/*! 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 101 A \(1.2 \mathrm{~kg}\) block sli... [FREE SOLUTION] | 91影视

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A \(1.2 \mathrm{~kg}\) block sliding on a horizontal frictionless surface is attached to a horizontal spring with \(k=480 \mathrm{~N} / \mathrm{m} .\) Let \(x\) be the displacement of the block from the position at which the spring is unstretched. At \(t=0\) the block passes through \(x=0\) with a speed of \(5.2 \mathrm{~m} / \mathrm{s}\) in the positive \(x\) direction. What are the (a) frequency and (b) amplitude of the block's motion? (c) Write an expression for \(x\) as a function of time.

Short Answer

Expert verified
(a) Frequency: 3.18 Hz. (b) Amplitude: 0.260 m. (c) Expression: \( x(t) = 0.260 \sin(20t) \).

Step by step solution

01

Determine the Angular Frequency

The angular frequency \( \omega \) of a spring-block system is given by the formula \( \omega = \sqrt{\frac{k}{m}} \). Here, \( k = 480 \, \mathrm{N/m} \) is the spring constant, and \( m = 1.2 \, \mathrm{kg} \) is the mass of the block. Substituting these values, we get: \[ \omega = \sqrt{\frac{480}{1.2}} = \sqrt{400} = 20 \, \mathrm{rad/s} \].
02

Calculate the Frequency

The frequency \( f \) is related to the angular frequency by the equation \( f = \frac{\omega}{2\pi} \). Using the previously calculated \( \omega = 20 \, \mathrm{rad/s} \), we find: \[ f = \frac{20}{2\pi} \approx 3.18 \, \mathrm{Hz} \].
03

Find the Amplitude Using Energy Conservation

The total mechanical energy \( E \) in the spring system is conserved and given as \( E = \frac{1}{2} m v_0^2 = \frac{1}{2} k A^2 \), where \( v_0 = 5.2 \, \mathrm{m/s} \) is the initial speed and \( A \) is the amplitude. Solve for \( A \) by setting the initial kinetic energy equal to the maximum potential energy: \[ \frac{1}{2} (1.2)(5.2)^2 = \frac{1}{2} (480) A^2 \]. Simplifying gives: \[ 16.224 = 240 A^2 \]\[ A^2 = \frac{16.224}{240} \]\[ A = \sqrt{0.0676} \approx 0.260 \, \mathrm{m} \].
04

Express Displacement as a Function of Time

The equation for simple harmonic motion is \( x(t) = A \cos(\omega t + \phi) \). At \( t = 0 \), \( x = 0 \) and the block is moving in the positive direction, indicating \( \phi = \frac{\pi}{2} \) for cosine. Thus, \[ x(t) = 0.260 \cos(20t + \frac{\pi}{2}) = 0.260 \sin(20t) \].

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Key Concepts

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

Angular Frequency
Angular frequency, denoted by \( \omega \), is an essential concept in simple harmonic motion. It describes how quickly the object oscillates back and forth. For a spring-block system, the angular frequency is determined by the equation: \[\omega = \sqrt{\frac{k}{m}} \]where \( k \) is the spring constant and \( m \) is the mass of the block.In this exercise, we're dealing with a spring constant \( k = 480 \,\mathrm{N/m} \) and a mass \( m = 1.2 \,\mathrm{kg} \). Plugging these values into the formula gives us:\[\omega = \sqrt{\frac{480}{1.2}} = \sqrt{400} = 20 \,\mathrm{rad/s}\]This result shows that the block completes its oscillation at a rate of 20 radians per second. Understanding angular frequency is crucial as it directly influences how the motion progresses over time.
Spring Constant
The spring constant \( k \) is a measure of the stiffness of a spring. It tells us how much force is needed to stretch or compress the spring by a unit length. In the realm of simple harmonic motion, this constant plays a pivotal role in determining both the angular frequency and the amplitude of motion.Typically, a higher spring constant indicates a stiffer spring, resulting in a higher angular frequency for a given mass. This is because it requires more force to displace the spring, leading to faster oscillations.In our exercise, the spring constant is \( k = 480 \,\mathrm{N/m} \). This value signifies that a force of 480 Newtons is required to stretch or compress the spring by one meter.The stiffer the spring, the quicker the system will oscillate back to its equilibrium position. This constant is crucial for understanding how the spring influences the motion's behavior and energy characteristics.
Mechanical Energy Conservation
Conservation of mechanical energy is a fundamental principle used to understand simple harmonic motion in spring systems. It states that the total mechanical energy in a closed system remains constant, provided no external forces like friction are present.In the context of a block-spring system, the mechanical energy is shared between potential energy in the compressed or stretched spring and kinetic energy of the moving block. Mathematically, it's expressed as:\[E = \frac{1}{2} m v_0^2 = \frac{1}{2} k A^2\]where \( E \) is the total mechanical energy, \( m \) is the mass, \( v_0 \) is the initial velocity, \( k \) is the spring constant, and \( A \) is the amplitude of motion.In the given problem, initially, when the block is moving with a speed of \( 5.2 \,\mathrm{m/s} \), all its energy is kinetic. As the block reaches its furthest displacement (amplitude), all this energy converts into potential energy stored in the spring. By equating the initial kinetic energy to the potential energy at maximum displacement, we find:\[\frac{1}{2} (1.2)(5.2)^2 = \frac{1}{2} (480) A^2 \]Solving this provides the amplitude \( A \approx 0.260 \,\mathrm{m} \). This principle highlights how energy transitions between forms during oscillations, explaining the sustained motion of the block.

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Most popular questions from this chapter

A \(2.00 \mathrm{~kg}\) block hangs from a spring. A \(300 \mathrm{~g}\) body hung below the block stretches the spring \(2.00 \mathrm{~cm}\) farther. (a) What is the spring constant? (b) If the 300 g body is removed and the block is set into oscillation, find the period of the motion.

What is the frequency of a simple pendulum \(2.0 \mathrm{~m}\) long (a) in a room, (b) in an elevator accelerating upward at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2},\) and \((\mathrm{c})\) in free fall?

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