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A 20 -kilogram mass is attached to a spring. If the frequency of simple harmonic motion is \(2 / \pi\) cycles/s, what is the spring constant \(k ?\) What is the frequency of simple harmonic motion if the original mass is replaced with an 80 -kilogram mass?

Short Answer

Expert verified
The spring constant is 320 N/m, and the new frequency is \( \frac{1}{\pi} \) cycles/s.

Step by step solution

01

Understand Frequency Formula

For a mass-spring system in simple harmonic motion, the angular frequency \( \omega \) is given by \( \omega = \sqrt{\frac{k}{m}} \). Here, \( k \) is the spring constant and \( m \) is the mass. The frequency \( f \) is related to angular frequency by \( f = \frac{\omega}{2\pi} \). We use \( f = \frac{2}{\pi} \) cycles/s from the problem statement.
02

Relate Frequency to Angular Frequency

Given the frequency \( f = \frac{2}{\pi} \) cycles/s, we find \( \omega \) using \( \omega = 2 \pi f \). Substituting the frequency, \( \omega = 2 \pi \left(\frac{2}{\pi}\right) = 4 \).
03

Solve for Spring Constant \(k\)

Substitute \( \omega = 4 \) and \( m = 20 \) kg into the formula \( \omega = \sqrt{\frac{k}{m}} \). So, we have \( 4 = \sqrt{\frac{k}{20}} \). Square both sides to remove the square root: \( 16 = \frac{k}{20} \). Then, solve for \( k \) by multiplying both sides by 20: \( k = 320 \) N/m.
04

Calculate New Frequency with 80 kg Mass

Repeat the process using the new mass \( m = 80 \) kg while keeping \( k = 320 \) N/m. Using \( \omega = \sqrt{\frac{k}{m}} \), substitute the values: \( \omega = \sqrt{\frac{320}{80}} = \sqrt{4} = 2 \). The new frequency is \( f = \frac{\omega}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi} \) cycles/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Spring Constant
The spring constant, denoted as \( k \), is a measure of a spring's stiffness. It plays a crucial role in determining how a spring will react to a force. In simple harmonic motion, the spring constant is essential for understanding the relationship between the force applied to a spring and the displacement it causes. A higher \( k \) value indicates a stiffer spring, which requires more force to stretch or compress.

Key points about the spring constant include:
  • Measured in newtons per meter (N/m).
  • Determines the restoring force exerted by the spring, calculated using Hooke's Law: \( F = -kx \), where \( F \) is the force, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium.
  • Critical in calculating the spring's behavior in a mass-spring system where the motion can be described by simple harmonic motion.

In our problem, the spring constant \( k \) was derived from the connection between angular frequency and the mass involved. Knowing the spring constant helps predict how the spring will behave when interacting with different masses.
Mass-Spring System
A mass-spring system is a fundamental model for demonstrating simple harmonic motion. This system comprises a mass attached to a spring, which can either stretch or compress, leading to oscillations. These oscillations are a classic example of simple harmonic motion, a type of periodic motion where the restoring force is directly proportional to the displacement and acts in the opposite direction.

Characteristics of a mass-spring system:
  • The motion is characterized by the mass moving symmetrically around an equilibrium position.
  • It serves as a great example to illustrate concepts like energy transformation and damping when realistic factors are considered.
  • Key equations involving this system include the angular frequency \( \omega = \sqrt{\frac{k}{m}} \), where \( m \) is the mass attached to the spring, and \( k \) is the spring constant.

Understanding a mass-spring system provides insight into how harmonic motion works and the factors affecting the frequency and oscillation pattern. In this problem, changing the mass impacts the system's frequency, illustrating how mass influences the system behavior.
Angular Frequency
Angular frequency, represented by \( \omega \), is a crucial concept in understanding oscillatory systems like the mass-spring system. It describes how quickly an object oscillates in simple harmonic motion. Angular frequency is linked to the spring constant and mass, enabling calculations involving oscillation properties.

Important aspects of angular frequency:
  • Expressed in radians per second (rad/s), distinguishing it from regular frequency measured in cycles per second.
  • Related to the frequency \( f \) by the formula \( \omega = 2\pi f \).
  • Critical in determining the system's oscillation speed and the dynamics of the motion through \( \omega = \sqrt{\frac{k}{m}} \).

In our exercise, calculating \( \omega \) was essential for determining changes in the system's behavior as mass varied. By understanding angular frequency, one gains valuable insights into the rate of oscillation, key for solving problems involving harmonic motion.

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Most popular questions from this chapter

(a) Experiment with acalculator to find an interval \(0 \leq \theta<\theta_{1}\), where \(\theta\) is measured in radians, for which you think \(\sin \theta \approx \theta\) is a fairly good estimate. Then use a graphing utility to plot the graphs of \(y=x\) and \(y=\sin x\) on the same coordinate axes for \(0 \leq x \leq \pi / 2\). Do the graphs confirm your observations with the calculator? (b) Use a numerical solver to plot the solutions curves of the initial-value problems $$ \begin{aligned} \quad \frac{d^{2} \theta}{d t^{2}}+\sin \theta=0, & \theta(0)=\theta_{0}, \theta^{\prime}(0)=0 \\ \text { and } \quad \frac{d^{2} \theta}{d t^{2}}+\theta=0, \quad \theta(0)=\theta_{0}, \theta^{\prime}(0)=0 \end{aligned} $$ for several values of \(\theta_{0}\) in the interval \(0 \leq \theta<\theta_{1}\) found in part (a). Then plot solution curves of the initialvalue problems for several values of \(\theta_{0}\) for which \(\theta_{0}>\theta_{1}\)

Use the substitution \(t=-x\) to solve the given initial-value problem on the interval \((-\infty, 0)\). $$ x^{2} y^{\prime \prime}-4 x y^{\prime}+6 y=0, y(-2)=8, y^{\prime}(-2)=0 $$

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The indefinite integrals of the equations in (5) are nonelementary. Use a CAS to find the first four nonzero terms of a Maclaurin series of each integrand and then integrate the result. Find a particular solution of the given differential equation. $$ y^{\prime \prime}+y=\sqrt{1+x^{2}} $$

In Problems, find the eigenvalues and eigenfunctions for the given boundary- value problem. Consider only the case \(\lambda=\alpha^{4}, \alpha>0\) $$ \begin{aligned} &y^{(4)}-\lambda y=0, \quad y^{\prime}(0)=0, y^{m \prime}(0)=0, y(x)=0, \\ &y^{\prime \prime}(\pi)=0 \end{aligned} $$

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