/*! 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 4 If the spring of a simple harmon... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 13: Problem 4

If the spring of a simple harmonic oscillator is cut in half, what happens to the frequency?

Short Answer

Expert verified
If the spring of a simple harmonic oscillator is cut in half, the frequency of oscillation increases by a factor of \(\sqrt{2}\).

Step by step solution

01

Identify the change in the spring constant

Since the spring is cut in half, its constant \(k\) will double. This is because the spring constant is a measure of the spring's stiffness, and when a spring is cut, the shorter spring is stiffer. So if the original spring constant was \(k\), after cutting, the new spring constant is \(2k\).
02

Substitute the new spring constant into the frequency formula

The frequency of the oscillator before the spring was cut was given by \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\). Now, with the new spring constant, the frequency can be calculated as \(f' = \frac{1}{2\pi} \sqrt{\frac{2k}{m}}\).
03

Compare the initial and final frequencies

To understand the change in frequency, it's helpful to compare the initial and final frequencies. Express \(f'\) in terms of \(f\): \(f' = \sqrt{2} \times f\). You can see that the frequency \(f'\) of the cut spring is \(\sqrt{2}\) times higher than the initial frequency \(f\).

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 91影视!

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

A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency?

The quartz crystal in a watch executes simple harmonic motion at \(32,768 \mathrm{Hz}\) (This is \(2^{15} \mathrm{Hz}\), chosen so that 15 divisions by 2 give a signal at \(1.00000 \mathrm{Hz}\) ) If each face of the crystal undergoes a maximum displacement of \(100 \mathrm{nm}\), find the maximum velocity and acceleration of the crystal faces.

An automobile suspension has an effective spring constant of \(26 \mathrm{kN} / \mathrm{m},\) and the car's suspended mass is \(1900 \mathrm{kg} .\) In the absence of damping, with what frequency and period will the car undergo simple harmonic motion?

Two mass-spring systems have the same mass and the same total energy. The amplitude of system 1 is twice that of system \(2 .\) How do (a) their frequencies and (b) their maximum accelerations compare?

A violin string playing the note A oscillates at \(440 \mathrm{Hz}\). What's its oscillation period?
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Particle Model of Matter

Read Explanation

Solid State Physics

Read Explanation

Radiation

Read Explanation

Magnetism

Read Explanation

Scientific Method Physics

Read Explanation
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.