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A particle with a mass of \(3.00 \times 10^{-20} \mathrm{~kg}\) is oscillating with simple harmonic motion with a period of \(2.00 \times 10^{-5} \mathrm{~s}\) 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) Angular frequency is \(3.14 \times 10^5\, \mathrm{rad/s}\). (b) Maximum displacement is \(3.18 \times 10^{-3}\, \mathrm{m}\).

Step by step solution

01

Calculate Angular Frequency

Angular frequency \( \omega \) is related to the period \( T \) by the formula \[ \omega = \frac{2\pi}{T} \]We are given that the period \( T = 2.00 \times 10^{-5} \mathrm{~s} \). Substituting the given period into the formula gives:\[ \omega = \frac{2\pi}{2.00 \times 10^{-5}} = \frac{2\pi}{2 \times 10^{-5}} \]Calculating the value:\[ \omega = \frac{2 \times 3.14}{2 \times 10^{-5}} = \frac{6.28}{2 \times 10^{-5}} = 3.14 \times 10^{5} \ \mathrm{rad/s} \]
02

Relate Maximum Speed to Maximum Displacement

The maximum speed \( v_{max} \) of a particle undergoing simple harmonic motion is given by\[ v_{max} = \omega A \]where \( A \) is the maximum displacement (amplitude). We are given that \( v_{max} = 1.00 \times 10^3 \mathrm{~m/s} \), and we previously found \( \omega = 3.14 \times 10^5 \ \mathrm{rad/s} \). Substituting these into the equation gives:\[ 1.00 \times 10^3 = 3.14 \times 10^5 \times A \]
03

Solve for Maximum Displacement

Rearrange the equation from Step 2 to solve for \( A \):\[ A = \frac{v_{max}}{\omega} = \frac{1.00 \times 10^3}{3.14 \times 10^5} \]Calculating this gives:\[ A = \frac{1}{3.14} \times \frac{10^3}{10^5} = \frac{1}{3.14} \times 10^{-2} \approx 3.18 \times 10^{-3} \mathrm{~m} \]

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Key Concepts

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

Angular Frequency
Angular frequency, denoted as \( \omega \), is a fundamental aspect of simple harmonic motion. It describes how quickly an object oscillates. Angular frequency is expressed in radians per second (\( \text{rad/s} \)). It helps us understand how fast a system can vibrate back and forth. In simple harmonic motion, it can be calculated by using the period \( T \) of the oscillation with the formula:\[\omega = \frac{2\pi}{T}\]- **\( 2\pi \)**: This factor represents one complete cycle of circular motion in radians.- **Period \( T \)**: The time taken for one complete cycle of oscillation, expressed in seconds.For example, if a particle oscillates with a period \( T = 2.00 \times 10^{-5} \text{s} \), its angular frequency would be calculated as \( \omega = \frac{2\pi}{2.00 \times 10^{-5}} \) resulting in approximately \( 3.14 \times 10^5 \text{ rad/s} \). With this knowledge, systems in sports, music, and engineering can be analyzed by their oscillation speeds.
Period of Oscillation
The period of oscillation, symbolized as \( T \), is a crucial characteristic of oscillatory systems. It indicates the time duration required for a particle to complete one full oscillation cycle. Understanding the period helps in predicting the system's future state during its repetitive motion. In practical terms:
  • **Relationship with Frequency**: Period \( T \) is the inverse of frequency \( f \), given by \( T = \frac{1}{f} \). Frequency explains how many cycles occur in a unit of time, usually per second.
  • **Real-World Examples**: A pendulum swinging back and forth in a clock, daily oscillations in stock market graphs, and vibrating strings of musical instruments all have defined periods.
In our exercise, the period \( T = 2.00 \times 10^{-5} \) seconds means that each oscillation takes this specific amount of time to complete. Knowing the period, designers and engineers can calibrate systems like timing devices or electronic circuits effectively.
Maximum Displacement
Maximum displacement in simple harmonic motion refers to the furthest point from the equilibrium position that an oscillating object reaches. It is commonly termed as amplitude and signifies the object's energy during its motion. In the context of mathematical relations:- **Amplitude \( A \)**: This is the peak distance from the central point, usually measured in meters.- **Relation with Maximum Speed**: The maximum speed \( v_{max} \) an object can achieve during oscillation is linked to amplitude through angular frequency, expressed as \( v_{max} = \omega A \).For example, if a particle has a maximum speed \( v_{max} = 1.00 \times 10^3 \text{ m/s} \) with an angular frequency \( \omega = 3.14 \times 10^5 \text{ rad/s} \), then the maximum displacement \( A \) would be \( A = \frac{v_{max}}{\omega} \approx 3.18 \times 10^{-3} \text{ m} \).Understanding maximum displacement is crucial in evaluating the liveliness of motion systems, such as designing safe roller coasters or analyzing vibrating mechanical structures.

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Most popular questions from this chapter

The suspension system of a \(2400 \mathrm{~kg}\) automobile "sags" \(10 \mathrm{~cm}\) when the chassis is placed on it. Also, the oscillation amplitude decreases by \(50 \%\) each cycle. Estimate the values of (a) the spring constant \(k\) and (b) the damping constant \(b\) for the spring and shock absorber system of one wheel, assuming each wheel supports \(600 \mathrm{~kg}\).

In an electric shaver, the blade moves back and forth over a distance of \(2.0 \mathrm{~mm}\) in simple harmonic motion, with frequency 100 \(\mathrm{Hz}\). Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.

A block rides on a piston (a squat cylin- drical piece) that is moving vertically with Figure 15-40 simple harmonic motion. (a) If the SHM has \mathrm{~s}\(, at what amplitude of motion will the block and piston separate? (b) If the piston has an amplitude of \)3.0 \mathrm{~cm}$, what is the maximum frequency for which the block and piston will be in contact continuously?

A thin uniform rod (mass \(=0.90\) \(\mathrm{kg}\) ) 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 \(1.5 \mathrm{~s}\) 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) What is the oscillation amplitude of a \(4.00 \mathrm{~kg}\) box oscillating on a spring with spring constant \(100 \mathrm{~N} / \mathrm{m}\) if at time \(t=1.00 \mathrm{~s}\) the position is \(x=0.129 \mathrm{~m}\) and the velocity is \(v=5.00 \mathrm{~m} / \mathrm{s}\) ? At \(t=0\), what are (a) the position and (b) the velocity?

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