Calculating Average Power for Sinusoidal Acceleration with an Accelerometer

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The discussion revolves around calculating the average power delivered to a mass experiencing sinusoidal acceleration measured by an accelerometer. Participants debate whether the average power can be zero due to oscillation, with some arguing that continuous energy input is necessary to maintain acceleration. The calculation of instantaneous power using force and velocity is explored, but confusion arises regarding the integration period and the implications of energy conservation. Ultimately, it is concluded that the average power cannot be derived simply from the RMS of instantaneous power, as energy in the system fluctuates between the mass and the source, potentially dissipating energy to the surrounding fluid. The conversation highlights the complexities of power calculations in mechanical systems compared to electrical oscillators.
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Hello,

I have a system of mass m that has an accelerometer strapped to it. The accelerometer is measuring a sinusoidal acceleration with peak amplitude of amax:

https://www.dropbox.com/s/y49pcrhqbrzk27t/Accel_System.png


How can I calculate the average power (in Watts) that is being delivered to that system to generate such acceleration?

Thanks!
 
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diemilio said:
The accelerometer is measuring a sinusoidal acceleration
If it's an oscillation, isn't the average delivered power zero?
 
I don't think so. No power, no acceleration.

I think it is something like this, but I am not 100% sure:

P(t) = F(t) * v(t),

where F(t) is the force and v(t) is the velocity

F(t) = m*amax*sin(ω0*t)
v(t) = ∫ amax*sin(ω0*t) = - amax0*cos(ω0*t)

P(t) = -m*amax20*sin(ω0*t)*cos(ω0*t)
P(t) = -m*amax2/(2*ω0)*sin(2*ω0*t)

I think the average power can be calculated taking the RMS of P(t) Pavg = √(1/T * ∫ P(t)2 * dt).

What I don't know is what should be the integration period? should it be T = 2*π/ω0 or only half of that because is a square function?

Thanks!
 
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If I am correct, then I am getting:

Pavg = amax2*m/(2*√2*ω0)

It would be great if someone could let me know if this is correct.

Thanks again!
 
diemilio said:
No power, no acceleration.
The average power can be zero, if positive and negative power cancel.
 
A.T. said:
The average power can be zero, if positive and negative power cancel.

To have negative power wouldn't the mass have to be sourcing energy back into the power source?

What is wrong with my calculation?
 
diemilio said:
To have negative power wouldn't the mass have to be sourcing energy back into the power source?
No, the energy can be dissipated as heat. If the average mechanical power delivered to the oscillating mass wasn't zero, it would accumulate more and more kinetic energy.

diemilio said:
What is wrong with my calculation?
P = F dot v which is negative when F and v oppose each other.
 
A.T. said:
No, the energy can be dissipated as heat. If the average mechanical power delivered to the oscillating mass wasn't zero, it would accumulate more and more kinetic energy.

P = F dot v which is negative when F and v oppose each other.

So you're basically saying the average delivered power by the source is zero; correct? It seems counter-intuitive to me

There is no loss in this system, so how can the energy be dissipated?

There is only a mass; there is no friction, no fluid damping, no thermo-elasticity, etc.

There is not even resistance to the motion (no spring constants), so there cannot even be a transfer of energy between two energy-storing components (typically spring and mass exchanging potential and kinetic energy).

All there is in this system is mass and a source of energy, so I think the average power cannot be zero. For example, if you want the acceleration to increase in one direction (during the first half of period the sine function), you need to put energy into the system. Similarly, if you want to change its direction (during the second half of the sine function), you need to keep putting energy to decelerate the mass (which has already has energy in the form of 1/2*m*v^2).

Is this reasoning incorrect? Sorry, I haven't done this in a while.

Thanks again.
 
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diemilio said:
There is no loss in this system
If there is no loss, why is a continuous power supply needed? A loss-less oscillator can oscillate forever, without power supply.

diemilio said:
you need to keep putting energy to decelerate the mass
A decelerating mass doesn't consume energy, it releases energy.
 
  • #10
A.T. said:
If there is no loss, why is a continuous power supply needed? A loss-less oscillator can oscillate forever, without power supply.

A decelerating mass doesn't consume energy, it releases energy.

It is not that the continuos power supply is needed; it is just there, delivering power continuously.

Anyway, it seems that my understanding of the problem I am dealing with is all wrong.

Thanks though.
 
  • #11
diemilio said:
To have negative power wouldn't the mass have to be sourcing energy back into the power source?
Yes, so whether that is the case or not depends on the details of the system. A spring-mass in gravity would have positive and negative power, for an average of zero and no loss. A wave power station, on the other hand, extracts power with both the up and down stroke. As does alternating current electricity.
What is wrong with my calculation?
Your order of operations looks wrong (in p(t) is cos on top or bottom?...should be on top), but assuming you just wrote it in calculator order, it may be ok. But for the last bit, you don't need to integrate to find rms manually: it's just 1/sqrt(2).
 
  • #12
diemilio said:
Anyway, it seems that my understanding of the problem I am dealing with is all wrong.
I wouldn't say your understanding is wong, I'd just say you haven't told us the purpose, so we don't know if it is or not. Calculating average power in this way may be valid -- heck, it is in electricity!
 
  • #13
russ_watters said:
Yes, so whether that is the case or not depends on the details of the system. A spring-mass in gravity would have positive and negative power, for an average of zero and no loss. A wave power station, on the other hand, extracts power with both the up and down stroke. As does alternating current electricity.

Thanks Russ. This can be viewed as a wave power station. As a matter of fact, in the problem I am dealing with, the mass is suspended in a fluid, and there is this wave generator sucking and pushing the mass back and forth in an oscillatory manner (viscosity is very small, so losses can be ignored).

russ_watters said:
Your order of operations looks wrong (in p(t) is cos on top or bottom?...should be on top), but assuming you just wrote it in calculator order, it may be ok. But for the last bit, you don't need to integrate to find rms manually: it's just 1/sqrt(2).

cos is on the top. Notice I do not have parenthesis after the division, so the only term in the denominator is ω0

Thanks again!
 
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  • #14
russ_watters said:
I wouldn't say your understanding is wong, I'd just say you haven't told us the purpose, so we don't know if it is or not.

Sorry, I didn't want to be too specific because I thought I simple diagram would be enough, but I guess the diagram can be very ambiguous.

I'm dealing with a component (of mass m) submerged in a fluid, which is used for something known as ultrasonic cleaning. The component has an accelerometer inside that can tell me the acceleration amplitude that it is object is undergoing. So far, I have the peak value of the measured acceleration, and I know the ultrasonic source has a frequency of 40 kHz, but I need to calculate the power at which its emitting to cause such an acceleration.

A fair assumption I can make is that most of the power is being delivered to the component, and very little is lost due to the fluid's viscosity. That's why I assumed that the best way to calculate the average power of the source was by calculating the instantaneous power of the object and then taking the RMS.

russ_watters said:
Calculating average power in this way may be valid -- heck, it is in electricity!

Yes, I have had a lot of exposure to electrical oscillators, so that's why I assumed it could be calculated this way. First time dealing with a mechanical component though.

Thanks!
 
  • #15
diemilio said:
it is just there, delivering power continuously.
If power goes in continuously, and there is no loss, where does all that energy go to? Unless it accumulates in the oscillator, making it oscillate more and more, this would violate energy conservation.
 
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  • #16
A.T. said:
If power goes in continuously, and there is no loss, where does all that energy go to? Unless it accumulates in the oscillator, making it oscillate more and more, this would violate energy conservation.

I agree. I think the way I am postulating the problem is incorrect. The energy in the system will just fluctuate back-and-forth between the mass and the source, so no power would be delivered.

After extensive searching, I realized that the average power cannot be calculated by using the RMS of the instantaneous power.

Thanks again.
 
  • #17
diemilio said:
The energy in the system will just fluctuate back-and-forth between the mass and the source, so no power would be delivered.
That is one possibility, depending on how the motor applies the force. If the mass is submerged in a fluid it will of course also dissipate energy to that fluid.
 
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