/*! 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] | 91Ó°ÊÓ

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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 is approximately 3.18 Hz, (b) Amplitude is 0.26 m, (c) Position function is \( x(t) = 0.26 \sin(20t) \).

Step by step solution

01

Understand the System

We have a block with mass \( m = 1.2 \text{ kg} \) attached to a spring with spring constant \( k = 480 \text{ N/m} \). The spring is horizontal and frictionless. At \( t = 0 \), the block is at the equilibrium position \( x = 0 \) with an initial speed \( v = 5.2 \text{ m/s} \). We need to find the frequency, amplitude, and an expression for \( x(t) \).
02

Calculate the Frequency

The frequency \( f \) of the simple harmonic motion can be found using the formula \[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}. \] Substituting the given values, \( k = 480 \text{ N/m} \) and \( m = 1.2 \text{ kg} \), we get: \[ f = \frac{1}{2\pi} \sqrt{\frac{480}{1.2}} = \frac{1}{2\pi} \sqrt{400} = \frac{1}{2\pi} \times 20 = \frac{10}{\pi} \approx 3.18 \text{ Hz}. \]
03

Determine the Amplitude

To find the amplitude, we use energy conservation. At \( x = 0 \), all energy is kinetic: \[ E_{\text{kinetic}} = \frac{1}{2} mv^2. \] At the maximum displacement, all energy is potential: \[ E_{\text{potential}} = \frac{1}{2} k A^2. \] Set the kinetic energy equal to the potential energy: \[ \frac{1}{2} mv^2 = \frac{1}{2} k A^2. \] Solving for \( A \), \[ A = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{1.2 \times (5.2)^2}{480}} = \sqrt{\frac{32.448}{480}} \approx 0.26 \text{ m}. \]
04

Write the Expression for x(t)

The general form for the position as a function of time in simple harmonic motion is \[ x(t) = A \cos{(\omega t + \phi)}. \] Since at \( t = 0 \), \( x = 0 \), and the motion is in the positive direction, \( \phi = \frac{\pi}{2} \) and the expression for \( x(t) \) becomes \[ x(t) = A \sin(\omega t), \] where \( \omega = 2\pi f \). Therefore, \( \omega = 20 \) rad/s (from \( \omega = \sqrt{\frac{k}{m}} = 20 \)). With the amplitude \( A = 0.26 \) m, \[ x(t) = 0.26 \sin(20t). \]

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Key Concepts

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

Frequency Calculation
Frequency in simple harmonic motion is a vital parameter that tells us how many oscillations occur per unit of time. For a block sliding on a frictionless surface attached to a spring, the frequency can be determined from the spring constant and the mass of the block.
To calculate the frequency (\( f \)), we use the formula:\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}.\]This equation indicates that frequency is directly influenced by the spring constant (\( k \)) and inversely by the mass (\( m \)). For our block with a given \( k = 480 \text{ N/m} \) and \( m = 1.2 \text{ kg} \), substituting the values gives us:\[ f = \frac{1}{2\pi} \sqrt{\frac{480}{1.2}} = \frac{10}{\pi} \approx 3.18 \text{ Hz}.\]
This means the block oscillates approximately 3.18 times per second.
Amplitude Determination
Amplitude in simple harmonic motion represents the maximum displacement from the equilibrium position. To find the amplitude (\( A \)), we rely on the principle of energy conservation. Initially, at the equilibrium position, all the energy is kinetic, given by:\[ E_{\text{kinetic}} = \frac{1}{2} mv^2.\]At the point of maximum displacement, the energy is entirely potential:\[ E_{\text{potential}} = \frac{1}{2} k A^2.\]By equating kinetic and potential energy and solving for \( A \), we find:\[ A = \sqrt{\frac{mv^2}{k}} = \sqrt{\frac{1.2 \times (5.2)^2}{480}} \approx 0.26 \text{ m}.\]
This calculation shows that the block reaches a maximum displacement of about 0.26 meters from the central position during its motion.
Energy Conservation
The concept of energy conservation is fundamental in analyzing simple harmonic motion. It implies that the total mechanical energy in the system remains constant, comprising both kinetic and potential energies. As the block moves back and forth, the energy continually exchanges between kinetic and potential forms.
  • At the equilibrium point (\( x = 0 \)), the energy is purely kinetic: \( E_{\text{kinetic}} = \frac{1}{2} mv^2 \).
  • At maximum displacement (\( x = A \)), the energy converts entirely to potential form: \( E_{\text{potential}} = \frac{1}{2} k A^2 \).
The reliability of energy conservation allows us to derive other properties, such as amplitude, by using the equality of kinetic and potential energies at different phases of motion.
Angular Frequency
Angular frequency (\( \omega \)) is another crucial parameter in simple harmonic motion. It showcases how fast the oscillating object moves through its cycle in terms of radians per second. Derived from the spring constant and mass, it is mathematically represented as:\[ \omega = \sqrt{\frac{k}{m}}.\]For our given problem, calculating \( \omega \) provides:\[ \omega = \sqrt{\frac{480}{1.2}} = 20 \text{ rad/s}.\]
Moreover, angular frequency relates to the frequency (\( f \)) via the relation \( \omega = 2\pi f \).This means that for this system, not only does \( \omega \) define the speed of oscillation in radians, but it also ties into the periodic nature that \( f \) represents, enhancing our understanding of the motion's rhythm.

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Most popular questions from this chapter

When a \(20 \mathrm{~N}\) can is hung from the bottom of a vertical spring, it causes the spring to stretch \(20 \mathrm{~cm} .\) (a) What is the spring constant? (b) This spring is now placed horizontally on a frictionless table. One end of it is held fixed, and the other end is attached to a \(5.0 \mathrm{~N}\) can. The can is then moved (stretching the spring) and released from rest. What is the period of the resulting oscillation?

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A simple harmonic oscillator consists of a block of mass \(2.00 \mathrm{~kg}\) attached to a spring of spring constant \(100 \mathrm{~N} / \mathrm{m} .\) When \(t=1.00 \mathrm{~s}\), the position and velocity of the block are \(x=0.129\) \(\mathrm{m}\) and \(v=3.415 \mathrm{~m} / \mathrm{s}\). (a) What is the amplitude of the oscillations? What were the (b) position and (c) velocity of the block at \(t=0 \mathrm{~s}\) ?

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