/*! 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 76 A \(55.0 \mathrm{~g}\) block osc... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(55.0 \mathrm{~g}\) block oscillates in SHM on the end of a spring with \(k=1500 \mathrm{~N} / \mathrm{m}\) according to \(x=x_{m} \cos (\omega t+\phi) .\) How long does the block take to move from position \(+0.800 x_{m}\) to (a) position \(+0.600 x_{m}\) and (b) position \(-0.800 x_{n}\) ?

Short Answer

Expert verified
The time from \(+0.800x_m\) to \(+0.600x_m\) is \(0.00171\ s\), and to \(-0.800x_m\) is \(0.0112\ s\).

Step by step solution

01

Identify the Key Variables

We have a block with mass \( m = 55.0 \, \text{g} = 0.055 \, \text{kg} \), and it is attached to a spring with spring constant \( k = 1500 \, \text{N/m} \). The position function is given by \( x = x_m \cos(\omega t + \phi) \). We need to find the positions \( +0.800x_m \) and \( +0.600x_m \) for part (a), and from \( +0.800x_m \) to \( -0.800x_m \) for part (b).
02

Calculate the Angular Frequency \( \omega \)

The angular frequency \( \omega \) is determined by the formula \( \omega = \sqrt{\frac{k}{m}} \). Substituting the given values, we have \( \omega = \sqrt{\frac{1500}{0.055}} \approx 165.6 \, \text{rad/s} \).
03

Set Up the Equation for Phase Angles

For position \( x = 0.800x_m \), \( x = x_m \cos(\omega t + \phi) \), thus \( \cos(\omega t_1 + \phi) = 0.800 \). Similar steps for position \( x = 0.600x_m \), gives \( \cos(\omega t_2 + \phi) = 0.600 \) for part (a). For part (b), position \( x = -0.800x_m \) gives \( \cos(\omega t_3 + \phi) = -0.800 \).
04

Calculate the Time Intervals for Part (a)

For the interval from \( +0.800x_m \) to \( +0.600x_m \), we calculate the difference in time. Solving \( \cos^{-1}(0.800) - \cos^{-1}(0.600) \), we find the angles \( \theta_1 = \cos^{-1}(0.800) \approx 0.6435 \, \text{rad} \) and \( \theta_2 = \cos^{-1}(0.600) \approx 0.9273 \, \text{rad} \). The time interval \( \Delta t = \frac{\theta_2 - \theta_1}{\omega} \approx \frac{0.9273 - 0.6435}{165.6} \approx 0.00171 \, \text{s} \).
05

Calculate the Time Intervals for Part (b)

For the interval from \( +0.800x_m \) to \(-0.800x_m \), we solve \( \cos^{-1}(-0.800) \approx 2.4981 \, \text{rad} \). The time interval is \( \Delta t = \frac{\theta_3 - \theta_1}{\omega} \approx \frac{2.4981 - 0.6435}{165.6} \approx 0.0112 \, \text{s} \).

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Key Concepts

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

Angular Frequency
In the study of simple harmonic motion, angular frequency is a key component that defines the oscillation rate of an object attached to a spring. Angular frequency, denoted as \(\omega\), describes how fast an object moves through its oscillatory cycle. It is measured in radians per second (rad/s).
To calculate \(\omega\), we use the formula: \(\omega = \sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass of the object. In the example of a 55.0 g block and a spring with \(k = 1500\) N/m, the angular frequency is computed as \(\omega = \sqrt{\frac{1500}{0.055}} \approx 165.6 \, \text{rad/s}\).
Understanding \(\omega\) is crucial because it tells us how rapidly the block moves back and forth, thus influencing how quickly it covers specific distances over time. A higher angular frequency means the object completes its oscillations faster, whereas a lower angular frequency implies a slower oscillation movement.
Phase Angle
The phase angle is a part of the equation used to describe the position of an oscillating object: \(x = x_m \cos(\omega t + \phi)\). In this equation:\(\phi\) represents the phase angle.
The phase angle indicates the initial angle of the cosine function when \(t = 0\). It essentially sets the start point of the oscillation cycle. This concept is significant because it helps describe not only the position but also the direction and timing of the oscillation at any given moment.
A phase angle is typically determined by the initial conditions of the problem, such as where the block starts its oscillation cycle. Changes in the phase angle translate to shifts in the waveform along the time axis, giving different scenarios for the beginning and subsequent path of motion.
Time Interval
The time interval in the context of simple harmonic motion refers to the time it takes for an object to move between two specified points in its oscillation cycle. Calculating this interval involves knowing the function of motion, \(x = x_m \cos(\omega t + \phi)\), and applying specific steps to determine when the object passes particular positions.
For the block moving from \(+0.800x_m\) to \(+0.600x_m\): we find the time difference \(\Delta t\) by first determining the angles using the inverse cosine function: \(\theta_1 = \cos^{-1}(0.800)\) and \(\theta_2 = \cos^{-1}(0.600)\). The formula \(\Delta t = \frac{\theta_2 - \theta_1}{\omega}\) helps us find the time interval, showing how slight changes in position correlate with the time taken.
Such calculations provide essential insights into the dynamics of oscillatory systems. By understanding the intervals, one can predict how long it will take for an object to reach another location in its oscillation path, enhancing our grasp of the time-based behavior of these systems.

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Most popular questions from this chapter

A loudspeaker produces a musical sound by means of the oscillation of a diaphragm whose amplitude is limited to \(1.00 \mu \mathrm{m}\). (a) At what frequency is the magnitude \(a\) of the diaphragm's acceleration equal to \(g\) ? (b) For greater frequencies, is \(a\) greater than or less than \(g\) ?

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?

A pendulum is formed by pivoting a long thin rod about a point on the rod. In a series of experiments, the period is measured as a function of the distance \(x\) between the pivot point and the rod's center. (a) If the rod's length is \(L=2.20 \mathrm{~m}\) and its mass is \(m=22.1 \mathrm{~g}\), what is the minimum period? (b) If \(x\) is chosen to minimize the period and then \(L\) is increased, does the period increase, decrease, or remain the same? (c) If, instead, \(m\) is increased without \(L\) increasing, does the period increase, decrease, or remain the same?

A thin uniform rod (mass \(=0.50 \mathrm{~kg}\) ) swings about an axis that passes through one end of the rod and is perpendicular to the plane of the swing. The rod swings with a period of \(1.5\) s and an angular amplitude of \(10^{\circ} .\) (a) What is the length of the rod? (b) What is the maximum kinetic energy of the rod as it swings?

A \(4.00 \mathrm{~kg}\) block hangs from a spring, extending it \(16.0 \mathrm{~cm}\) from its unstretched position. (a) What is the spring constant? (b) The block is removed, and a \(0.500 \mathrm{~kg}\) body is hung from the same spring. If the spring is then stretched and released, what is its period of oscillation?

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