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A \(0.12 \mathrm{~kg}\) body undergoes 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?

Short Answer

Expert verified
(a) The maximum force is 3.22 N. (b) The spring constant is 118.80 N/m.

Step by step solution

01

Convert Units

First, let's convert the amplitude from centimeters to meters since SI units are required. Amplitude \(A = 8.5 \mathrm{~cm} = 0.085 \mathrm{~m}\).
02

Calculate Maximum Acceleration

For simple harmonic motion, the maximum acceleration \(a_{max}\) can be calculated using the formula \(a_{max} = \omega^2 A\), where \(\omega\) is the angular frequency. First, find \(\omega\) using \(\omega = \frac{2\pi}{T}\), where \(T = 0.20 \mathrm{~s}\). Thus, \(\omega = \frac{2\pi}{0.20} = 10\pi\) rad/s. Therefore, \(a_{max} = (10\pi)^2 \times 0.085 = 26.85 \mathrm{~m/s^2}\).
03

Calculate Maximum Force

The maximum force \(F_{max}\) can be determined using Newton's second law: \(F_{max} = m \cdot a_{max}\), where \(m = 0.12 \mathrm{~kg}\). So, \(F_{max} = 0.12 \cdot 26.85 = 3.22 \mathrm{~N}\).
04

Calculate Spring Constant

The spring constant \(k\) can be calculated using the formula relating period and spring constant: \(k = \frac{4\pi^2 m}{T^2}\). Substitute \(m = 0.12 \mathrm{~kg}\) and \(T = 0.20 \mathrm{~s}\) into the equation: \(k = \frac{4\pi^2 \times 0.12}{0.20^2} = 118.80 \mathrm{~N/m}\).

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Key Concepts

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

Understanding Maximum Force in Simple Harmonic Motion
When an object is oscillating in simple harmonic motion (SHM), the maximum force that acts on the object is at the points of maximum displacement from the equilibrium position. This is known as the maximum force, and it is a key component for understanding the dynamics of SHM.
To calculate the maximum force (\( F_{max} \)) on the object, we use Newton's second law of motion, which states that force is the product of mass and acceleration. For SHM, the maximum acceleration (\( a_{max} \)) is calculated using \( a_{max} = \omega^2 A \), where \( \omega \) is the angular frequency and \( A \) is the amplitude of oscillation.
Subsequently, the maximum force is determined by the formula:
  • \( F_{max} = m \cdot a_{max} \)
Here, \( m \) is the mass of the object. This relationship helps us determine how much force is involved at the most extreme points of the motion, which is crucial in applications like designing springs and damping systems.
Exploring the Spring Constant
The spring constant, denoted by \( k \), is a fundamental property of a spring that describes its stiffness. In the context of SHM, this constant is crucial for understanding how much force is needed to stretch or compress the spring by a certain amount.
A higher spring constant indicates a stiffer spring, meaning it requires more force to achieve the same displacement compared to a spring with a lower constant. It's calculated using the formula
  • \( k = \frac{4\pi^2 m}{T^2} \)
where \( m \) is the mass attached to the spring and \( T \) is the period of oscillation. This formula emerges from setting the equations of motion for harmonic oscillators, showing us how naturally the system oscillates. Understanding \( k \) is essential for designing systems like suspension in vehicles or measuring forces in lab instruments.
Unraveling Angular Frequency
Angular frequency (\( \omega \)) is a measure of how rapidly an object oscillates in a circular motion, and in the context of SHM, it is pivotal for determining the speed of oscillation.
Angular frequency is given by the formula
  • \( \omega = \frac{2\pi}{T} \)
where \( T \) is the period of oscillation. It is expressed in radians per second. This relationship ties the periodic nature of oscillation to the movement in radians, giving a clear view of how fast each cycle of motion occurs.
Understanding angular frequency not only enables us to calculate maximum speed and acceleration in SHM systems but also finds its applications in wave phenomena and electrical circuits, where frequency components need to be analyzed.
The Period of Oscillation Explained
The period of oscillation (\( T \)) is the time it takes to complete one full cycle of motion in SHM. It is a fundamental concept that defines how long it will take for an object to return to its original position after a complete cycle.
The period is inversely related to the frequency of oscillation, meaning that a longer period corresponds to a slower frequency and vice versa. This relationship is crucial because it allows for the determination of angular frequency, which in turn aids in computing maximum velocities and accelerations.
Periods are used in numerous applications, not just in engineering but also in natural sciences, where understanding the periodic nature of processes is vital. Through its relation with time, the period helps in synchronization of systems and in the analysis of wave patterns.

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Most popular questions from this chapter

A \(4.00 \mathrm{~kg}\) block hangs from a spring, extending it \(16.0 \mathrm{~cm}\) from its unstretched position. (a) What is the spring constant? (b) The block is removed, and a \(0.500 \mathrm{~kg}\) body is hung from the same spring. If the spring is then stretched and released, what is its period of oscillation?

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 \(50.0 \mathrm{~cm}\) 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 \(1450 \mathrm{~kg}\) and the mass is evenly distributed over the springs? (b) What will be the oscillation frequency if five passengers, averaging \(73.0 \mathrm{~kg}\) each, ride in the car with an even distribution of mass?

The balance wheel of an old-fashioned watch oscillates with angular amplitude \(\pi\) rad and period \(0.500 \mathrm{~s}\). Find (a) the maximum angular speed of the wheel, (b) the angular speed at displacement \(\pi / 2 \mathrm{rad}\), and \((\mathrm{c})\) the magnitude of the angular acceleration at displacement \(\pi / 4\) rad.

A \(1000 \mathrm{~kg}\) car carrying four \(82 \mathrm{~kg}\) people travels over a "washboard" dirt road with corrugations \(4.0 \mathrm{~m}\) apart. The car bounces with maximum amplitude when its speed is \(16 \mathrm{~km} / \mathrm{h}\). When the car stops, and the people get out, by how much does the car body rise on its suspension?

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