Calculating Viscous Damper Energy in Spring-Mass Systems

In summary, the equation for calculating the energy absorbed by a damper is E = C * ∫ (v^2) dt, where C is the damper constant, t is the elapsed time, and v is the velocity difference between the ends of the damper. There may be other equations such as W = 1/2C * dV * dx, but the first equation is the most commonly used and correct way to calculate the energy absorbed by a damper. The damper converts mechanical energy to thermal energy, so to measure the energy absorbed directly, one would need to measure changes in temperature or internal energy of the damper's components.
  • #1
physea
211
3
Hello,

In problems that involve springs and viscous dampers, eg a mass falls on a spring that also has a viscous damper, how can we calculate the energy absorbed by the damper?

I mean, we know the energy absorbed by the spring (1/2*k*x^2), we know the kinetic energy of the mass (1/2*m*u^2), but what is the energy absorbed by the damper?

thanks
 
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  • #2
If there were no viscous damping, what would be the change in the sum of kinetic energy of the mass and elastic energy stored in the spring?
 
  • #3
Chestermiller said:
If there were no viscous damping, what would be the change in the sum of kinetic energy of the mass and elastic energy stored in the spring?

Zero
 
  • #4
So what does that tell you about the change in the sum of kinetic energy of the mass and elastic energy stored in the spring if there is viscous damping also present?
 
  • #5
Chestermiller said:
So what does that tell you about the change in the sum of kinetic energy of the mass and elastic energy stored in the spring if there is viscous damping also present?

Zero again, we wasted all that time and these posts to check if energy conservation still applies to systems with dampers? Ofcourse it does but you have not addressed my question: what is the equation for the damper energy absorption?
 
  • #6
physea said:
Zero again, we wasted all that time and these posts to check if energy conservation still applies to systems with dampers? Ofcourse it does but you have not addressed my question: what is the equation for the damper energy absorption?
The answer is not zero when viscous damping is present. So, do you still feel I wasted your time?
 
  • #7
Chestermiller said:
The answer is not zero when viscous damping is present.

The energy lost in the damper and the energy stored in the spring equals to the kinetic energy of the mass. But you do not answer my question.
I don't want to calculate the damper energy indirectly, via the kinetic energy of the mass and the spring energy. I want to calculate it DIRECTLY with an equation.
 
  • #8
physea said:
The energy lost in the damper and the energy stored in the spring equals to the kinetic energy of the mass. But you do not answer my question.
I don't want to calculate the damper energy indirectly, via the kinetic energy of the mass and the spring energy. I want to calculate it DIRECTLY with an equation.
The change in kinetic energy plus the change in stored elastic energy is equal to minus the energy dissipated in the damper.
 
  • #9
physea said:
Zero again, we wasted all that time and these posts to check if energy conservation still applies to systems with dampers? Ofcourse it does but you have not addressed my question: what is the equation for the damper energy absorption?
A little more politeness would be in order, I think. The customer is only always right when they are actually paying money for a service. Play nicely!
 
  • #10
sophiecentaur said:
A little more politeness would be in order, I think. The customer is only always right when they are actually paying money for a service. Play nicely!
Whoever said “the customer is always right” has never met the customer.
 
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  • #11
physea said:
But you do not answer my question.
I don't want to calculate the damper energy indirectly, via the kinetic energy of the mass and the spring energy. I want to calculate it DIRECTLY with an equation.
Well, based on your posts up to now, how could I possibly have guessed that this is what you wanted? Do you think I can read your mind? If you want to calculate the energy dissipated by the damper DIRECTLY, then you can use this equation:
$$E=C\int_0^t{v^2}dt$$where C is the damper constant, t is the elapsed time, and v is the velocity difference between the ends of the damper.
 
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  • #12
Chestermiller said:
Well, based on your posts up to now, how could I possibly have guessed that this is what you wanted? Do you think I can read your mind? If you want to calculate the energy dissipated by the damper DIRECTLY, then you can use this equation:
$$E=C\int_0^t{v^2}dt$$where C is the damper constant, t is the elapsed time, and v is the velocity difference between the ends of the damper.
Yeah, that's what I wanted thanks.

Btw, I derived it on my own like this:
F=cV (force of the damper)
So W=Fdx=c(Vdx)
We know that dx=Vdt
So W=c(V^2dt)

But I see some other equations like W=1/2c*dV*dx

Which is correct? I know this is maths, but I am not sure how to derive it myself.
 
  • #13
physea said:
Yeah, that's what I wanted thanks.

Btw, I derived it on my own like this:
F=cV (force of the damper)
So W=Fdx=c(Vdx)
We know that dx=Vdt
So W=c(V^2dt)

But I see some other equations like W=1/2c*dV*dx

Which is correct? I know this is maths, but I am not sure how to derive it myself.
Your derivation is correct.
 
  • #14
Chestermiller said:
Your derivation is correct.

So W=c(V^2dt) is correct
and W=1/2c*dV*dx is not correct?

If both are correct, how do we go from the first to the second?
 
  • #15
physea said:
So W=c(V^2dt) is correct
and W=1/2c*dV*dx is not correct?

If both are correct, how do we go from the first to the second?
The first one should be an integral with respect to t. Otherwise you have a finite quantity equated to a differential. The second one makes no sense to me. To demonstrate that the first one is correct, obtain the dissipated damper energy from this equation , and then compare it with the change in the sum of kinetic- and elastic potential energy.
 
  • #16
physea said:
The energy lost in the damper and the energy stored in the spring equals to the kinetic energy of the mass. But you do not answer my question.
I don't want to calculate the damper energy indirectly, via the kinetic energy of the mass and the spring energy. I want to calculate it DIRECTLY with an equation.

The damper converts mechanical energy to thermal energy. You would need to measure things like temperature changes of the internal components of the damper and relate those to the increase in thermal energy of those components. (Often authors will use the term internal energy rather than thermal energy.)
 

1. What is a viscous damper in a spring-mass system?

A viscous damper is a mechanical device that is used to dissipate energy in a spring-mass system. It consists of a piston that moves through a fluid-filled cylinder, converting kinetic energy into heat as it moves.

2. How does a viscous damper affect the energy of a spring-mass system?

A viscous damper absorbs energy from the system, reducing its amplitude and damping its oscillations. This helps to prevent the system from oscillating uncontrollably and dissipates energy that could damage the system.

3. How is the energy dissipated by a viscous damper calculated?

The energy dissipated by a viscous damper can be calculated by multiplying the damping coefficient (c) by the square of the velocity (v) of the damper. This can be represented by the equation E = cv2.

4. What factors affect the energy dissipated by a viscous damper?

The amount of energy dissipated by a viscous damper depends on several factors, including the damping coefficient, the velocity of the damper, and the frequency of oscillations in the system. It also depends on the material and design of the damper.

5. How can the energy dissipated by a viscous damper be optimized?

The energy dissipation of a viscous damper can be optimized by adjusting the damping coefficient to match the frequency of oscillations in the system. This can help to reduce the amount of energy that is dissipated and improve the overall efficiency of the system.

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