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Magazine Chapter 13: Problem 18
A 0.500 \(\mathrm{kg}\) mass on a spring has velocity as a function of time given by \(v_{x}(t)=(3.60 \mathrm{cm} / \mathrm{s}) \sin \left[\left(4.71 \mathrm{s}^{-1}\right) t-\pi / 2\right]\) . What are (a) the period; (b) the amplitude; (c) the maximum acceleration of the mass; (d) the force constant of the spring?
Short Answer
Expert verified
Period: 1.333 s; Amplitude: 0.764 cm; Max Acceleration: 16.9 cm/s虏; Force Constant: 11.1 N/m.
Step by step solution
01
Identify the Angular Frequency
The given velocity function is \( v_x(t) = (3.60 \ \mathrm{cm/s}) \sin [(4.71 \,\mathrm{s}^{-1}) \cdot t - \pi/2] \). The term in front of \( t \), which is 4.71 s\(^{-1}\), is the angular frequency \( \omega \).
02
Calculate the Period
The period \( T \) of the motion is related to the angular frequency by the formula \( T = \frac{2\pi}{\omega} \). Substituting \( \omega = 4.71 \,\mathrm{s}^{-1} \), we get:\[ T = \frac{2\pi}{4.71\,\mathrm{s}^{-1}} \approx 1.333\,\mathrm{s} \]
03
Determine the Amplitude of Velocity
The amplitude of the velocity, given by the sine term, is 3.60 cm/s, as it is the coefficient in front of the sine function in the velocity expression.
04
Relate Velocity Amplitude to Amplitude of Motion
The amplitude of motion \( A \) is related to the amplitude of velocity \( v_{max} \) by the formula \( v_{max} = \omega A \). Solving for \( A \), we substitute \( v_{max} = 3.60 \,\mathrm{cm/s} \) and \( \omega = 4.71 \,\mathrm{s}^{-1} \):\[ A = \frac{v_{max}}{\omega} = \frac{3.60\,\mathrm{cm/s}}{4.71\,\mathrm{s}^{-1}} \approx 0.764\,\mathrm{cm} \]
05
Calculate Maximum Acceleration
The maximum acceleration \( a_{max} \) is given by \( a_{max} = \omega^2 A \). Using \( \omega = 4.71 \,\mathrm{s}^{-1} \) and \( A = 0.764 \,\mathrm{cm} \), we find:\[ a_{max} = (4.71\,\mathrm{s}^{-1})^2 \times 0.764\,\mathrm{cm} = 16.9\,\mathrm{cm/s}^2 \]
06
Determine the Force Constant of the Spring
The force constant \( k \) is related to the mass and angular frequency by \( k = m\omega^2 \). Using \( m = 0.500 \,\mathrm{kg} \) and \( \omega = 4.71\,\mathrm{s}^{-1} \), we calculate:\[ k = 0.500\,\mathrm{kg} \times (4.71\,\mathrm{s}^{-1})^2 \approx 11.1\,\mathrm{N/m} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Angular Frequency
Angular frequency is a key concept in harmonic motion as it describes how quickly an object oscillates over time. In the given problem, the angular frequency is represented by the term **4.71 s**-1 in the sine function. This value indicates the rate of oscillation. The unit, **s**-1, also known as radians per second, measures how many full cycles the motion completes each second.
Angular frequency is mathematically represented by the symbol \( \omega \). It plays a crucial role in determining other features of the motion, such as the period and acceleration.
Angular frequency is mathematically represented by the symbol \( \omega \). It plays a crucial role in determining other features of the motion, such as the period and acceleration.
- Higher angular frequency implies more oscillations per second.
- Lower angular frequency means fewer oscillations per second.
Period of Motion
The period of motion refers to the time it takes for an object in harmonic motion to complete one full cycle of movement. It is symbolized by \( T \) and is directly related to the angular frequency by the formula \( T = \frac{2\pi}{\omega} \).
In this exercise, we calculate the period using the angular frequency **4.71 s**-1:
\[T = \frac{2\pi}{4.71\,\mathrm{s}^{-1}} \approx 1.333\,\mathrm{s}\]
This calculation shows that it takes approximately 1.333 seconds for the mass-spring system to complete one oscillation.
This concept is fundamental because it defines how long one cycle lasts in a repeating sequence of movement. Changes in period affect how fast or slow the motion appears over time, providing insights into the dynamical properties of the system.
In this exercise, we calculate the period using the angular frequency **4.71 s**-1:
\[T = \frac{2\pi}{4.71\,\mathrm{s}^{-1}} \approx 1.333\,\mathrm{s}\]
This calculation shows that it takes approximately 1.333 seconds for the mass-spring system to complete one oscillation.
This concept is fundamental because it defines how long one cycle lasts in a repeating sequence of movement. Changes in period affect how fast or slow the motion appears over time, providing insights into the dynamical properties of the system.
Maximum Acceleration
Maximum acceleration is the peak rate of change in velocity for the object in motion, representing the greatest extent to which the object speeds up or slows down in its path. In harmonic motion, this maximum is reached when the displacement is also at a maximum, due to the restoring force's influence.
To calculate maximum acceleration \( a_{max} \), we use the formula:
\[a_{max} = \omega^2 A\]where \( A \) is the amplitude of motion and \( \omega \) is the angular frequency. From the problem, substituting values gives:
\[a_{max} = (4.71\,\mathrm{s}^{-1})^2 \times 0.764\,\mathrm{cm} \approx 16.9\,\mathrm{cm/s}^2\]
This indicates that the greatest acceleration experienced by the mass occurs when it is in the extremities of its motion path.
Understanding the maximum acceleration helps in analyzing the forces that act upon the system, especially in predicting how vigorously or lightly the system responds to displacement.
To calculate maximum acceleration \( a_{max} \), we use the formula:
\[a_{max} = \omega^2 A\]where \( A \) is the amplitude of motion and \( \omega \) is the angular frequency. From the problem, substituting values gives:
\[a_{max} = (4.71\,\mathrm{s}^{-1})^2 \times 0.764\,\mathrm{cm} \approx 16.9\,\mathrm{cm/s}^2\]
This indicates that the greatest acceleration experienced by the mass occurs when it is in the extremities of its motion path.
Understanding the maximum acceleration helps in analyzing the forces that act upon the system, especially in predicting how vigorously or lightly the system responds to displacement.
Force Constant
The force constant, symbolized as \( k \), is a measure of the stiffness of the spring in a harmonic oscillator. It describes how much force is required to stretch or compress the spring by a unit distance. A larger force constant means a stiffer spring.
For this system, the force constant is determined using the formula:
\[k = m\omega^2\]where \( m \) is the mass attached to the spring, and \( \omega \) is the angular frequency. The given mass is **0.500 kg** and the angular frequency is **4.71 s**-1. The calculation is as follows:
\[k = 0.500\,\mathrm{kg} \times (4.71\,\mathrm{s}^{-1})^2 \approx 11.1\,\mathrm{N/m}\]
This value tells us about the resistance of the spring to deformation. A higher value of \( k \) implies that the spring is less easily compressed or extended, which directly influences how the overall system vibrates.
For this system, the force constant is determined using the formula:
\[k = m\omega^2\]where \( m \) is the mass attached to the spring, and \( \omega \) is the angular frequency. The given mass is **0.500 kg** and the angular frequency is **4.71 s**-1. The calculation is as follows:
\[k = 0.500\,\mathrm{kg} \times (4.71\,\mathrm{s}^{-1})^2 \approx 11.1\,\mathrm{N/m}\]
This value tells us about the resistance of the spring to deformation. A higher value of \( k \) implies that the spring is less easily compressed or extended, which directly influences how the overall system vibrates.
