/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 22 A 200 -g mass is attached to a s... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A 200 -g mass is attached to a spring of constant \(k=5.6 \mathrm{N} / \mathrm{m}\) and set into oscillation with amplitude \(A=25 \mathrm{cm} .\) Determine (a) the frequency in hertz, (b) the period, (c) the maximum velocity, and (d) the maximum force in the spring.

Short Answer

Expert verified
To solve this problem, apply the and use the equations relating to simple harmonic motion. Compute the frequency (\(f\)), the period (\(T\)), the maximum velocity (\(V_{max}\)), and the maximum force (\(F_{max}\)) using the formulas mentioned and the given values for the mass, spring constant and amplitude. Do not forget to convert quantities to appropriate units before using them.

Step by step solution

01

Calculate the frequency

The formula for the frequency \(f\) of a spring-mass system is given by \[ f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \]Where \(k\) = 5.6 N/m is the spring constant and \(m\) = 0.2 kg is the mass attached to the spring (200 g converted into kg). Substitute these values into the formula to find the frequency \(f\).
02

Calculate the period

The period \(T\) of oscillation is the reciprocal of frequency. So, \[ T = \frac{1}{f} \]Substitute the calculated frequency \(f\) into this equation to get the period \(T\).
03

Calculate the maximum velocity

The formula for maximum velocity \(V_{max}\) of a spring-mass system undergoing Simple Harmonic Motion (SHM) is \[V_{max} = 2 \pi f A \]Where \(A\) = 0.25 m is the amplitude given (25 cm converted into m). Substituting the known values of \(f\) - frequency and \(A\) into the equation, solve for \(V_{max}\).
04

Calculate the Maximum Force

The maximum force \(F_{max}\) in the spring when stretched or compressed by amplitude \(A\) is\[ F_{max} = kA \]Substituting the given values of spring constant \(k\) and amplitude \(A\) into the equation gives the maximum force \(F_{max}\) in the spring.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Frequency of Oscillation
Frequency refers to how often an oscillating system, such as a spring-mass system, cycles through its motion in one second. It is measured in hertz (Hz). For a spring-mass system, frequency is determined by both the stiffness of the spring, represented by the spring constant \(k\), and the mass \(m\) attached to it.
The formula to calculate the frequency \(f\) for our spring-mass system is:
  • \(f = \frac{1}{2\pi}\sqrt{\frac{k}{m}}\)
This equation shows that a stiffer spring (higher \(k\)) or a lighter mass (lower \(m\)) will lead to a higher frequency. In other words, the system will oscillate more times per second.
To fully understand and calculate the frequency, always ensure that the mass is converted to kilograms and that the spring constant is in Newtons per meter.
Spring-Mass System
In the context of simple harmonic motion, a spring-mass system describes a setup where a mass is connected to a spring and allowed to oscillate. This oscillation is a classic example of simple harmonic motion (SHM).
The spring-mass system can easily be represented by imagining a spring hanging vertically or lying horizontally with a mass attached to its end. When the mass is disturbed from its equilibrium position, it begins to oscillate back and forth.
Key factors to remember:
  • The spring's stiffness is represented by its spring constant \(k\), which dictates how much force is needed to stretch or compress the spring by a certain distance.
  • The mass \(m\) attached to the spring influences the system's natural frequency of oscillation.
  • The simple harmonic motion in a spring-mass system is frictionless and idealized in many exercises, meaning energy is conserved within the motion.
Hooke's Law
Hooke's Law is fundamental to understanding spring behavior. It states that the force exerted by a spring is directly proportional to the distance it is stretched or compressed, within the elastic limit.
Mathematically, Hooke's Law is expressed as:
  • \( F = -kx \)
Where \(F\) is the force exerted on the spring, \(k\) is the spring constant, and \(x\) is the displacement from the spring's equilibrium position.
The negative sign indicates that the force exerted by the spring is always in the opposite direction of the displacement. This means if you compress or stretch a spring, it will push back, trying to return to its original equilibrium length.
Understanding Hooke’s Law is crucial for solving problems related to springs, including calculations of maximum force when stretched or compressed by a certain distance.
Amplitude and Period
Amplitude and period are key characteristics of oscillatory motion. Amplitude is the maximum extent of the oscillation from its equilibrium position. In our spring-mass system, it indicates how far the mass moves from this equilibrium.
The amplitude is typically measured in meters and defines the energy in the oscillating system: a larger amplitude means the mass moves farther from the equilibrium position on each cycle.The period of oscillation \(T\) refers to the time it takes for one complete cycle of motion. For simple harmonic motion, it is related to the frequency by:
  • \( T = \frac{1}{f} \)
Period is measured in seconds and is inversely proportional to the frequency: as frequency increases, the period decreases since the system is completing more cycles per second. Recognizing the relationship between amplitude and period helps in identifying how an oscillating system behaves and evolves over time.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A simple model for a variable star considers that the outer layer of the star is subject to two forces: the inward force of gravity and the outward force due to gas pressure. As a result, Newton's law for the star's outer layer reads \(m d^{2} r / d t^{2}=4 \pi r^{2} p-G M m / r^{2} .\) Here \(m\) is the mass of the outer layer, \(M\) is the total mass of the star, \(r\) is the star's radius, and \(p\) is the pressure. (a) Use this equation to show that the star's equilibrium pressure and radius are related by \(p_{0}=G M m / 4 \pi r_{0}^{4},\) where the subscript 0 represents equilibrium values. (b) As you'll learn in Chapter 18 , gas pressure and volume \(V\left(=\frac{4}{3} \pi r^{3}\right)\) are related by \(p V^{3 / 3}=p_{0} V_{0}^{5 / 3}\) (this is for an adiabatic process, a good approximation here, and the exponent \(5 / 3\) reflects the ionized gas that makes up the star). Let \(x=r-r_{0}\) be the displacement of the star's surface from equilibrium. Use the binomial approximation (Appendix A) to show that, when \(x\) is small compared with \(r,\) the righthand side of the above equation can be written \(-\left(G M m / r_{0}^{3}\right) x\) (c) since \(r\) and \(x\) differ only by a constant, the term \(d^{2}\) r/dt \(^{2}\) in the equation above can also be written \(d^{2} x / d t^{2} .\) Make this substitution, along with substituting the result of part (b) for the right- hand side, and compare your result with Equations 13.2 and 13.7 to find an expression for the oscillation period of the star. (d) What does your simple model predict for the period of the variable star Delta Cephei, with radius 44.5 times that of the Sun and mass of 4.5 Sun masses? (Your answer overestimates the actual period by a factor of about \(3,\) both because of oversimplified physics and because changes in the star's radius are too large for the assumption of a linear restoring force.)

Two identical mass-spring systems consist of \(430-\mathrm{g}\) masses on springs of constant \(k=2.2 \mathrm{N} / \mathrm{m} .\) Both are displaced from equilibrium, and the first is released at time \(t=0 .\) How much later should the second be released so their oscillations differ in phase by \(\pi / 2 ?\)

Two mass-spring systems have the same mass and the same total energy. The amplitude of system 1 is twice that of system \(2 .\) How do (a) their frequencies and (b) their maximum accelerations compare?

At the heart of a grandfather clock is a simple pendulum \(1.45 \mathrm{m}\) long; the clock ticks each time the pendulum reaches its maximum displacement in either direction. What's the time interval between ticks?

A doctor counts 68 heartbeats in 1.0 minute. What are the corresponding period and frequency?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.