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A massless spring is hanging vertically and unloaded, from the ceiling. A mass is attached to the bottom end and released. How close to its final resting position is the mass after 1 second, given that it finally comes to rest 0.5 meters below the point of release and that the motion is critically damped?

Short Answer

Expert verified
The mass is close to rest but precise calculations require measured constants for numerics.

Step by step solution

01

Understanding Critically Damped Motion

In a critically damped system, the damping force is just enough to prevent oscillations, and the system returns to equilibrium as quickly as possible without oscillating. The position function for motion that is critically damped is given by:x(t)=(A+Bt)ewtwhere A and B are constants determined by initial conditions and w is the angular frequency related to the system's parameters.
02

Apply Initial Conditions

Since the mass is released from rest, we have the initial conditions: x(0)=0 and x˙(0)=0. Substituting these into the equation:- At t=0, x(0)=AA=0.- x˙(t)=(BBwt)ewt gives x˙(0)=B=0. So, the position equation simplifies to:x(t)=Btewt.
03

Determine the Final Position Constants

The final rest position of the mass is 0.5 meters below the point of release. Thus, we know x()=0.5. Since the exponential decays to zero as t, the constant shift here would normally suggest long-term behavior, but we ensure exponents balance for short-term playoffs.
04

Evaluate Position at 1 second

To find how close it is to the final position after 1 second, evaluate the expression derived. Without explicit damping frequency and precise measurements, this positions a general progression of time depending on natural unresolved constants, requiring experimental review for validation across unstated constants directly elsewhere verified over time.
05

Approximation of Position

Given the physical setup and resolved critical damping feasibility, it typically nears substantial completion but requires practical setting compensations for derivations and approximated motion resolving fine closure limits at designated experimentation evaluations.

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Key Concepts

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

Damping Force
When discussing a critically damped motion, understanding the damping force is key. This force opposes the motion of the mass and works to bring it to a stop.

A damping force is crucial because it prevents the system from oscillating. It achieves this by balancing the natural energy of the system:
  • Too much damping force, and the system takes forever to return to the rest position.
  • Too little, and the system oscillates back and forth before finally settling.
  • A critical damping force perfectly tunes the system to return to rest quickly, without overshooting.
In engineering and physics, calculating the exact amount of damping force required for critical damping ensures that the system achieves maximum efficiency and stability when returning to equilibrium. It’s akin to perfect braking in a car, stopping swiftly without skidding or creeping.
Equilibrium Position
The equilibrium position is the natural resting place of the system when all forces are balanced. In the case of the spring-mass system described, the equilibrium position is 0.5 meters below the point of release.

This is notable because, when released, the mass initially drops due to gravity, pulling it away from this natural resting spot. However, due to the critically damped motion:
  • The system is designed to quickly find this equilibrium without oscillating around it.
  • The damping force acts precisely to counteract the velocity that would otherwise cause oscillation.
In practical terms, reached equilibrium allows us to measure and predict the mass's behavior over time. For engineers and designers, equilibrium positions help them understand how long a system will take to settle and adjust components to optimize these characteristics.
Angular Frequency
Angular frequency, represented by the symbol w, plays a significant role in the behavior of a damped system. It is associated with the natural dynamics of how quickly the system would oscillate if there were no damping.

In a critically damped system, while it doesn’t cause oscillation, the angular frequency influences how swiftly the system can return to its equilibrium position:
  • It helps determine the rate at which the exponential term ewt reduces the effects of the initial conditions and motion over time.
  • A higher angular frequency implies the potential for rapid settling, given that critical damping is maintained.
Understanding angular frequency aids engineers and scientists in setting up systems that need precise and quick responses, such as in automotive suspensions or electronic signal processing.

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Most popular questions from this chapter

Consider a simple harmonic oscillator with period τ. Let f denote the average value of any variable f(t), averaged over one complete cycle: f=1τ0τf(t)dt Prove that T=U=12E where E is the total energy of the oscillator. [Hint: Start by proving the more general, and extremely useful, results that sin2(ωtδ)=cos2(ωtδ)=12. Explain why these two results are almost obvious, then prove them by using trig identities to rewrite sin2θ and cos2θ in terms of cos(2θ).]

Another interpretation of the Q of a resonance comes from the following: Consider the motion of a driven damped oscillator after any transients have died out, and suppose that it is being driven close to resonance, so you can set ω=ωo. (a) Show that the oscillator's total energy (kinetic plus potential) is E=12mω2A2. (b) Show that the energy ΔEdis dissipated during one cycle by the damping force  Fdmp  is 2πmβωA2. (Remember that the rate at which a force does work is Fv. ) (c) Hence show that Q is 2π times the ratio E/ΔEdis: 

Suppose that you have found a particular solution xp(t) of the inhomogeneous equation (5.48) for a driven damped oscillator, so that Dxp=f in the operator notation of (5.49). Suppose also that x(t) is any other solution, so that Dx=f. Prove that the difference xxp must satisfy the corresponding homogeneous equation, D(xxp)=0. This is an alternative proof that any solution x of the inhomogeneous equation can be written as the sum of your particular solution plus a homogeneous solution; that is, x=xp+xh.

The potential energy of a one-dimensional mass m at a distance r from the origin is U(r)=U0(rR+λ2Rr) for 0<r<, with Uo,R, and λ all positive constants. Find the equilibrium position ro. Let x be the distance from equilibrium and show that, for small x, the PE has the form U= const +12kx2. What is the angular frequency of small oscillations?

(a) If a mass m=0.2kg is tied to one end of a spring whose force constant k=80N/m and whose other end is held fixed, what are the angular frequency ω, the frequency f, and the period τ of its oscillations? (b) If the initial position and velocity are xo=0 and vo=40m/s, what are the constants A and δ in the expression x(t)=Acos(ωtδ)?

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