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Chapter 16: Problem 7

7\. A Particle of Mass A particle with a mass of \(1.00 \times 10^{-20} \mathrm{~kg}\) is oscillating with simple harmonic motion with a period of \(1.00 \times 10^{-5} \mathrm{~s}\) and a maximum speed of \(1.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\). Calculate (a) the angular frequency and (b) the maximum displacement of the particle.

Short Answer

Expert verified
(a) \(6.28 \times 10^{5} \, \text{rad/s}\), (b) \(1.59 \times 10^{-3} \, \text{m}\)

Step by step solution

01

- Calculate the Angular Frequency

The angular frequency \(\omega\) can be calculated using the relation between period T and angular frequency. The formula is:\[ \omega = \frac{2\pi}{T} \]Given the period T is \(1.00 \times 10^{-5} \mathrm{~s}\), we substitute this value into the formula to get:\[ \omega = \frac{2\pi}{1.00 \times 10^{-5}} \approx 6.28 \times 10^{5} \text{ rad/s} \]
02

- Calculate the Maximum Displacement

The maximum displacement \(A\) can be determined using the relationship between maximum speed \(v_{max}\) and angular frequency \(\omega\). The formula is:\[ v_{max} = \omega A \]Rearranging this formula to solve for \(A\):\[ A = \frac{v_{max}}{\omega} \]Given the maximum speed \(v_{max}\) is \(1.00 \times 10^{3} \mathrm{~m/s}\) and we calculated \(\omega\) as \(6.28 \times 10^{5} \mathrm{~rad/s}\), we substitute these values into the formula to get:\[ A = \frac{1.00 \times 10^{3}}{6.28 \times 10^{5}} \approx 1.59 \times 10^{-3} \text{ m} \]

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Key Concepts

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

Angular Frequency
Angular frequency, usually denoted as \( \omega \), describes how rapidly an object oscillates in radians per second. It is a central concept in simple harmonic motion. To calculate angular frequency, we use the relationship:

  • The formula is \( \omega = \frac{2\pi}{T} \)

  • Here, \( T \) is the period, the time it takes to complete one oscillation


In the given problem, the period \( T \) is \( 1.00 \times 10^{-5} \) seconds. Substituting this into the formula, we find:

\[ \omega = \frac{2\pi}{1.00 \times 10^{-5}} \approx 6.28 \times 10^{5} \text{ rad/s} \]

Knowing the angular frequency helps us understand the speed of the oscillation cycles.
Period
The period, denoted as \( T \), is the duration of time it takes for one complete cycle of oscillation. It is inversely related to the angular frequency:

  • Period (\( T \)) is measured in seconds

  • The formula, \( T = \frac{2\pi}{\omega} \), shows this relationship mathematically


For the given particle, the period is \( 1.00 \times 10^{-5} \) seconds. This small value indicates a very fast oscillation frequency. Simply put, a shorter period means more cycles of oscillation per second.

Understanding period gives us an insight into how quickly or slowly a particle is moving back and forth in simple harmonic motion.
Maximum Displacement
Maximum displacement, also known as the amplitude and denoted as \( A \), is the peak value of the particle's displacement from its equilibrium (center) position. For simple harmonic motion, this can be calculated using the relationship between maximum speed and angular frequency:

  • The formula is \( v_{max} = \omega A \)

  • Rearranging to solve for \( A \), we get \[ A = \frac{v_{max}}{\omega} \]


Given:

  • Maximum speed \( v_{max} = 1.00 \times 10^{3} \text{ m/s} \)

  • Angular frequency \( \omega = 6.28 \times 10^{5} \text{ rad/s} \)

Substituting in the values, we get:

\[ A = \frac{1.00 \times 10^{3}}{6.28 \times 10^{5}} \approx 1.59 \times 10^{-3} \text{ m} \]

This value represents how far the particle moves from its central position at the extremes of its oscillatory motion.

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Most popular questions from this chapter

12\. A Body Oscillates A body oscillates with simple harmonic motion according to the equation $$ x=(6.0 \mathrm{~m}) \cos [(3 \pi \mathrm{rad} / \mathrm{s}) t+\pi / 3 \mathrm{rad}] $$ At \(t=2.0 \mathrm{~s}\), what are (a) the displacement, (b) the velocity, (c) the acceleration, and (d) the phase of the motion? Also, what are (e) the frequency and (f) the period of the motion?

6\. Spring Balance The scale of a spring balance that reads from 0 to \(15.0 \mathrm{~kg}\) is \(12.0 \mathrm{~cm}\) long. A package suspended from the balance is found to oscillate vertically with a frequency of \(2.00 \mathrm{~Hz} .\) (a) What is the spring constant? (b) How much does the package weigh?

8\. A Small Body A small body of mass \(0.12 \mathrm{~kg}\) is undergoing simple harmonic motion of amplitude \(8.5 \mathrm{~cm}\) and period \(0.20 \mathrm{~s}\) (a) What is the magnitude of the maximum force acting on it? (b) If the oscillations are produced by a spring, what is the spring constant?

21\. Massless Spring A massless spring hangs from the ceiling with a small object attached to its lower end. The object is initially held at rest in a position \(y_{1}\) such that the spring is at its rest length. The object is then released from \(y_{1}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{1}\). (a) What is the frequency of the oscillation? (b) What is the speed of the object when it is \(8.0 \mathrm{~cm}\) below the initial position? (c) An object of mass \(300 \mathrm{~g}\) is attached to the first object, after which the system oscillates with half the original frequency. What is the mass of the first object? (d) Relative to \(y_{1}\) where is the new equilibrium (rest) position with both objects attached to the spring?

55\. Block Suspended from Spring A \(4.0 \mathrm{~kg}\) block is suspended from a spring with a spring constant of \(500 \mathrm{~N} / \mathrm{m}\). A \(50 \mathrm{~g}\) bullet is fired into the block from directly below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and becomes embedded in the block. (a) Find the amplitude of the resulting simple harmonic motion. (b) What fraction of the original kinetic energy of the bullet is transferred to mechanical energy of the harmonic oscillator?
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