/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 29 29\. Unstretched Length A unifor... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

29\. Unstretched Length A uniform spring with unstretched length \(L\) and spring constant \(k\) is cut into two pieces of unstretched lengths \(L_{1}\) and \(L_{2}\), with \(L_{1}=n L_{2}\). What are the corresponding spring constants (a) \(k_{1}\) and (b) \(k_{2}\) in terms of \(n\) and \(k ?\) If a block is attached to the original spring, as in Fig. \(16-10\), it oscillates with frequency \(f\). If the spring is replaced with the piece \(L_{1}\) or \(L_{2}\), the corresponding frequency is \(f_{1}\) or \(f_{2}\). Find (c) \(f_{1}\) and \((\mathrm{d}) f_{2}\) in terms of \(f\).

Short Answer

Expert verified
k_1 = \frac{k}{n+1}, k_2 = \frac{kn}{n+1}. f_1 = \frac{f}{\sqrt{n+1}}, f_2 = \sqrt{\frac{n}{n+1}} f.

Step by step solution

01

- Understand the Problem

We have a spring with original unstretched length L and spring constant k. This spring is cut into two pieces with lengths L_1 and L_2 such that L_1 = n * L_2. We need to find the spring constants k_1 and k_2 for these new pieces, and the oscillation frequencies f_1 and f_2 for each piece if they are attached to the same block.
02

- Use Hooke's Law

Hooke's law states that the spring constant of a spring is inversely proportional to its length when the material of the spring remains the same. Therefore, if L_1 = n * L_2, then the spring constants will be related by: \[ k_1 = \frac{k}{n+1} \] and \[ k_2 = k - k_1 = k - \frac{k}{n+1} \]
03

- Simplify the Spring Constants

Given k_1 and k_2, we express them as: \[ k_1 = \frac{k}{n+1} \] \[ k_2 = k - \frac{k}{n+1} = \frac{kn}{n+1} \]
04

- Frequency Relation

The frequency of oscillation for a spring-mass system is given by \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]. Using this relation for each segment, we have: \[ f_1 = \frac{1}{2\pi} \sqrt{\frac{k_1}{m}} = \frac{1}{2\pi} \sqrt{\frac{k/(n+1)}{m}} = \frac{f}{\sqrt{n+1}} \] and \[ f_2 = \frac{1}{2\pi} \sqrt{\frac{k_2}{m}} = \frac{1}{2\pi} \sqrt{\frac{kn/(n+1)}{m}} = \sqrt{\frac{n}{n+1}} f \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Spring Constant
The spring constant, denoted by the letter \( k \), is a measure of a spring's stiffness. It tells us how much force is required to stretch or compress the spring by a certain distance. According to Hooke's Law, the force \( F \) exerted by a spring is proportional to the displacement \( x \) from its equilibrium position. Mathematically, it is described as:
\[ F = -kx \]
Here, the negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement.
When a spring is cut into two pieces with lengths \( L_1 \) and \( L_2 \) where \( L_1 = nL_2 \), the spring constants of the new pieces \( k_1 \) and \( k_2 \) change accordingly. The new spring constants are derived using the inverse proportionality principle:
  • \( k_1 = \frac{k}{n+1} \)
  • \( k_2 = k - k_1 = \frac{kn}{n+1} \)
Understanding these relationships helps in analyzing the behavior of each spring piece after they are cut.
Oscillation Frequency
The oscillation frequency \( f \) of a spring-mass system refers to how many times the system oscillates back and forth in a unit of time, typically measured in Hertz (Hz). The frequency is determined by both the spring constant \( k \) and the mass \( m \) attached to it. The formula for calculating the frequency is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}} \]
When the original spring is cut into two pieces with new spring constants \( k_1 \) and \( k_2 \), the corresponding new frequencies \( f_1 \) and \( f_2 \) are affected. The new frequencies can be derived as:
  • \( f_1 = \frac{f}{\sqrt{n+1}} \)
  • \( f_2 = \sqrt{\frac{n}{n+1}} f \)
This shows how the cutting of the spring alters the oscillation characteristics of the system.
Spring-Mass System
A spring-mass system consists of a spring attached to a mass. When the mass is displaced from its equilibrium position and then released, it oscillates due to the restoring force exerted by the spring. This simple harmonic motion (SHM) is a fundamental concept in physics.
Parameters influencing the behavior and characteristics of a spring-mass system include:
  • The spring constant \( k \)
  • The mass \( m \)
  • The initial displacement
When considering a system where a spring is cut into two pieces, each piece \( L_1 \) and \( L_2 \) with new spring constants \( k_1 \) and \( k_2 \), the system's behavior and characteristics such as frequency and amplitude will change accordingly. These changes follow the principles discussed in the oscillation frequency section. Understanding these changes helps in analyzing real-world scenarios such as car suspensions, seismic detectors, and any system where springs are used to absorb or release energy.
Proportionality
Proportionality is a core concept in physics that describes the relationship between two quantities. In the context of springs, two key types of proportionality arise:
  • Direct Proportionality
  • Inverse Proportionality
For a spring's force, Hooke's Law states that force is directly proportional to displacement: \( F = -kx \).

When considering the spring constant in segments of a cut spring, we use inverse proportionality. If a spring of length \( L \) with constant \( k \) is cut into segments \( L_1 \) and \( L_2 \) such that \( L_1 = n L_2 \), the resulting spring constants are inversely proportional to their lengths:
  • \( k_1 = \frac{k}{n+1} \)
  • \( k_2 = k - k_1 = \frac{kn}{n+1} \)
Understanding proportionality helps in predicting and calculating changes in system behavior when variables are altered. This concept is vital for various engineering and scientific applications where precise control of physical properties is required.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

7\. A Particle of Mass A particle with a mass of \(1.00 \times 10^{-20} \mathrm{~kg}\) is oscillating with simple harmonic motion with a period of \(1.00 \times 10^{-5} \mathrm{~s}\) and a maximum speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Calculate (a) the angular frequency and (b) the maximum displacement of the particle.

48\. Large Slingshot A (hypothetical) large slingshot is stretched \(1.50 \mathrm{~m}\) to launch a \(130 \mathrm{~g}\) projectile with speed sufficient to escape from Earth \((11.2 \mathrm{~km} / \mathrm{s})\). Assume the elastic bands of the slingshot obey Hooke's law. (a) What is the spring constant of the device, if all the elastic potential energy is converted to kinetic energy? (b) Assume that an average person can exert a force of \(220 \mathrm{~N}\). How many people are required to stretch the elastic bands?

36\. Trapeze A performer seated on a trapeze is swinging back and forth with a period of \(8.85 \mathrm{~s}\). If she stands up, thus raising the center of mass of the trapeze \(+\) performer system by \(35.0 \mathrm{~cm}\), what will be the new period of the system? Treat trapeze \(+\) performer as a simple pendulum.

37\. Pivoting Long Rod A pendulum is formed by pivoting a long thin rod of length \(L\) and mass \(m\) about a point on the rod that is a distance \(d\) above the center of the rod. (a) Find the period of this pendulum in terms of \(d, L, m\), and \(g\), assuming small-amplitude swinging. What happens to the period if (b) \(d\) is decreased, (c) \(L\) is increased, or (d) \(m\) is increased?

51\. Horizontal Frictionless A \(5.00 \mathrm{~kg}\) object on a horizontal frictionless surface is attached to a spring with spring constant 1000 \(\mathrm{N} / \mathrm{m} .\) The object is displaced from equilibrium \(50.0 \mathrm{~cm}\) horizontally and given an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) back toward the equilibrium position. (a) What is the frequency of the motion? What are (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the hmolitude of the oscillation?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.