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A linear oscillator with mass \(m\), spring constant \(k\) and viscous damping coefficient \(c\) is set into motion when released from a displaced position. Derive an expression for the energy loss \(Q\) during one complete cycle in terms of the amplitude \(x_{1}\) at the start of the cycle. (See Fig. \(8 / 7\).)

Short Answer

Expert verified
The energy loss per cycle is given by \(Q = c x_1^2 \omega \pi\).

Step by step solution

01

Identify the Energy Dissipated

The energy dissipation in each cycle is due to the damping force, which is proportional to the velocity and the damping coefficient. The energy dissipation over a period for a damping force is given by:\[Q = \int_0^T c \left( \frac{dx(t)}{dt} \right)^2 dt\]where \(T\) is the time period of one complete cycle.
02

Express Velocity in Terms of Amplitude

For a harmonic oscillator with displacement \(x(t) = x_1 \cos(\omega t)\), the velocity is the derivative of displacement with respect to time:\[\frac{dx(t)}{dt} = -x_1 \omega \sin(\omega t)\]where \(\omega\) is the damped angular frequency. In a lightly damped system, \(\omega \approx \sqrt{\frac{k}{m}}\).
03

Substitute Velocity into Energy Dissipation Equation

Substitute the expression for velocity into the energy dissipation equation:\[Q = \int_0^T c (-x_1 \omega \sin(\omega t))^2 dt = c x_1^2 \omega^2 \int_0^T \sin^2(\omega t) dt\]
04

Simplify the Integral

Use the trigonometric identity \(\sin^2(\omega t) = \frac{1}{2} (1 - \cos(2\omega t))\) to simplify the integral:\[\int_0^T \sin^2(\omega t) dt = \frac{1}{2} \int_0^T (1 - \cos(2\omega t)) dt\]The integral evaluates to:\[\frac{1}{2} \left[ T - \frac{1}{2\omega} \sin(2\omega t) \right]_0^T = \frac{T}{2}\] because \(\sin(2\omega T)\) completes a full cycle, thus equals zero.
05

Calculate Energy Dissipation per Cycle

Substitute \(\frac{T}{2}\) back into the energy dissipation expression:\[Q = c x_1^2 \omega^2 \left( \frac{T}{2} \right)\]For a lightly damped system, the period \(T \approx \frac{2\pi}{\omega}\). Substitute this into the expression:\[Q = c x_1^2 \omega^2 \left( \frac{2\pi}{2\omega} \right) = c x_1^2 \omega \pi\]Thus, the energy loss per cycle is directly proportional to the square of the amplitude, the damping coefficient, and the angular frequency.

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Key Concepts

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

Damped Harmonic Oscillator
A damped harmonic oscillator is a system where a mass is attached to a spring and experiences a force that opposes its motion, in addition to the restoring spring force. This opposition to motion is due to damping, which is essentially a result of friction or resistance within the system. The damping force acts in such a way to reduce the amount of oscillation over time.

This type of oscillator doesn't oscillate indefinitely at constant amplitude because some of the system's energy is dissipated in each cycle. In an ideal frictionless environment, all energy would be conserved and the oscillator would continue oscillating indefinitely. However, in real-world scenarios, damping causes the amplitude to gradually decrease until motion ceases.
  • Energy loss in damping is usually due to resistive forces like air resistance or internal material friction.
  • Damping can vary in intensity and can be classified into underdamping, critical damping, and overdamping.
Viscous Damping
Viscous damping is a common mechanism for dissipating energy in an oscillator, where the damping force is proportional to the velocity of the mass. This means the faster the object moves, the stronger the damping force pushing against it. The proportionality constant in viscous damping is known as the damping coefficient, denoted by "c."

In the formula for energy dissipation, the damping force can be written as \(-c \frac{dx}{dt}\). The presence of this force impacts the system's motion, described by the equation \(m \frac{d^2x}{dt^2} + c \frac{dx}{dt} + kx = 0\). Here, the damping term \(c \frac{dx}{dt}\) is crucial for calculating energy loss per cycle.

Viscous damping is beneficial for controlling oscillations in systems such as car suspensions and building structures to reduce undue vibrations.
Angular Frequency
Angular frequency, represented by \(\omega\), is a measure of how quickly a system oscillates in terms of radians per second. In a lightly damped harmonic oscillator, the angular frequency is close to the natural frequency, which is \(\sqrt{\frac{k}{m}}\), where \(k\) is the spring constant and \(m\) is the mass.

With damping, the effective angular frequency is slightly modified but often still approximates the undamped frequency, particularly in systems where damping is minimal.

Angular frequency is central to determining the oscillator's speed and energy characteristics:
  • A higher angular frequency corresponds to faster oscillations.
  • It influences the energy dissipation rate per cycle.
  • Typically remains constant within lightly damped systems but can change with significant damping fluctuations.
Energy Loss per Cycle
Energy loss per cycle in a damped harmonic oscillator is an important measure of how much energy a system dissipates on each oscillation. This is pivotal for understanding the damping effect: how quickly the amplitude of the system reduces over time.

In our system, energy loss \(Q\) is derived by integrating the force due to damping over one full cycle. For a harmonic oscillator with viscous damping, the force into velocity at each instant is \(c \left( \frac{dx}{dt} \right)^2\), which gets integrated over the period of oscillation. The resulting expression \(Q = c x_1^2 \omega \pi\) shows how \(Q\) is directly dependent on:
  • The square of the amplitude \(x_1\) (energy loss increases with amplitude).
  • The damping coefficient \(c\) (greater damping leads to more energy dissipated).
  • The angular frequency \(\omega\) (determines cycle length and energy characteristics).
Understanding this energy dissipation helps in engineering and physics to design effective oscillatory systems with desired damping levels.

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Most popular questions from this chapter

A viscously damped spring-mass system is forced harmonically at the undamped natural frequency \(\left(\omega / \omega_{n}=1\right) .\) If the damping ratio \(\zeta\) is doubled from 0.1 to \(0.2,\) compute the percentage reduction \(R_{1}\) in the steady-state amplitude. Compare with the result \(R_{2}\) of a similar calculation for the condition \(\omega / \omega_{n}=2\) Verify your results by inspecting Fig. \(8 / 11\)

The owner of a 3400 -lb pickup truck tests the action of his rear-wheel shock absorbers by applying a steady 100 -lb force to the rear bumper and measuring a static deflection of 3 in. Upon sudden release of the force, the bumper rises and then falls to a maximum of \(\frac{1}{2}\) in. below the unloaded equilibrium position of the bumper on the first rebound. Treat the action as a one-dimensional problem with an equivalent mass of half the truck mass. Find the viscous damping factor \(\zeta\) for the rear end and the viscous damping coefficient \(c\) for each shock absorber assuming its action to be vertical.

Determine the amplitude of vertical vibration of the car as it travels at a velocity \(v=40 \mathrm{km} / \mathrm{h}\) over the wavy road whose contour may be expressed as a sine or cosine function with a double amplitude \(2 b=50 \mathrm{mm} .\) The mass of the car is \(1800 \mathrm{kg}\) and the stiffness of each of the four car springs is \(35 \mathrm{kN} / \mathrm{m}\) Assume that all four wheels are in continuous con- tact with the road, and neglect damping. Note that the wheelbase of the car and the spatial period of the road are the same at \(L=3 \mathrm{m},\) so that it may be assumed that the car translates but does not rotate. At what critical speed \(v_{c}\) is the vertical vibration of the car at its maximum?

The block of weight \(W=100\) lb is suspended by two springs each of stiffness \(k=200 \mathrm{lb} / \mathrm{ft}\) and is acted upon by the force \(F=75 \cos 15 t\) lb where \(t\) is the time in seconds. Determine the amplitude \(X\) of the steady-state motion if the viscous damping coefficient \(c\) is \((a) 0\) and (b) 60 lb-sec/ft. Compare these amplitudes to the static spring deflection \(\delta_{\mathrm{st}}\)

A uniform rod of mass \(m\) and length \(l\) is welded at one end to the rim of a light circular hoop of radius \(l .\) The other end lies at the center of the hoop. Determine the period \(\tau\) for small oscillations about the vertical position of the bar if the hoop rolls on the horizontal surface without slipping.

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