Math Problem Statement

Opgave 4: En fjeder belastes med forskellige lodder i fysiklokalet. I tabel 1 ses sammenhørende målte værdier for massen af loddet og fjederens længde. Massen måles i gram, og længden af fjederen i centimeter. a) Bestem en lineær model L(x) = αx + b, der beskriver længden af fjederen som funktion af massen af loddet, og begrund hvor godt denne model passer ud fra målingerne. b) Forklar betydningen af konstanten α i denne situation.

Solution

In this problem, we are given data about a spring being stretched by different weights, and the task is to find a linear model that describes the length of the spring as a function of the mass of the weight. Let's go through the parts step-by-step:

Given Data:

Mass mm (grams)4060100200500
Length LL (cm)15.519.824.131.751.9

Task (a):

We are asked to determine a linear model of the form:

L(m)=αm+βL(m) = \alpha \cdot m + \beta

Where LL is the length of the spring (in cm), mm is the mass of the weight (in grams), and α\alpha and β\beta are constants to be determined from the data.

Approach:

  1. We need to use linear regression or the least squares method to fit the best line to the given points.
  2. This involves solving for α\alpha (slope) and β\beta (intercept).

Task (b):

After finding the linear equation, we are asked to explain the meaning of α\alpha, which represents how much the length of the spring increases per unit mass, i.e., the "spring constant" in a simplified form.

I will now calculate the linear model and return with the result.The linear model for the length L(m)L(m) of the spring as a function of the mass mm is:

L(m)=0.0752m+15.065L(m) = 0.0752 \cdot m + 15.065

Explanation:

  • The slope α=0.0752\alpha = 0.0752 indicates that for every additional gram of mass added to the spring, the length of the spring increases by approximately 0.0752 cm.
  • The intercept β=15.065\beta = 15.065 represents the initial length of the spring when no mass is applied (assuming m=0m = 0, though not directly measured in the table).
  • The coefficient of determination R2=0.985R^2 = 0.985, which is very close to 1, suggests that this linear model fits the data extremely well.

Task (b):

In this context, the constant α=0.0752\alpha = 0.0752 can be interpreted as the rate at which the spring stretches per unit of mass. It is related to the stiffness of the spring: a smaller value of α\alpha would indicate a stiffer spring, and a larger value would indicate a more elastic spring.

Would you like any more details or further clarification on these points?

Here are five additional questions to deepen your understanding:

  1. How would the model change if we used a nonlinear fit, such as a quadratic?
  2. What would the physical interpretation of the intercept β\beta be in real-world terms?
  3. Can you predict the length of the spring for a mass of 300 grams using the model?
  4. How would the model change if you included more data points for smaller masses?
  5. Could you explore how changing the spring material might affect the slope α\alpha?

Tip: Understanding R2R^2 can help assess the accuracy of linear models. The closer it is to 1, the better the fit.

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Math Problem Analysis

Mathematical Concepts

Linear Regression
Physics of Springs
Algebra

Formulas

Linear equation L(m) = α * m + β

Theorems

Hooke's Law
Least Squares Method

Suitable Grade Level

Grades 10-12