Mixing
Differential equation - mass spring system
$begingroup$
So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$Mddot{x}+cdot{x}+kx=0$$ where $M=4times 10^5text{kg}, c=5times 10^4text{kg/s}, k=10^7text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.
Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$Mddot{x}+cdot{x}+kx=300dot{N}.$$
Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?
I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.
ordinary-differential-equations
asked Nov 9 '16 at 0:38
AnonAnon
408516
$endgroup$
add a comment |
$begingroup$
So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$Mddot{x}+cdot{x}+kx=0$$ where $M=4times 10^5text{kg}, c=5times 10^4text{kg/s}, k=10^7text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.
Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$Mddot{x}+cdot{x}+kx=300dot{N}.$$
Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?
I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.
ordinary-differential-equations
asked Nov 9 '16 at 0:38
AnonAnon
408516
$endgroup$
$begingroup$
What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
$endgroup$
– Chee Han
Nov 9 '16 at 8:24
add a comment |
$begingroup$
So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$Mddot{x}+cdot{x}+kx=0$$ where $M=4times 10^5text{kg}, c=5times 10^4text{kg/s}, k=10^7text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.
Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$Mddot{x}+cdot{x}+kx=300dot{N}.$$
Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?
I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.
ordinary-differential-equations
asked Nov 9 '16 at 0:38
AnonAnon
408516
$endgroup$
So we have a mass spring system, which is supposed to be modelling a bridge, and the equation that the displacement of the bridge $x$ satisfies is given by $$Mddot{x}+cdot{x}+kx=0$$ where $M=4times 10^5text{kg}, c=5times 10^4text{kg/s}, k=10^7text{kg} $, where $m$ is the mass, $k$ is the 'spring constant'and $c$ is the friction or level of damping or something, I'm not sure - this equation is pretty standard for this type of problem so you probably know what it means more than me anyway.
Now there is a part of the question where another factor is introduced, so that $x$ now satisfies the equation $$Mddot{x}+cdot{x}+kx=300dot{N}.$$
Now my question is, is the damping level a constant value, so is it still $c$ or has it changed due to the new factor affecting $dot{x}$ so would the damping now be $c-300N$? or does it just stay at $c$?
I think the main problem here is that I don't really know what is damping (what actually is damping )in general, so if anyone could just answer the question I asked above or say something about damping in general in these types of problems that would explain my confusion.
ordinary-differential-equations
ordinary-differential-equations
asked Nov 9 '16 at 0:38
AnonAnon
408516
asked Nov 9 '16 at 0:38
AnonAnon
408516
asked Nov 9 '16 at 0:38
AnonAnon
408516
asked Nov 9 '16 at 0:38
AnonAnon
408516
asked Nov 9 '16 at 0:38
AnonAnon
408516
408516
$begingroup$
What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
$endgroup$
– Chee Han
Nov 9 '16 at 8:24
add a comment |
$begingroup$
What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
$endgroup$
– Chee Han
Nov 9 '16 at 8:24
$begingroup$
What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
$endgroup$
– Chee Han
Nov 9 '16 at 8:24
$begingroup$
What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
$endgroup$
– Chee Han
Nov 9 '16 at 8:24
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$Mx''+cx'+kx=0$
$x = Ae^{-frac c{2M}t}sinbig(phi + tsqrt {frac {k}{M} -(frac{c}{2M})^2} big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + K$
$x'$ and $x''$ are unchanged.
$kK = 300\
K = frac {300}{k}$
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + frac {300}{k}$
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{rm damping} = -c frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{rm external}$. Applying Newton's second law you find
begin{eqnarray}
Ma = sum_i F_i &=& F_{rm restoring} + F_{rm damping} + F_{rm external} \
M ddot{x} + adot{x} + kx &=& F_{rm external}
end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
add a comment |
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2 Answers
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2 Answers
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$begingroup$
$Mx''+cx'+kx=0$
$x = Ae^{-frac c{2M}t}sinbig(phi + tsqrt {frac {k}{M} -(frac{c}{2M})^2} big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + K$
$x'$ and $x''$ are unchanged.
$kK = 300\
K = frac {300}{k}$
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + frac {300}{k}$
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
$Mx''+cx'+kx=0$
$x = Ae^{-frac c{2M}t}sinbig(phi + tsqrt {frac {k}{M} -(frac{c}{2M})^2} big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + K$
$x'$ and $x''$ are unchanged.
$kK = 300\
K = frac {300}{k}$
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + frac {300}{k}$
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
$endgroup$
add a comment |
$begingroup$
$Mx''+cx'+kx=0$
$x = Ae^{-frac c{2M}t}sinbig(phi + tsqrt {frac {k}{M} -(frac{c}{2M})^2} big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + K$
$x'$ and $x''$ are unchanged.
$kK = 300\
K = frac {300}{k}$
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + frac {300}{k}$
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
$endgroup$
$Mx''+cx'+kx=0$
$x = Ae^{-frac c{2M}t}sinbig(phi + tsqrt {frac {k}{M} -(frac{c}{2M})^2} big) $
I hope I have that right, I eyeballed it.
Now we change the equation.
$Mx''+cx'+kx=300$
What happens to $x?$ $x$ adjusts by a constant.
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + K$
$x'$ and $x''$ are unchanged.
$kK = 300\
K = frac {300}{k}$
$x = Ae^{-frac c2t}sin(phi + tsqrt {Mk -frac {c^2}4} ) + frac {300}{k}$
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
answered Nov 9 '16 at 0:58
Doug MDoug M
45.4k31954
45.4k31954
add a comment |
add a comment |
$begingroup$
Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{rm damping} = -c frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{rm external}$. Applying Newton's second law you find
begin{eqnarray}
Ma = sum_i F_i &=& F_{rm restoring} + F_{rm damping} + F_{rm external} \
M ddot{x} + adot{x} + kx &=& F_{rm external}
end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
add a comment |
$begingroup$
Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{rm damping} = -c frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{rm external}$. Applying Newton's second law you find
begin{eqnarray}
Ma = sum_i F_i &=& F_{rm restoring} + F_{rm damping} + F_{rm external} \
M ddot{x} + adot{x} + kx &=& F_{rm external}
end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
$endgroup$
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
add a comment |
$begingroup$
Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{rm damping} = -c frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{rm external}$. Applying Newton's second law you find
begin{eqnarray}
Ma = sum_i F_i &=& F_{rm restoring} + F_{rm damping} + F_{rm external} \
M ddot{x} + adot{x} + kx &=& F_{rm external}
end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
$endgroup$
Damping in this model is a force that resists motion and that is proportional to the instantaneous velocity,
$$
F_{rm damping} = -c frac{dx}{dt}
$$
This is a very typical force when you're immersed in a fluid. In this case $c$ is a constant, the larger $c$ the stronger is the force opposing the motion. So for instance between oil and water, oil has a larger $c$.
Now, there is also a restoring force in your model that is taken proportional to the deformation of the bridge
$$
F_{rm restoring} = -k x
$$
And finally an external force (e.g. wind) $F_{rm external}$. Applying Newton's second law you find
begin{eqnarray}
Ma = sum_i F_i &=& F_{rm restoring} + F_{rm damping} + F_{rm external} \
M ddot{x} + adot{x} + kx &=& F_{rm external}
end{eqnarray}
$c$ is still constant, regardless of the fact that the bridge is being externally forced to move!
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
answered Nov 9 '16 at 1:03
caveraccaverac
14.8k31130
14.8k31130
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
add a comment |
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
A later part of the question says that additional damping is added, does this mean that the value of $c$ increases since the level of damping has been increased artificially?
$endgroup$
– Anon
Nov 9 '16 at 1:38
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
Correct, $c$ increases $Rightarrow$ damping increases
$endgroup$
– caverac
Nov 9 '16 at 1:42
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
That makes sense, so in the case of a bridge, $c$ is set to a level of whatever the designers of the bridge wanted, so they could just increase the value of $c$ if they wanted to by adding some structure to the bridge (I'm not engineer)?
$endgroup$
– Anon
Nov 9 '16 at 2:16
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
$begingroup$
This is a very simple model, but indeed, the geometry of the bridge impacts the effective value of $c$. The materials used to build it usually go into $k$
$endgroup$
– caverac
Nov 9 '16 at 2:19
add a comment |
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What is $N$ or $dot N$ on the RHS? Damping essentially means that the amplitude of oscillations gets smaller and smaller as time evolves. Imagine the ODE without the term $cdot x$, then you will have a harmonic oscillator where the amplitude doesn't decrease as time evolves; once you include the term $cdot x$, the solution is multiplied by an exponential decaying function (see below), which is exactly what I just explained about damping.
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– Chee Han
Nov 9 '16 at 8:24