/*! 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 30 An oscillating block-spring syst... [FREE SOLUTION] | 91Ó°ÊÓ

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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) k = 200 N/m; (b) m ≈ 1.39 kg; (c) f ≈ 1.91 Hz.

Step by step solution

01

Understanding the Problem

We are given a block-spring system with a known mechanical energy, amplitude, and maximum speed. We need to find the spring constant (\(k\)), the mass of the block (\(m\)), and the frequency of oscillation (\(f\)).
02

Calculating the Spring Constant

The total mechanical energy (\(E\)) in a block-spring system is given by \(E = \frac{1}{2} k A^2\), where \(A\) is the amplitude. Given \(E = 1.00\,\text{J}\) and \(A = 0.10\,\text{m}\), we have:\[1.00 = \frac{1}{2} k (0.10)^2\]Solving for \(k\), we multiply both sides by 2:\[2.00 = k (0.01)\]\[k = \frac{2.00}{0.01} = 200\,\text{N/m}\]
03

Finding the Mass of the Block

The maximum speed (\(v_{max}\)) is related to the mass and the spring constant by \(v_{max} = \sqrt{\frac{k}{m}} A\). We need to rearrange this to find the mass \(m\).\[ (v_{max})^2 = \frac{k}{m} A^2\]Given \(v_{max} = 1.20\,\text{m/s}\), \(A = 0.10\,\text{m}\), and \(k = 200\,\text{N/m}\):\[(1.20)^2 = \frac{200}{m} (0.10)^2\]\[1.44 = \frac{200 \times 0.01}{m}\]\[1.44 = \frac{2.00}{m}\]\[m = \frac{2.00}{1.44} \approx 1.39\,\text{kg}\]
04

Determining the Frequency of Oscillation

The frequency of oscillation \(f\) is related to the spring constant and mass by the formula \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\). We have \(k = 200\,\text{N/m}\) and \(m \approx 1.39\,\text{kg}\). Substitute these values:\[ f = \frac{1}{2\pi} \sqrt{\frac{200}{1.39}}\]\[ f \approx \frac{1}{2\pi} \sqrt{143.88}\]\[ f \approx \frac{1}{2\pi} (11.99)\]\[ f \approx \frac{11.99}{6.28}\]\[ f \approx 1.91\,\text{Hz}\]

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Key Concepts

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

Spring Constant
In an oscillating block-spring system, the spring constant (\(k\)) 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 certain distance. A stiffer spring has a higher spring constant.
The spring constant is crucial in determining the dynamics of the oscillating system. It's calculated using the formula for mechanical energy in a spring-mass system, which is \(E = \frac{1}{2} k A^2\). Here, \(E\) is the mechanical energy and \(A\) is the amplitude.
In the given problem, we solved for \(k\) by rearranging the equation to get:- \(E = \frac{1}{2} k A^2\)By substituting the known values:- \(1.00 = \frac{1}{2} k (0.10)^2\)We multiplied both sides by 2 and divided by \(0.01\) to find:- \(k = 200\,\text{N/m}\)This result shows the spring constant, telling us how robust the spring is against deformation.
Mechanical Energy
Mechanical energy in a block-spring system refers to the total energy due to both kinetic and potential energy.
This energy is constant in an ideal system with no external forces or energy losses. The potential energy is stored in the spring when it is compressed or stretched, and the kinetic energy is possessed by the block due to its motion.
The formula for mechanical energy when considering the amplitude \(A\) of the oscillation is- \(E = \frac{1}{2} k A^2\)This formula helps clarify how energy relates to the displacement of the spring.
It also shows that the mechanical energy depends on both the spring constant and amplitude. Since energy is conserved in the system, you can use it to solve for other properties if some values are known, just like finding the spring constant in this problem.
Amplitude
Amplitude is the maximum extent of a vibration or oscillation, measured from the center position. In the context of the block-spring system, it represents the furthest point the block moves from the equilibrium position.
It is a key variable in calculating the mechanical energy, given by- \(E = \frac{1}{2} k A^2\)Here, the amplitude \(A\) (given as 10.0 cm or 0.10 m) directly influences how much energy the system holds. A larger amplitude means more energy is stored.
The amplitude doesn't change with time in an ideal system, as energy is conserved. Instead, it represents the scale of oscillation. Understanding amplitude helps visualize the system's overall motion and energetic properties.
Frequency of Oscillation
Frequency of oscillation \(f\) refers to how many complete cycles of motion occur in a given time period, usually a second. In the block-spring system, frequency describes how quickly the system oscillates back and forth.
Frequency is important because it tells you how fast the system vibrates. The formula to calculate frequency in a block-spring system is:- \(f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\)Where:- \(k\) is the spring constant- \(m\) is the mass of the block
In this problem, we used the calculated spring constant and mass to find the frequency:- \(f \approx 1.91\,\text{Hz}\)This means the block completes about 1.91 full oscillations per second. Frequency gives insights into the natural oscillation characteristics of the system, impacted by both the spring's stiffness and the block's mass.

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Most popular questions from this chapter

A simple harmonic oscillator consists of an \(0.80 \mathrm{~kg}\) block attached to a spring \((k=200 \mathrm{~N} / \mathrm{m})\). The block slides on a horizontal frictionless surface about the equilibrium point \(x=0\) with a total mechanical energy of \(4.0 \mathrm{~J}\). (a) What is the amplitude of the oscillation? (b) How many oscillations does the block complete in \(10 \mathrm{~s} ?(\mathrm{c})\) What is the maximum kinetic energy attained by the block? (d) What is the speed of the block at \(x=0.15 \mathrm{~m} ?\)

A block rides on a piston (a squat cylindrical piece) 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?

A \(4.00 \mathrm{~kg}\) block is suspended from a spring with \(k=500 \mathrm{~N} / \mathrm{m} . \mathrm{A}\) \(50.0 \mathrm{~g}\) bullet is fired into the block from directly below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and becomes embedded in the block. (a) Find the amplitude of the resulting SHM. (b) What percentage of the original kinetic energy of the bullet is transferred to mechanical energy of the oscillator?

Although California is known for earthquakes, it has large regions dotted with precariously balanced rocks that would be easily toppled by even a mild earthquake. Apparently no major earthquakes have occurred in those regions. If an earthquake were to put such a rock into sinusoidal oscillation (parallel to the ground) with a frequency of \(2.2 \mathrm{~Hz}\), an oscillation amplitude of \(1.0 \mathrm{~cm}\) would cause the rock to topple. What would be the magnitude of the maximum acceleration of the oscillation, in terms of \(g ?\)

A block weighing \(20 \mathrm{~N}\) oscillates at one end of a vertical spring for which \(k=100 \mathrm{~N} / \mathrm{m} ;\) the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched \(0.30 \mathrm{~m}\) beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

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