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Two particles oscillate in simple harmonic motion along a common straight-line segment of length \(A\). Each particle has a period of \(1.5 \mathrm{~s}\), but they differ in phase by \(\pi / 6 \mathrm{rad}\). (a) How far apart are they (in terms of A) \(0.50 \mathrm{~s}\) after the lagging particle leaves one end of the path? (b) Are they then moving in the same direction, toward each other, or away from each other?

Short Answer

Expert verified
(a) Distance is \(A/4\). (b) Particles are moving in the same direction.

Step by step solution

01

Determine Angular Frequency

The angular frequency \( \omega \) for a simple harmonic oscillator is given by the formula \( \omega = \frac{2\pi}{T} \), where \( T \) is the period. For this problem, \( T = 1.5 \) seconds. Thus,\[\omega = \frac{2\pi}{1.5} = \frac{4\pi}{3}\, \text{rad/s}.\]
02

Write Position Equations

For two particles with simple harmonic motion, the position of each particle as a function of time can be expressed as follows:- First particle: \( x_1(t) = \frac{A}{2} \cos\left(\omega t \right) \)- Second particle (lagging by phase \( \pi/6 \) rad): \( x_2(t) = \frac{A}{2} \cos\left(\omega t - \frac{\pi}{6}\right) \) Here, \( A/2 \) is used as each particle moves from \( -A/2 \) to \( A/2 \), covering a total distance of \( A \).
03

Calculate Particle Positions at \( t = 0.5 \) s

Substitute \( t = 0.5 \) s and \( \omega = \frac{4\pi}{3} \) rad/s into the position equations: - First particle:\[x_1(0.5) = \frac{A}{2} \cos\left(\frac{4\pi}{3} \times 0.5 \right) = \frac{A}{2} \cos\left(\frac{2\pi}{3} \right)\]- Second particle:\[x_2(0.5) = \frac{A}{2} \cos\left(\frac{4\pi}{3} \times 0.5 - \frac{\pi}{6} \right) = \frac{A}{2} \cos\left(\frac{2\pi}{3} - \frac{\pi}{6}\right)\]
04

Simplify Cosine Terms

We calculate the trigonometric expressions for cosine:- \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \)- \( \frac{2\pi}{3} - \frac{\pi}{6} = \frac{4\pi}{6} - \frac{\pi}{6} = \frac{3\pi}{6} = \frac{\pi}{2} \)- \( \cos\left(\frac{\pi}{2}\right) = 0 \)
05

Calculate Distance Between Particles

Use the results from Step 4 to find the distance between the two particles:- First particle's position:\[x_1(0.5) = \frac{A}{2} \times \left(-\frac{1}{2}\right) = -\frac{A}{4}\]- Second particle's position:\[x_2(0.5) = \frac{A}{2} \times 0 = 0\]Distance:\[| x_1(0.5) - x_2(0.5) | = | -\frac{A}{4} - 0 | = \frac{A}{4}\]
06

Determine Particles' Direction of Motion

Differentiate the position equations to find velocities:- \( v_1(t) = -\frac{A}{2} \omega \sin\left(\omega t \right) \)- \( v_2(t) = -\frac{A}{2} \omega \sin\left(\omega t - \frac{\pi}{6}\right) \)At \( t = 0.5 \)s: - \( \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2} \), thus \( v_1(0.5) < 0 \) (moving left). - \( \sin\left(\frac{\pi}{2}\right) = 1 \), thus \( v_2(0.5) < 0 \) (also moving left).Both particles are moving in the same direction (to the left).

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Key Concepts

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

Angular Frequency
In simple harmonic motion, angular frequency is a crucial concept. It is the rate at which the particles oscillate back and forth through their equilibrium positions. It is denoted by the Greek letter \(\omega\) and is calculated using the expression \(\omega = \frac{2\pi}{T}\), where \(T\) is the oscillation period. This formula reveals that the angular frequency shows how quickly a particle completes one full cycle of motion.
For example, with a period of 1.5 seconds, the angular frequency becomes \( \omega = \frac{4\pi}{3} \text{ rad/s}\). This means that the particles complete their cycles sooner or later depending on this rate, which is fundamental in describing any oscillatory behavior, like waves, sound, and even light.
Phase Difference
Phase difference is another important aspect of simple harmonic motion. When two particles are oscillating along the same path, it's the angle by which one particle's position lags or leads the position of another. It's like saying both runners are on the same track but one's starting line is ahead of the other by a specific distance turned into an angle, measured in radians.
In the exercise, a phase difference of \( \pi/6 \) radians means one particle is "behind" the other by this amount of the oscillation path. This difference manifests as a shift in the wave representation of their motion and affects how often and when they cross paths or are apart. It's an essential feature to consider when predicting and calculating their positions at any given time during their shared oscillation path.
Particle Motion
Particle motion in simple harmonic motion can conveniently be tracked using equations of motion. Each particle's position as a function of time can be represented using the formula \( x(t) = X_0 \cos(\omega t + \phi) \), where \( X_0 \) is the amplitude, \(\omega\) the angular frequency, and \(\phi\) the phase angle.
In this case, the two particles move back and forth along a straight path from \( -\frac{A}{2} \) to \( \frac{A}{2} \). By substituting the angular frequency and the desired time into these equations, the exact positions of the particles can be determined. In our problem, after 0.5s, calculations showed one particle at \(-\frac{A}{4}\) while the other was at 0, meaning they were a distance of \(\frac{A}{4}\) apart. Understanding motion forms the basis of predicting how far apart or close the particles will be at any time.
Trigonometric Identities
Trigonometric identities play a significant role in simplifying and solving problems in simple harmonic motion. They allow us to manipulate equations involving functions like sine and cosine, which represent oscillating motions.
For example, in the exercise, we used the identity \( \cos\left(\theta - \phi\right) \) and solved expressions at specific angles such as \( \pi/2 \) and \( 2\pi/3 \). Identifying that \( \cos(\pi/2) = 0 \) and \( \cos(2\pi/3) = -1/2 \) allowed us to determine that one particle was at a position of \(-\frac{A}{4}\), while the other was at 0. Understanding these identities aids in breaking down more complex trigonometric relationships into manageable pieces, making the analysis of simple harmonic motion more straightforward.

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Most popular questions from this chapter

At a certain harbor, the tides cause the ocean surface to rise and fall a distance \(d\) (from highest level to lowest level) in simple harmonic motion, with a period of \(12.5 \mathrm{~h}\). How long does it take for the water to fall a distance \(0.250 d\) from its highest level?

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.

A block weighing \(10.0 \mathrm{~N}\) is attached to the lower end of a vertical spring \((k=200.0 \mathrm{~N} / \mathrm{m})\), the other end of which is attached to a ceiling. The block oscillates vertically and has a kinetic energy of \(2.00 \mathrm{~J}\) as it passes through the point at which the spring is unstretched. (a) What is the period of the oscillation? (b) Use the law of conservation of energy to determine the maximum distance the block moves both above and below the point at which the spring is unstretched. (These are not necessarily the same.) (c) What is the amplitude of the oscillation? (d) What is the maximum kinetic energy of the block as it oscillates?

A \(3.0 \mathrm{~kg}\) particle is in simple harmonic motion in one dimension and moves according to the equation $$ x=(5.0 \mathrm{~m}) \cos [(\pi / 3 \mathrm{rad} / \mathrm{s}) t-\pi / 4 \mathrm{rad}] $$ with \(t\) in seconds. (a) At what value of \(x\) is the potential energy of the particle equal to half the total energy? (b) How long does the particle take to move to this position \(x\) from the equilibrium position?

A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance \(d\) from the 50 \(\mathrm{cm}\) mark. The period of oscillation is \(2.5 \mathrm{~s}\). Find \(d\).

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