constant of the spring?



Homework Equations


[tex]\sum[/tex] F = 0 = -kx + mg



The Attempt at a Solution



From what I know, the force required to stretch a spring is not linear, so I am guessing that plugging just the 40 grams and the 5 cm into the equation won't work... What I got when I did that was k = 0.040kg * 9.8 / 0.05m = 7.84

Im guessing this is wrong?

Help would be appreciated.
 
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The usual assumption about springs is that the force exerted by them is indeed linear (as in #2 above). All your solution needs is units. (Not all information given in the problem is necessary to solve it.)
 
ConstableZiM said:
From what I know, the force required to stretch a spring is not linear, so I am guessing that plugging just the 40 grams and the 5 cm into the equation won't work...
Even for a real-world spring, as long as you don't stretch the spring so much that it plastically deforms or collapses onto itself, the relationship between force and stretching distance can be approximated quite well using Hooke's law, F = kx (some authors define it as F = -kx). Here F is directly proportional to x, so it is linear.
What I got when I did that was k = 0.040kg * 9.8 / 0.05m = 7.84
'Looks okay to me. :approve:

[Edit: SEngstrom beat me to the answer. And as SEngstrom says, don't forget your units. :smile:]
 
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