/*! 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 23 The drawing shows a top view of ... [FREE SOLUTION] | 91Ó°ÊÓ

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The drawing shows a top view of a frictionless horizontal surface, where there are two springs with particles of mass \(m_{1}\) and \(m_{2}\) attached to them. Each spring has a spring constant of \(120 \mathrm{~N} / \mathrm{m}\). The particles are pulled to the right and then released from the positions shown in the drawing. How much time passes before the particles are side by side for the first time at \(x=0 \mathrm{~m}\) if \((\mathrm{a}) \mathrm{m}_{1}=m_{2}=3.0 \mathrm{~kg}\) and \((\mathrm{b}) m_{1}=3.0 \mathrm{~kg}\) and \(m_{2}=27 \mathrm{~kg} ?\)

Short Answer

Expert verified
(a) 0.497 seconds (b) 2.98 seconds

Step by step solution

01

Determine the Angular Frequency

The formula for the angular frequency of a spring-mass system is given by \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass of the particle attached to the spring. Given that \( k = 120 \, \text{N/m} \), we can calculate \( \omega_1 \) for particle 1 and \( \omega_2 \) for particle 2 based on their respective masses.
02

Calculate Angular Frequency for Each Case

(a) For \( m_1 = m_2 = 3.0 \, \text{kg} \), the angular frequency for both particles is \( \omega = \sqrt{\frac{120}{3.0}} = \sqrt{40} = 6.32 \, \text{rad/s} \). (b) For \( m_1 = 3.0 \, \text{kg} \) and \( m_2 = 27 \, \text{kg} \), \( \omega_1 = 6.32 \, \text{rad/s} \) and \( \omega_2 = \sqrt{\frac{120}{27}} = \sqrt{4.44} = 2.11 \, \text{rad/s} \).
03

Understand the System's Behavior

Since both springs are ideal and the particles start oscillating opposite to each other, they will be side by side each time one completes a full oscillation from extreme to extreme. We calculate the periods of oscillation for each particle.
04

Calculate the Period of Oscillation

The period of oscillation \( T \) is given by \( T = \frac{2\pi}{\omega} \). (a) For both masses being equal, \( T_1 = T_2 = \frac{2\pi}{6.32} = 0.995 \, \text{s} \). (b) For different masses: - \( T_1 = \frac{2\pi}{6.32} = 0.995 \, \text{s} \)- \( T_2 = \frac{2\pi}{2.11} = 2.98 \, \text{s} \).
05

Find the First Common Moment Side by Side

(a) For identical masses, they have the same period, meaning they are side by side every half period, so the first time will be at \( 0.497 \, \text{s} \) (half of 0.995). (b) For different masses, seek the smallest positive time \( t \) where both \( \cos(\omega_1 t) = \cos(\omega_2 t) = 0 \). This implies finding the common multiple of their periods or other side-by-side condition: - Only at a common frequency cycle, i.e., first side-by-side at \( T \). Finding least common multiple (LCM) of periods directly needs precise judgment which solves to \( 3 \times 0.995 \approx 2.98 \).

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Key Concepts

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

Spring Constant
The spring constant, often denoted as \( k \), is a parameter that defines how stiff or flexible a spring is. It is measured in units of Newtons per meter (N/m). This parameter is crucial in determining how much force is needed to compress or stretch a spring by a certain distance. In simple terms, the larger the spring constant, the stiffer the spring. Conversely, a smaller spring constant means a more flexible spring.

In a mass-spring system, the spring constant directly affects the system's oscillatory behavior. For instance, in the given exercise, both springs have a constant \( k = 120 \, \text{N/m} \). This value is used in calculating the angular frequency, which directly influences the oscillation period. Therefore, understanding the spring constant helps predict how quickly the system will oscillate back to its equilibrium position after displacement.
Angular Frequency
Angular frequency, denoted by \( \omega \), tells us how quickly an object in a vibrating or oscillatory system completes its cycle. It is represented in radians per second (rad/s). For a mass-spring system, the formula to find the angular frequency is \( \omega = \sqrt{\frac{k}{m}} \), where \( k \) is the spring constant and \( m \) is the mass. This formula illustrates the dependency of angular frequency on these two variables.

  • If the mass \( m \) increases, the angular frequency decreases, indicating slower oscillation.
  • If the spring constant \( k \) increases, so does the angular frequency, indicating faster oscillation.

In the exercise, calculations for two scenarios are given: When masses are identical and when they differ. With \( m_1 = m_2 = 3.0 \, \text{kg} \), the angular frequency is \( 6.32 \, \text{rad/s} \). Alternatively, with \( m_1 = 3.0 \, \text{kg} \) and \( m_2 = 27 \, \text{kg} \), a marked difference in angular frequencies, \( 6.32 \, \text{rad/s} \) versus \( 2.11 \, \text{rad/s} \) is observed.
Oscillation Period
The oscillation period, symbolized as \( T \), defines how long it takes for an oscillating system to complete one full cycle. This is the reciprocal of the frequency \( f \). The period is calculated using \( T = \frac{2\pi}{\omega} \), where \( \omega \) is the angular frequency.

An oscillation period reflects:
  • For identical masses: the period \( T_1 = T_2 = 0.995 \, \text{s} \) suggests they are in sync.
  • For different masses, a notable variation: \( T_1 = 0.995 \, \text{s} \) and \( T_2 = 2.98 \, \text{s} \).
The differing periods indicate how quickly or slowly each mass reaches the equilibrium line (\( x = 0 \)). A shorter period means frequent returns to \( x = 0 \) per second, whereas a longer period denotes less frequent returns.
Mass-Spring System
A mass-spring system is a classic physics model used to study harmonic motion. It consists of a mass connected to a spring, where the mass can move back and forth while the spring either compresses or stretches. This system is often used to illustrate simple harmonic motion characterized by sinusoidal oscillations.

The behavior of a mass-spring system is governed by Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. Consequently, the system's oscillatory behavior depends on:
  • Spring constant \( k \): Determines the spring's stiffness.
  • Mass \( m \): Influences how inertia affects the return to equilibrium.
In the exercise, the mass-spring system is analyzed under two conditions: with identical and different values for mass \( m_1 \) and \( m_2 \). Under these scenarios, the particles oscillate to return to \( x = 0 \), shedding light on how both mass and spring constant dictate the dynamics of motion within a frictionless medium.

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Most popular questions from this chapter

A vertical spring with a spring constant of \(450 \mathrm{~N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is unstrained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by \(2.5 \mathrm{~cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm}\) ) above the compressed spring was the block dropped?

A spring is hung from the ceiling. A \(0.450-\mathrm{kg}\) block is then attached to the free end of the spring. When released from rest, the block drops \(0.150 \mathrm{~m}\) before momentarily coming to rest. (a) What is the spring constant of the spring? (b) Find the angular frequency of the block's vibrations.

A \(30.0-\mathrm{kg}\) block is resting on a flat horizontal table. On top of this block is resting a \(15.0-\) kg block, to which a horizontal spring is attached, as the drawing illustrates. The spring constant of the spring is \(325 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the lower block and the table is 0.600 , and the coefficient of static friction between the two blocks is 0.900 . A horizontal force \(\overrightarrow{\mathbf{F}}\) is applied to the lower block as shown. This force is increasing in such a way as to keep the blocks moving at a constant speed. At the point where the upper block begins to slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of the force \(\overrightarrow{\mathbf{F}}\).

Atoms in a solid are not stationary, but vibrate about their equilibrium positions. Typically, the frequency of vibration is about \(f=2.0 \times 10^{12} \mathrm{~Hz},\) and the amplitude is about \(1.1 \times 10^{-11} \mathrm{~m}\). For a typical atom, what is its (a) maximum speed and (b) maximum acceleration?

The length of a simple pendulum is \(0.79 \mathrm{~m}\) and the mass of the particle (the "bob") at the end of the cable is \(0.24 \mathrm{~kg}\). The pendulum is pulled away from its equilibrium position by an angle of \(8.50^{\circ}\) and released from rest. Assume that friction can be neglected and that the resulting oscillatory motion is simple harmonic motion. (a) What is the angular frequency of the motion? (b) Using the position of the bob at its lowest point as the reference level, determine the total mechanical energy of the pendulum as it swings back and forth. (c) What is the bob's speed as it passes through the lowest point of the swing?

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