/*! 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 48 Damping is negligible for a \(0.... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Damping is negligible for a \(0.150-\mathrm{kg}\) object hanging from a light \(6.30-\mathrm{N} / \mathrm{m}\) spring. A sinusoidal force with an amplitude of \(1.70 \mathrm{N}\) drives the system. At what frequency will the force make the object vibrate with an amplitude of \(0.440 \mathrm{m} ?\)

Short Answer

Expert verified
The frequency at which the force will make the object vibrate with an amplitude of \(0.440 \mathrm{m}\) is approximately \(1.42 \mathrm{Hz}\)

Step by step solution

01

Identify the given

Check the given values from the problem: mass \(m = 0.150 \mathrm{kg}\), spring constant \(k = 6.30 \mathrm{N/m}\), the amplitude of force applied \(F_{0} = 1.70 \mathrm{N}\), and the amplitude of vibration \(a = 0.440 \mathrm{m}\)
02

Substitute the knowns into the formula to solve for frequency

Insert the given quantities to the formula. Therefore, \(f = \frac{1}{2\pi}\sqrt{\frac{6.30\, \mathrm{N/m}}{0.150\, \mathrm{kg}} - \left(\frac{1.70\, \mathrm{N}}{0.150\, \mathrm{kg} \cdot 0.440\, \mathrm{m}}\right)^2}\)
03

Solve the equation

Calculating the above expression, we get \(f \approx 1.42 \mathrm{Hz}\)

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.

Sinusoidal Force
The application of a sinusoidal force in the study of physics relates to a type of periodic force that can lead to harmonic motion. This force has the unique property of being able to vary in a sine or cosine wave pattern over time. In the context of our exercise, the sinusoidal force is what drives the system—oscillating back and forth, thereby imparting energy onto the mass-spring system.

The effect of a sinusoidal force on the system's behavior can be visualized when you imagine pushing a child on a swing. If you apply pushes at a rate that matches the natural swing of the pendulum, the amplitude of the swing increases significantly. This concept is similar to what happens with the object attached to the spring when subjected to the sinusoidal force with the specified amplitude of 1.70 N. The synchronization of the force with the object's natural frequency is what we explore when seeking the 'resonance frequency' that maximizes the amplitude of vibration.
Frequency of Vibration
Understanding the frequency of vibration is crucial when dealing with harmonic motion. It represents how often the object completes a full cycle of motion in one second and is measured in hertz (Hz). In damped harmonic motion, we typically look for the frequency at which the system naturally vibrates when disturbed from its equilibrium position—the natural frequency.

When an external sinusoidal force has the same frequency as the system’s natural frequency, resonance occurs, leading to a significant increase in the amplitude of vibration. This effect can be particularly strong when damping forces, which resist motion, are negligible—as stated in the original problem. Nevertheless, the resonance frequency must be carefully calculated to ensure the object vibrates with the desired amplitude.
Spring Constant
The spring constant, denoted as 'k' in physics equations, is a measure of a spring's stiffness—how much force is required to stretch or compress it by a certain amount. It is a fundamental characteristic in the study of springs and harmonic motion because it determines the system's natural frequency. The spring constant is expressed in units of newtons per meter (N/m).

A higher spring constant indicates a stiffer spring that requires more force to achieve the same displacement. Conversely, a lower spring constant points to a more compliant spring. In our exercise, we're given a spring constant of 6.30 N/m, essential for calculating the natural frequency of the mass-spring system and ultimately solving for the frequency that matches the desired amplitude of vibration.
Amplitude of Vibration
The amplitude of vibration reflects the maximum displacement from the equilibrium position in a system undergoing harmonic motion. In the context of our exercise, it's of particular interest because it signifies the distance the mass moves from its rest position at the peak of its oscillation. It's directly related to the energy within the system; greater amplitude translates to higher levels of potential energy stored in the spring at maximum compression or extension.

When the sinusoidal force is at resonance with the system's natural frequency, this amplitude can reach a maximum based on the applied force's amplitude. The fact that the desired amplitude of vibration, 0.440 m in our case, can be achieved by tuning the frequency of the sinusoidal force underlines the delicate balance and interplay between all these variables in damped harmonic motion.

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 50.0 -g object connected to a spring with a force constant of \(35.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless surface with an amplitude of \(4.00 \mathrm{cm} .\) Find (a) the total energy of the system and (b) the speed of the object when the position is \(1.00 \mathrm{cm} .\) Find \((\mathrm{c})\) the kinetic energy and \((\mathrm{d})\) the potential energy when the position is \(3.00 \mathrm{cm} .\)

A simple pendulum has a mass of \(0.250 \mathrm{kg}\) and a length of \(1.00 \mathrm{m} .\) It is displaced through an angle of \(15.0^{\circ}\) and then released. What are (a) the maximum speed, (b) the maximum angular acceleration, and (c) the maximum restoring force? What If? Solve this problem by using the simple harmonic motion model for the motion of the pendulum, and then solve the problem more precisely by using more general principles.

After a thrilling plunge, bungee-jumpers bounce freely on the bungee cord through many cycles (Fig. P15.22). After the first few cycles, the cord does not go slack. Your little brother can make a pest of himself by figuring out the mass of each person, using a proportion which you set up by solving this problem: An object of mass \(m\) is oscillating freely on a vertical spring with a period \(T\). An object of unknown mass \(m^{\prime}\) on the same spring oscillates with a period.\(T^{\prime} .\) Determine (a) the spring constant and (b) the unknown mass.

Consider a bob on a light stiff rod, forming a simple pendulum of length \(L=1.20 \mathrm{m} .\) It is displaced from the vertical by an angle \(\theta_{\max }\) and then released. Predict the subsequent angular positions if \(\theta_{\max }\) is small or if it is large. Proceed as follows: Set up and carry out a numerical method to integrate the equation of motion for the simple pendulum: $$\frac{d^{2} \theta}{d t^{2}}=-\frac{g}{L} \sin \theta$$,Take the initial conditions to be \(\theta=\theta_{\max }\) and \(d \theta / d t=0\) at \(t=0 .\) On one trial choose \(\theta_{\max }=5.00^{\circ},\) and on another trial take \(\theta_{\max }=100^{\circ} .\) In each case find the position \(\theta\) as a function of time. Using the same values of \(\theta_{\max },\) compare your results for \(\theta\) with those obtained from \(\theta(t)=\) \(\theta_{\max } \cos \omega t .\) How does the period for the large value of \(\theta_{\max }\) compare with that for the small value of \(\theta_{\max } ?\) Note:Using the Euler method to solve this differential equation, you may find that the amplitude tends to increase with time. The fourth-order Runge-Kutta method would be a better choice to solve the differential equation. However, if you choose \(\Delta t\) small enough, the solution using Euler's method can still be good.

The mass of the deuterium molecule \(\left(\mathrm{D}_{2}\right)\) is twice that of the hydrogen molecule \(\left(\mathrm{H}_{2}\right) .\) If the vibrational frequency of \(\mathrm{H}_{2}\) is \(1.30 \times 10^{14} \mathrm{Hz},\) what is the vibrational frequency of \(\mathrm{D}_{2} ?\) Assume that the "spring constant" of attracting forces is the same for the two molecules.

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.