Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 3.0 -kg mass is vibrating on a spring. It has a resonant angular speed of \(2.4 \mathrm{rad} / \mathrm{s}\) and a damping angular speed of \(0.14 \mathrm{rad} / \mathrm{s}\). If the driving force is \(2.0 \mathrm{~N}\), find the maximum amplitude if the driving angular speed is (a) \(1.2 \mathrm{rad} / \mathrm{s},\) (b) \(2.4 \mathrm{rad} / \mathrm{s}\), and \((\) c) \(4.8 \mathrm{rad} / \mathrm{s}\).

Short Answer

Expert verified
Answer: The maximum amplitudes for the given driving angular speeds are approximately 0.187 m for 1.2 rad/s, 0.924 m for 2.4 rad/s, and 0.083 m for 4.8 rad/s.

Step by step solution

01

Identify the equation to use for maximum amplitude

The equation for the maximum amplitude of a damped harmonically driven mass on a spring is as follows: \(A_{max} = \frac{F_0}{m \sqrt{(\omega_0^2 - \omega^2)^2 + (2\beta\omega)^2}}\) Here, \(F_0\) is the driving force, \(m\) is the mass, \(\omega_0\) is the resonant angular speed, \(\omega\) is the driving angular speed, and \(\beta\) is the damping angular speed.
02

Plug in the given values

We have been given: - Mass, m = 3.0 kg - Resonant angular speed, \(\omega_0 = 2.4 \, \mathrm{rad/s}\) - Damping angular speed, \(\beta = 0.14 \, \mathrm{rad/s}\) - Driving force, \(F_0 = 2.0 \, \mathrm{N}\) We need to find the maximum amplitude for three driving angular speeds: 1. \(\omega = 1.2 \, \mathrm{rad/s}\) 2. \(\omega = 2.4 \, \mathrm{rad/s}\) 3. \(\omega = 4.8 \, \mathrm{rad/s}\)
03

Calculate the maximum amplitude for each driving angular speed

Now we can plug in the values and calculate the maximum amplitude for each driving angular speed: 1. For \(\omega = 1.2 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {1.2}^2)^2 + (2 \times 0.14 \times 1.2)^2}}\) \(A_{max} \approx 0.187 \, \mathrm{m}\) 2. For \(\omega = 2.4 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {2.4}^2)^2 + (2 \times 0.14 \times 2.4)^2}}\) \(A_{max} \approx 0.924 \, \mathrm{m}\) 3. For \(\omega = 4.8 \, \mathrm{rad/s}\): \(A_{max} = \frac{2.0}{3.0 \sqrt{({2.4}^2 - {4.8}^2)^2 + (2 \times 0.14 \times 4.8)^2}}\) \(A_{max} \approx 0.083 \, \mathrm{m}\)
04

Conclude the solution

Hence, the maximum amplitude for the given driving angular speeds are: 1. \(1.2 \, \mathrm{rad/s}\): \(\approx 0.187 \, \mathrm{m}\) 2. \(2.4 \, \mathrm{rad/s}\): \(\approx 0.924 \, \mathrm{m}\) 3. \(4.8 \, \mathrm{rad/s}\): \(\approx 0.083 \, \mathrm{m}\)

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 Vaia!

Key Concepts

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

Resonant Angular Speed
In the context of damped harmonic motion, the concept of resonant angular speed is vital. This is the frequency at which a system naturally oscillates if not disturbed by any external forces. When a system is driven at its resonant angular speed, it reaches its maximum amplitude. The resonant angular speed, denoted by \( \omega_0 \), is determined using the formula \( \omega_0 = \sqrt{\frac{k}{m}} \), where \(k\) is the spring constant and \(m\) is the mass of the oscillator.

Understanding this helps us predict how much energy is needed to achieve maximum amplitude. In our exercise, we observe that the resonant angular speed is 2.4 rad/s, which signifies that the system would naturally vibrate at this speed without any added damping or driving force.

Resonance can be both beneficial and potentially destructive. The efficiency of energy transfer is highest at resonance, making it useful in amplifying signals in musical instruments and electrical circuits. However, it can also cause catastrophic failures in structures due to excessive vibrations, so considering resonance properties is crucial in engineering.
Damping Angular Speed
Damping angular speed, represented by \(\beta\), describes how quickly a system loses energy due to damping forces. These forces might include friction or air resistance acting on the system. The damping angular speed is a measure of how fast the amplitude of oscillation decreases over time. To calculate this, we use \(\beta = \frac{b}{2m}\), where \(b\) is the damping coefficient and \(m\) is the mass.

In our example with a damping angular speed of 0.14 rad/s, it tells us how significantly damping affects the motion of the mass-spring system. As the damping speed increases, the system takes less time to come to rest after being disturbed, which means the oscillations die out faster.

Damping is essential in controlling systems. While it reduces motion, preventing resonance from causing damage, it also brings stability by keeping systems like car suspensions and building structures in check during environmental disturbances like strong winds or earthquakes.
Driving Force
The driving force is an external force applied to a system to keep it oscillating at a specific angular speed. It ensures that the oscillations continue, despite natural losses due to damping. In our scenario, the driving force is 2.0 N applied to keep the 3.0 kg mass vibrating at different angular speeds for maximum amplitude.

The concept of driving force comes into play when you need to sustain motion against damping effects. The balance between the driving force and the damping force determines the steady-state amplitude of the oscillatory system.
  • For slower angular speeds, such as 1.2 rad/s, the need for power from the driving force is lower, resulting in smaller amplitudes.
  • At resonant speed of 2.4 rad/s, the system achieves maximum efficiency and amplitude.
  • Beyond resonant speed, increasing to 4.8 rad/s, the amplitude drops as the system can no longer efficiently utilize the input energy.
This concept is visible in everyday devices like electric oscillators and pendulum clocks, where consistent external energy keeps the system in continuous, balanced motion without succumbing to damping.

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 massive object of \(m=5.00 \mathrm{~kg}\) oscillates with simple harmonic motion. Its position as a function of time varies according to the equation \(x(t)=2 \sin ([\pi / 2] t+\pi / 6)\). a) What is the position, velocity, and acceleration of the object at \(t=0 \mathrm{~s}^{2}\) b) What is the kinetic energy of the object as a function of time? c) At which time after \(t=0 \mathrm{~s}\) is the kinetic energy first at a maximum?

Two pendulums of identical length of \(1.000 \mathrm{~m}\) are suspended from the ceiling and begin swinging at the same time. One is at Manila, in the Philippines, where \(g=9.784 \mathrm{~m} / \mathrm{s}^{2}\) and the other is at Oslo, Norway, where \(g=9.819 \mathrm{~m} / \mathrm{s}^{2},\) After how many oscillations of the Manila pendulum will the two pendulums be in phase again? How long will it take for them to be in phase again?

Two springs, each with \(k=125 \mathrm{~N} / \mathrm{m}\), are hung vertically, and \(1.00-\mathrm{kg}\) masses are attached to their ends. One spring is pulled down \(5.00 \mathrm{~cm}\) and released at \(t=0\); the othen is pulled down \(4.00 \mathrm{~cm}\) and released at \(t=0.300 \mathrm{~s}\). Find the phase difference, in degrees, between the oscillations of the two masses and the equations for the vertical displacements of the masses, taking upward to be the positive direction.

The motion of a planet in a circular orbit about a star obeys the equations of simple harmonic motion. If the orbit is observed edge-on, so the planet's motion appears to be onedimensional, the analogy is quite direct: The motion of the planet looks just like the motion of an object on a spring a) Use Kepler's Third Law of planetary motion to determine the "spring constant" for a planet in circular orbit around a star with period \(T\) b) When the planet is at the extremes of its motion observed edge-on, the analogous "spring" is extended to its largest displacement. Using the "spring" analogy, determine the orbital velocity of the planet.

When the amplitude of the oscillation of a mass on a stretched string is increased, why doesn't the period of oscil lation also increase?

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.

Sign-up for free