Determining the Spring Constant

Oct 13, 2025 by magicalceylontours.com 132 views

Introduction

In this article, we will explore the behavior of a mass-spring system that is disturbed from its rest state by an external force. The system consists of a 10-kilogram object suspended from the end of a vertically hanging spring, which stretches the spring by 9.8 centimeters. At time $$t=0$$, the system is disturbed by the force $$F(t)=40 \cos (10 t)$$. We will use the principles of physics to analyze the motion of the system and determine its behavior over time.

The Mass-Spring System

A mass-spring system is a simple harmonic oscillator that consists of a mass attached to a spring. The spring exerts a force on the mass that is proportional to the displacement of the mass from its equilibrium position. The equation of motion for a mass-spring system is given by:

m \frac{d^2x}{dt^2} + kx = F(t)

where $$m$$ is the mass, $$k$$ is the spring constant, $$x$$ is the displacement of the mass from its equilibrium position, and $$F(t)$$ is the external force.

Determining the Spring Constant

To determine the spring constant $$k$$, we can use the fact that the spring stretches by 9.8 centimeters when the 10-kilogram object is suspended from it. We can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement:

F = kx

We can rearrange this equation to solve for the spring constant:

k = \frac{F}{x}

Since the spring stretches by 9.8 centimeters, we can use this value as the displacement $$x$$. However, we need to convert this value to meters to ensure that our units are consistent. There are 100 centimeters in 1 meter, so:

x = 9.8 \text{ cm} = 0.098 \text{ m}

We can now use this value to determine the spring constant:

k = \frac{F}{x} = \frac{mg}{x}

where $$mg$$ is the weight of the object, which is equal to the force exerted by the spring when it is stretched by 9.8 centimeters. We can use the value of the acceleration due to gravity $$g = 9.8 \text{ m/s}^2$$ to determine the weight of the object:

mg = 10 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N}

We can now use this value to determine the spring constant:

k = \frac{mg}{x} = \frac{98 \text{ N}}{0.098 \text{ m}} = 1000 \text{ N/m}

The Equation of Motion

Now that we have determined the spring constant $$k$$, we can use the equation of motion for a mass-spring system to analyze the behavior of the system. The equation of motion is given by:

m \frac{d^2x}{dt^2} + kx = F(t)

We can substitute the values of the mass $$m$$, the spring constant $$k$$, and the external force $$F(t)$$ into this equation to obtain:

10 \frac{d^2x}{dt^2} + 1000x = 40 \cos (10 t)

Solving the Equation of Motion

To solve this equation, we can use the method of undetermined coefficients. We can assume that the solution has the form:

x(t) = A \cos (\omega t) + B \sin (\omega t)

where $$A$$ and $$B$$ are constants, and $$\omega$$ is the angular frequency of the system. We can substitute this expression into the equation of motion and use the trigonometric identity:

\cos (2 \omega t) = 2 \cos^2 (\omega t) - 1

to simplify the equation. After some algebra, we can obtain:

\omega^2 = \frac{k}{m} = \frac{1000}{10} = 100

We can now use this value to determine the angular frequency $$\omega$$:

\omega = \sqrt{100} = 10 \text{ rad/s}

We can now use this value to determine the constants $$A$$ and $$B$$ by substituting the expression for $$x(t)$$ into the equation of motion and using the initial conditions.

The General Solution

After some algebra, we can obtain the general solution to the equation of motion:

x(t) = 2 \cos (10 t) + \frac{1}{5} \sin (10 t)

This solution represents the displacement of the mass from its equilibrium position as a function of time.

The Particular Solution

To determine the particular solution, we need to use the initial conditions. We are given that the system is disturbed from its rest state at time $$t=0$$, so we can use the initial conditions:

x(0) = 0

\frac{dx}{dt}(0) = 0

We can use these conditions to determine the constants $$A$$ and $$B$$ in the general solution.

The Particular Solution with Initial Conditions

After some algebra, we can obtain the particular solution with initial conditions:

x(t) = 2 \cos (10 t)

This solution represents the displacement of the mass from its equilibrium position as a function of time, subject to the initial conditions.

Conclusion

In this article, we have analyzed the behavior of a mass-spring system that is disturbed from its rest state by an external force. We have used the principles of physics to determine the spring constant $$k$$, the equation of motion, and the general solution to the equation of motion. We have also used the initial conditions to determine the particular solution with initial conditions. The results of this analysis can be used to design and analyze mass-spring systems in a variety of applications.

References

[1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of physics (10th ed.). John Wiley & Sons.
[2] Serway, R. A., & Jewett, J. W. (2018). Physics for scientists and engineers (10th ed.). Cengage Learning.

Appendix

The following is a list of the equations used in this article:

Equation of motion: $$m \frac{d^2x}{dt2} + kx = F(t)$$
Hooke's Law: $$F = kx$$
Equation for the spring constant: $$k = \frac{F}{x}$$
Equation for the weight of the object: $$mg = 10 \text{ kg} \times 9.8 \text{ m/s}^2 = 98 \text{ N}$$
Equation for the spring constant: $$k = \frac{mg}{x} = \frac{98 \text{ N}}{0.098 \text{ m}} = 1000 \text{ N/m}$$
Equation of motion for the mass-spring system: $$10 \frac{d^2x}{dt2} + 1000x = 40 \cos (10 t)$$
General solution to the equation of motion: $$x(t) = 2 \cos (10 t) + \frac{1}{5} \sin (10 t)$$
Particular solution with initial conditions: $$x(t) = 2 \cos (10 t)$$
A 10-kilogram object suspended from the end of a vertically hanging spring stretches the spring 9.8 centimeters. At time $$t=0$$, the resulting mass-spring system is disturbed from its rest state by the force $$F(t)=40 \cos (10 t)$$. Q&A ===========================================================

Q: What is the mass-spring system?

A: A mass-spring system is a simple harmonic oscillator that consists of a mass attached to a spring. The spring exerts a force on the mass that is proportional to the displacement of the mass from its equilibrium position.

Q: What is the equation of motion for a mass-spring system?

A: The equation of motion for a mass-spring system is given by:

m \frac{d^2x}{dt^2} + kx = F(t)

where $$m$$ is the mass, $$k$$ is the spring constant, $$x$$ is the displacement of the mass from its equilibrium position, and $$F(t)$$ is the external force.

Q: How do you determine the spring constant $$k$$?

A: To determine the spring constant $$k$$, you can use Hooke's Law, which states that the force exerted by a spring is proportional to its displacement:

F = kx

You can rearrange this equation to solve for the spring constant:

k = \frac{F}{x}

Q: What is the angular frequency $$\omega$$ of the system?

A: The angular frequency $$\omega$$ of the system is given by:

\omega = \sqrt{\frac{k}{m}}

Q: How do you determine the general solution to the equation of motion?

A: To determine the general solution to the equation of motion, you can use the method of undetermined coefficients. You can assume that the solution has the form:

x(t) = A \cos (\omega t) + B \sin (\omega t)

where $$A$$ and $$B$$ are constants, and $$\omega$$ is the angular frequency of the system.

Q: What is the particular solution with initial conditions?

A: The particular solution with initial conditions is given by:

x(t) = 2 \cos (10 t)

This solution represents the displacement of the mass from its equilibrium position as a function of time, subject to the initial conditions.

Q: What are the initial conditions?

A: The initial conditions are:

x(0) = 0

\frac{dx}{dt}(0) = 0

Q: How do you use the initial conditions to determine the particular solution?

A: You can use the initial conditions to determine the constants $$A$$ and $$B$$ in the general solution. After some algebra, you can obtain the particular solution with initial conditions.

Q: What is the significance of the mass-spring system?

A: The mass-spring system is a simple harmonic oscillator that is used to model a wide range of physical systems, including springs, pendulums, and oscillating circuits. It is an important concept in physics and engineering, and is used to analyze the behavior of systems that exhibit simple harmonic motion.

Q: How do you apply the mass-spring system to real-world problems?

A: You can apply the mass-spring system to real-world problems by using it to model the behavior of systems that exhibit simple harmonic motion. For example, you can use the mass-spring system to analyze the behavior of a spring-mass system that is subjected to an external force, or to design a system that exhibits simple harmonic motion.

Q: What are some common applications of the mass-spring system?

A: Some common applications of the mass-spring system include:

Designing springs and other elastic systems
Analyzing the behavior of pendulums and other oscillating systems
Modeling the behavior of systems that exhibit simple harmonic motion
Designing systems that exhibit simple harmonic motion, such as oscillating circuits and mechanical systems.

Q: What are some common mistakes to avoid when working with the mass-spring system?

A: Some common mistakes to avoid when working with the mass-spring system include:

Failing to account for the effects of friction and other dissipative forces
Failing to use the correct equation of motion for the system
Failing to use the correct initial conditions for the system
Failing to account for the effects of external forces on the system.

Q: How do you troubleshoot problems with the mass-spring system?

A: You can troubleshoot problems with the mass-spring system by:

Checking the equation of motion for the system
Checking the initial conditions for the system
Checking the external forces acting on the system
Using numerical methods to solve the equation of motion for the system.

Q: What are some common tools and techniques used to analyze the mass-spring system?

A: Some common tools and techniques used to analyze the mass-spring system include:

Mathematical software, such as Mathematica or MATLAB
Numerical methods, such as the finite difference method or the finite element method
Analytical methods, such as the method of undetermined coefficients or the method of variation of parameters.
Experimental methods, such as measuring the displacement and velocity of the mass as a function of time.