Power of a driven oscillating spring

In summary, the conversation discusses a driven mass-spring system with viscous friction and provides formulas for the rate of energy loss due to friction force and the rate at which the driving force does work on the system. It also mentions finding the total energy loss and work done over one period and how to use the integral of a sine or cosine function to do so. The conversation also includes a question about separating the integral, which is resolved by using the fact that both r and F depend on time.
  • #1
jdc15
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Homework Statement



a) Consider a driven mass-spring system with viscous friction using the notation of the lecture of Oct. 29, available on Vista. [The driving frequency is ω, the natural frequency is ω0, the friction force is -cv, the mass is m, the spring constant is k, the driving force is kD sin (ω t). Note that the phase Φ is negative, between 0 and π .] Write a formula for the rate of energy loss due to the friction force, once the steady state has been reached, as a function of time, t.

b) Write a formula for the rate at which the driving force is doing work on the mass-spring system, once the steady state has been reached, as a function of time, t.

c) Find the total energy loss due to friction over one period of the oscillations and also the total work done by the driving force over one period. Check whether or not they are equal. To do this problem you need to use the fact that the integral of a sine or cosine function over one period is zero. This integral is simply the average value over one period divided by the period. This is clearly zero if you think about what a cosine or sine function looks like.

Homework Equations



We have the equation for work: [tex]W=\int{F}\cdot dr[/tex]
Power: [tex]dW/dt[/tex]
Position: [tex]r=Asin(\omega t + \phi )[/tex]
Velocity: [tex]v=dr/dt[/tex]
Frictional force: [tex]F_f=-cv[/tex]
Driving force: [tex]F_D=kDsin(\omega t)[/tex]
Amplitude: A is a complex formula involving [tex]\omega, \omega 0, k, m, c[/tex] (will post later if needed)

The Attempt at a Solution



For a, I went r=Asin(wt+phi) so dr=Awcos(wt+phi)dt and v=Awcos(wt+phi). Thus work is [tex]\int{-cvA\omega cos(\omega t + \phi )} dt [/tex], so power is [tex]p=-cA^2\omega^2cos^2(\omega t + phi)[/tex]. Now this answer makes sense to me, I know I'm missing a phi term that I need to solve for, but it seems to match the simulations we've been shown. However, my TA told me I should separate the integral [tex]\int{F}dr=F\int{}dr=Fr[/tex] but this doesn't make sense to me because r and F both depend on time. Does anyone have an explanation?

Thanks in advance.

PS. This is a continuation of another post that originally was just a calc question: https://www.physicsforums.com/showthread.php?p=2967731&posted=1#post2967731"
 
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  • #2
Nevermind I figured it out.
 

1. What is the power of a driven oscillating spring?

The power of a driven oscillating spring is the rate at which work is done or energy is transferred by the spring as it moves back and forth.

2. How is the power of a driven oscillating spring calculated?

The power of a driven oscillating spring is calculated by multiplying the force exerted by the spring with the velocity of the object attached to it.

3. What factors affect the power of a driven oscillating spring?

The power of a driven oscillating spring is affected by the amplitude of the oscillations, the frequency of the oscillations, and the mass of the object attached to the spring.

4. Can the power of a driven oscillating spring be negative?

Yes, the power of a driven oscillating spring can be negative when work is being done on the spring, such as when it is being compressed or stretched.

5. How does the power of a driven oscillating spring relate to its potential and kinetic energy?

The power of a driven oscillating spring is directly related to its potential and kinetic energy. When the spring is at its maximum potential energy, the power is zero, and when it is at its maximum kinetic energy, the power is at its maximum value.

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