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Chapter 15: Problem 42

A body undergoes simple harmonic motion of amplitude \(4.25 \mathrm{~cm}\) and period \(0.200 \mathrm{~s}\). The magnitude of the maximum force acting on it is \(10.0 \mathrm{~N}\). (a) What is the mass? (b) If the oscillations are produced by a spring, what is the spring constant?

Short Answer

Expert verified
(a) Mass is approximately 0.02498 kg. (b) Spring constant is 24.750 N/m.

Step by step solution

01

Understanding Simple Harmonic Motion

Simple harmonic motion (SHM) can be described by the equation of motion: \[ x(t) = A \cos(\omega t + \phi) \]where \( A \) is the amplitude, \( \omega \) is the angular frequency, and \( \phi \) is the phase constant. The \( \omega \) is related to the period \( T \) of motion by \( \omega = \frac{2\pi}{T} \). For this problem, \( A = 4.25 \mathrm{~cm} = 0.0425 \mathrm{~m} \) and \( T = 0.200 \mathrm{~s} \).
02

Calculate Angular Frequency

Using the relationship between period and angular frequency:\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{0.200 \mathrm{~s}} = 31.4159 \mathrm{~rad/s} \]
03

Establish the Equation for Maximum Force

The maximum force \( F_{\text{max}} \) in SHM can be given by the equation:\[ F_{\text{max}} = m \omega^2 A \]where \( m \) is the mass we want to find.
04

Solve for Mass

We are given \( F_{\text{max}} = 10.0 \mathrm{~N} \). Substitute the known values:\[ 10.0 \mathrm{~N} = m \cdot (31.4159 \mathrm{~rad/s})^2 \cdot 0.0425 \mathrm{~m} \]Solving for \( m \), we have:\[ m = \frac{10.0}{31.4159^2 \cdot 0.0425} \approx 0.02498 \mathrm{~kg} \]
05

Use Hooke's Law to Determine the Spring Constant

If the oscillations are produced by a spring, we use Hooke's law for spring constant \( k \):\[ k = m \omega^2 \]Substitute the values:\[ k = 0.02498 \mathrm{~kg} \cdot (31.4159 \mathrm{~rad/s})^2 = 24.750 \mathrm{~N/m} \]

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Key Concepts

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

Amplitude
Amplitude in simple harmonic motion (SHM) refers to the maximum extent of oscillation measured from the equilibrium position. It is a crucial parameter that dictates how far the oscillating body moves on either side of its mean position. In the given exercise, the amplitude is specified as 4.25 cm, which equates to 0.0425 m.
Understanding amplitude helps in setting the boundaries of motion for systems like springs and pendulums in SHM.
  • Amplitude is a positive value and is often denoted by the letter \( A \).
  • The larger the amplitude, the more energetic the motion.
  • It plays a key role in determining the maximum potential energy stored in the system.
Angular Frequency
Angular frequency, denoted by \( \omega \), is a measure of how quickly an oscillating object moves through its motion cycle. It is fundamentally different from linear frequency, which measures cycles per second. Instead, angular frequency is expressed in radians per second (rad/s).
In SHM, angular frequency is directly related to the period \( T \) of the oscillation, which is the time to complete one cycle, through the equation \( \omega = \frac{2\pi}{T} \).
  • For the problem at hand, the period \( T \) is given as 0.200 s, leading to an angular frequency of 31.4159 rad/s.
  • Angular frequency helps determine how fast the system oscillates and affects the dynamic behavior of the system.
  • It is pivotal for calculating other parameters like velocity and acceleration in SHM.
Hooke's Law
Hooke's Law establishes a relationship between the force exerted by a spring and its displacement. It is a cornerstone concept in mechanics, especially within simple harmonic motion (SHM).
The mathematical expression for Hooke's Law is \( F = -kx \), where \( F \) is the force applied by the spring, \( k \) is the spring constant, and \( x \) is the displacement from the equilibrium position.
  • This law indicates that the force required to compress or extend a spring is proportional to the distance moved.
  • The negative sign signifies that the force exerted by the spring is in the opposite direction of the displacement.
  • Hooke's Law is applicable until the elastic limit of the material is reached.
Spring Constant
The spring constant, \( k \), in Hooke's Law and simple harmonic motion, quantifies the stiffness of a spring. It is expressed as force per unit length (N/m). The higher the value of \( k \), the stiffer the spring.
In the context of this exercise, the spring constant can be determined if we have knowledge of the mass \( m \) and the angular frequency \( \omega \) by using \( k = m \omega^2 \).
  • Given the mass and angular frequency, the spring constant was calculated to be 24.750 N/m.
  • The spring constant is crucial in understanding how much force is needed to achieve a certain displacement in the spring.
  • It also plays a significant role in the energy dynamics of SHM, contributing to both potential and total energy within the system.

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Most popular questions from this chapter

(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?

A \(5.00 \mathrm{~kg}\) object on a horizontal frictionless surface is attached to a spring with \(k=1000 \mathrm{~N} / \mathrm{m}\). The object is displaced from equilibrium \(40.0 \mathrm{~cm}\) horizontally and given an initial velocity of \(10.0 \mathrm{~m} / \mathrm{s}\) back toward the equilibrium position. What are (a) the motion's frequency, (b) the initial potential energy of the block-spring system, (c) the initial kinetic energy, and (d) the motion's amplitude?

An automobile can be considered to be mounted on four identical springs as far as vertical oscillations are concerned. The springs of a certain car are adjusted so that the oscillations have a frequency of \(3.00 \mathrm{~Hz}\). (a) What is the spring constant of each spring if the mass of the car is \(2110 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(85.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

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 \(y_{i}\) such that the spring is at its rest length. The object is then released from \(y_{i}\) and oscillates up and down, with its lowest position being \(10 \mathrm{~cm}\) below \(y_{i}\). (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 \(600 \mathrm{~g}\) 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) How far below \(y_{i}\) is the new equilibrium (rest) position with both objects attached to the spring?

What is the maximum acceleration of a platform that oscillates at amplitude \(2.50 \mathrm{~cm}\) and frequency \(6.60 \mathrm{~Hz}\) ?
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