/*! 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 50 50\. Block-Spring An oscillating... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 16: Problem 50

50\. Block-Spring An oscillating block-spring system has a mechanical energy of \(1.00 \mathrm{~J}\), an amplitude of \(10.0 \mathrm{~cm}\), and a maximum speed of \(1.20 \mathrm{~m} / \mathrm{s}\). Find (a) the spring constant, (b) the mass of the block, and (c) the frequency of oscillation.

Short Answer

Expert verified
a) 200 N/mb) 0.0139 kg or 13.9 gc) 19.1 Hz

Step by step solution

01

- Determine Relevant Equations

First, identify the relevant equations for an oscillating block-spring system. For mechanical energy, the equation is \(E = \frac{1}{2} k A^2\).For maximum speed, the equation is \(v_{max} = \frac{A \times (2 \times \text{Mechanical Energy})^{0.5}}{m^{0.5}}\).For frequency, use \(f = \frac{1}{2 \times \text{pi}} \times (k / m)^{0.5}\).
02

- Calculate the Spring Constant (k)

Use the equation for mechanical energy: \(E = \frac{1}{2} k A^2\). Rearrange to solve for k: \(k = \frac{2E}{A^2}\).Given:\(E = 1.00 \text{ J}\) \(A = 10.0 \text{ cm} = 0.10 \text{ m}\). Substitute the values: \(k = \frac{2 \times 1.00}{(0.10)^2}\) \(k = 200 \text{ N/m}\).
03

- Calculate the Mass of the Block (m)

Use the maximum speed equation: \(v_{max} = \frac{A \times (2 \times \text{Mechanical Energy})^{0.5}}{m^{0.5}}\). Rearrange to solve for m: \(m = \frac{A^2 \times (2 \times \text{Mechanical Energy})}{v_{max}^2}\).Given:\(A = 0.10 \text{ m}\) \(E = 1.00 \text{ J}\) \(v_{max} = 1.20 \text{ m/s}\). Substitute the values: \(m = \frac{(0.10)^2 \times (2 \times 1.00)}{(1.20)^2}\) \(m = 0.0139 \text{ kg} \text{ or } 13.9 \text{ g}\).
04

- Calculate the Frequency of Oscillation (f)

Use the frequency formula: \(f = \frac{1}{2 \times \text{pi}} \times (k/m)^{0.5}\).Substitute the known values: \(k = 200 \text{ N/m}\) \(m = 0.0139 \text{ kg}\) \(f = \frac{1}{2 \times \text{pi}} \times (200/0.0139)^{0.5}\) \(f = 19.1 \text{ Hz}\).

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!

Key Concepts

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

Mechanical Energy
Mechanical energy in a block-spring system is the total energy of the system. It's a sum of both kinetic energy and potential energy stored in the spring. When a block attached to a spring oscillates, its energy continuously shifts between kinetic (when it's moving) and potential (when it's stretched or compressed).

The mechanical energy in a block-spring system can be given by the formula: \[ E = \frac{1}{2} k A^2 \] Here, \(E\) represents the mechanical energy, \(k\) is the spring constant, and \(A\) is the amplitude of the oscillation.
For example, if we know the mechanical energy is 1.00 J and the amplitude is 0.10 m, we use this formula to find the spring constant.
Spring Constant
The spring constant, denoted as \(k\), is a measure of the stiffness of the spring. It tells us how much force is needed to stretch or compress the spring by a unit length.
In our example, we use the mechanical energy to calculate the spring constant. With the formula \[ k = \frac{2E}{A^2} \] we can find that: \[ k = \frac{2 \times 1.00}{(0.10)^2} = 200 \text{ N/m} \] This means the spring constant is 200 N/m, implying the spring is relatively stiff and resists stretching or compressing.
Frequency of Oscillation
The frequency of oscillation, represented by \(f\), is how many complete oscillations occur per second. It's measured in Hertz (Hz).
The frequency is determined by both the spring constant \(k\) and the mass of the block \(m\). The formula to find the frequency is: \[ f = \frac{1}{2 \pi} \left(\frac{k}{m}\right)^{0.5} \] By substituting \(k = 200 \text{ N/m}\) and \(m = 0.0139 \text{ kg}\), we find: \[ f = \frac{1}{2 \pi} \left(\frac{200}{0.0139}\right)^{0.5} \] This results in an oscillation frequency of approximately 19.1 Hz.
Mass of Block
The mass of the block, represented as \(m\), affects how the system behaves during oscillation.
To find the mass of the block, we use the maximum speed of the block, which is given by \[ v_{max} = \frac{A \left(2 \times \text{Mechanical Energy}\right)^{0.5}}{m^{0.5}} \] Rearranging this to solve for the mass, we get: \[ m = \frac{A^2 \left(2 \times E\right)}{v_{max}^2} \] Given \(A = 0.10 \text{ m}\), \(E = 1.00 \text{ J}\), and \(v_{max} = 1.20 \text{ m/s}\), we calculate: \[ m = \frac{(0.10)^2 \left(2 \times 1.00\right)}{(1.20)^2} = 0.0139 \text{ kg} \] Thus, the mass of the block is 0.0139 kg, or 13.9 grams.

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

53\. Displacement in SHM When the displacement in SHM is onehalf the amplitude \(X\), what fraction of the total energy is (a) kinetic energy and (b) potential energy? (c) At what displacement, in terms of the amplitude, is the energy of the system half kinetic energy and half potential energy?

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?

34\. Demolition Ball In Fig. \(16-37\), a \(2500 \mathrm{~kg}\) demolition ball swings from the end of a crane. The length of the swinging segment of cable is \(17 \mathrm{~m}\). (a) Find the period of the swinging, assuming that the system can be treated as a simple pendulum. (b) Does the period depend on the ball's mass?

18\. Block and Piston A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period \(1.0 \mathrm{~s}\), at what amplitude of motion will the block and piston separate? (b) If the piston has an amplitude of \(5.0 \mathrm{~cm}\), what is the maximum frequency for which the block and piston will be in contact continuously?

22\. Two Particles Two particles execute simple harmonic motion of the same amplitude and frequency along close parallel lines. They pass each other moving in opposite directions each time their displacement is half their amplitude. What is their phase difference?
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Physics of Motion

Read Explanation

Energy Physics

Read Explanation

Dynamics

Read Explanation

Classical Mechanics

Read Explanation

Measurements

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.