Springs with some finite mass The 2019 Stack Overflow Developer Survey Results Are InSprings...
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Springs with some finite mass
The 2019 Stack Overflow Developer Survey Results Are InSprings yet again, this time with a picture. Infinite displacement, makes no senseSpring-mass system with two springs and three massesModeling linear spring deformation in timeIs mass proportional to the displacement from equilibrium in Hooke's law?Can Hooke's law go along with Newton's 3rd law & how is momentum conserved in case of spring?On the definition of elastic restoring force in a springSpring pulled with one end fixedMomentum is conserved without the application of Newton's third law in cases of non-rigid bodies like springSimple Harmonic Motion - spring with massWhy does different part of a spring having mass expand proportional to their distance while the spring has some mass hanged in the bottom?
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Let us consider a spring which is having some finite mass. By the help of some external agent the spring has been extended by some distance $x$.
Will the restoring force produced in the spring still be directly proportional to the extension just like any massless spring?
newtonian-mechanics mass spring oscillators elasticity
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add a comment |
$begingroup$
Let us consider a spring which is having some finite mass. By the help of some external agent the spring has been extended by some distance $x$.
Will the restoring force produced in the spring still be directly proportional to the extension just like any massless spring?
newtonian-mechanics mass spring oscillators elasticity
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If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
$endgroup$
– Bill Watts
yesterday
add a comment |
$begingroup$
Let us consider a spring which is having some finite mass. By the help of some external agent the spring has been extended by some distance $x$.
Will the restoring force produced in the spring still be directly proportional to the extension just like any massless spring?
newtonian-mechanics mass spring oscillators elasticity
$endgroup$
Let us consider a spring which is having some finite mass. By the help of some external agent the spring has been extended by some distance $x$.
Will the restoring force produced in the spring still be directly proportional to the extension just like any massless spring?
newtonian-mechanics mass spring oscillators elasticity
newtonian-mechanics mass spring oscillators elasticity
edited yesterday
Qmechanic♦
107k121991241
107k121991241
asked yesterday
Rifat SafinRifat Safin
1268
1268
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If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
$endgroup$
– Bill Watts
yesterday
add a comment |
$begingroup$
If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
$endgroup$
– Bill Watts
yesterday
$begingroup$
If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
$endgroup$
– Bill Watts
yesterday
$begingroup$
If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
$endgroup$
– Bill Watts
yesterday
add a comment |
6 Answers
6
active
oldest
votes
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For a spring that suffers a uniform stretching (i.e., without acceleration), the force-elongation equation is $$F=ku_L$$ where u is the displacement of the end of the spring at x = L relative to its undeformed length of L. This equation can also be expressed in the form $$F=kLfrac{u_L}{L}$$
For a spring that suffers a non-uniform stretching (such as is present with acceleration), the corresponding relationship is $$F=kLfrac{partial u}{partial x}$$ where u = u(x) is the displacement of the spring cross section that was at axial position x in the undeformed configuration of the spring.
If we apply a 2nd law force balance on the section of spring (with acceleration and oscillation) between x and $x+Delta x$, we obtain: $$F(x+Delta x)-F(x)=frac{m}{L}Delta xfrac{partial ^2 u}{partial t^2}$$where m is the total mass of the spring and the partial derivative on the right hand side is the acceleration of the cross section at location x. If we divide by $Delta x$ and take the limit as $Delta x$ approaches zero, we obtain:
$$frac{partial F}{partial x}=kLfrac{partial u^2}{partial x^2}=frac{m}{L}frac{partial ^2 u}{partial t^2}$$or, equivalently, $$frac{kL^2}{m}frac{partial u^2}{partial x^2}=frac{partial ^2 u}{partial t^2}$$This is the wave equation for the oscillating spring, with the wave speed c given by $$c^2=frac{kL^2}{m}$$
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add a comment |
$begingroup$
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.
$endgroup$
add a comment |
$begingroup$
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.
Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.
Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.
Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.
Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.
The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.
This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.
Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).
We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).
In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$
This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.
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As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
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– Rifat Safin
13 hours ago
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As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
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– france95
5 hours ago
add a comment |
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The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.
Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.
We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.
Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.
However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.
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2
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In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
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– leftaroundabout
yesterday
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@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
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– lesnik
18 hours ago
add a comment |
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Hooke's law differs from other physical laws in an important respect:
Springs closely obey Hooke's law because they are carefully manufactured to closely obey Hooke's law.
Hooke's law is not a fundamental law - this should be obvious when you consider that a spring is not a naturally occurring object, but a manufactured piece of engineering.
It is possible to buy springs which do not obey Hooke's law, which are manufactured for specific purposes. In this respect Hooke's law is more akin to an engineering specification than a physical law, though for the purposes of teaching physics it is treated as one.
So in truth the answer to "will a spring obey Hooke's law under some specific conditions", is that you should consult the manufacturer's data sheet. This will give information such as the maximum extension before the spring is permanently deformed and so on.
New contributor
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Why do I have a patronising "new contributor" flag? I've been here ages.
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– Ben
yesterday
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Because it's your first answer.
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– BPP
yesterday
add a comment |
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Real springs have several properties that are not modeled by Hooke's linear law. You can find comparisons pretty easily. Experiments with springs might be one of the easiest to perform at home:
I don't see any experimental data on evaluating effect of spring's mass on force produced, probably because in most experiments mass of the spring is much smaller than the force applied to the spring to deform it. After all, mass of spring is linked with its spring constant k
. Thicker wire of a spring will increase its mass, but also will increase spring constant, i.e. it will require more force to deform it to the same length
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6 Answers
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active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a spring that suffers a uniform stretching (i.e., without acceleration), the force-elongation equation is $$F=ku_L$$ where u is the displacement of the end of the spring at x = L relative to its undeformed length of L. This equation can also be expressed in the form $$F=kLfrac{u_L}{L}$$
For a spring that suffers a non-uniform stretching (such as is present with acceleration), the corresponding relationship is $$F=kLfrac{partial u}{partial x}$$ where u = u(x) is the displacement of the spring cross section that was at axial position x in the undeformed configuration of the spring.
If we apply a 2nd law force balance on the section of spring (with acceleration and oscillation) between x and $x+Delta x$, we obtain: $$F(x+Delta x)-F(x)=frac{m}{L}Delta xfrac{partial ^2 u}{partial t^2}$$where m is the total mass of the spring and the partial derivative on the right hand side is the acceleration of the cross section at location x. If we divide by $Delta x$ and take the limit as $Delta x$ approaches zero, we obtain:
$$frac{partial F}{partial x}=kLfrac{partial u^2}{partial x^2}=frac{m}{L}frac{partial ^2 u}{partial t^2}$$or, equivalently, $$frac{kL^2}{m}frac{partial u^2}{partial x^2}=frac{partial ^2 u}{partial t^2}$$This is the wave equation for the oscillating spring, with the wave speed c given by $$c^2=frac{kL^2}{m}$$
$endgroup$
add a comment |
$begingroup$
For a spring that suffers a uniform stretching (i.e., without acceleration), the force-elongation equation is $$F=ku_L$$ where u is the displacement of the end of the spring at x = L relative to its undeformed length of L. This equation can also be expressed in the form $$F=kLfrac{u_L}{L}$$
For a spring that suffers a non-uniform stretching (such as is present with acceleration), the corresponding relationship is $$F=kLfrac{partial u}{partial x}$$ where u = u(x) is the displacement of the spring cross section that was at axial position x in the undeformed configuration of the spring.
If we apply a 2nd law force balance on the section of spring (with acceleration and oscillation) between x and $x+Delta x$, we obtain: $$F(x+Delta x)-F(x)=frac{m}{L}Delta xfrac{partial ^2 u}{partial t^2}$$where m is the total mass of the spring and the partial derivative on the right hand side is the acceleration of the cross section at location x. If we divide by $Delta x$ and take the limit as $Delta x$ approaches zero, we obtain:
$$frac{partial F}{partial x}=kLfrac{partial u^2}{partial x^2}=frac{m}{L}frac{partial ^2 u}{partial t^2}$$or, equivalently, $$frac{kL^2}{m}frac{partial u^2}{partial x^2}=frac{partial ^2 u}{partial t^2}$$This is the wave equation for the oscillating spring, with the wave speed c given by $$c^2=frac{kL^2}{m}$$
$endgroup$
add a comment |
$begingroup$
For a spring that suffers a uniform stretching (i.e., without acceleration), the force-elongation equation is $$F=ku_L$$ where u is the displacement of the end of the spring at x = L relative to its undeformed length of L. This equation can also be expressed in the form $$F=kLfrac{u_L}{L}$$
For a spring that suffers a non-uniform stretching (such as is present with acceleration), the corresponding relationship is $$F=kLfrac{partial u}{partial x}$$ where u = u(x) is the displacement of the spring cross section that was at axial position x in the undeformed configuration of the spring.
If we apply a 2nd law force balance on the section of spring (with acceleration and oscillation) between x and $x+Delta x$, we obtain: $$F(x+Delta x)-F(x)=frac{m}{L}Delta xfrac{partial ^2 u}{partial t^2}$$where m is the total mass of the spring and the partial derivative on the right hand side is the acceleration of the cross section at location x. If we divide by $Delta x$ and take the limit as $Delta x$ approaches zero, we obtain:
$$frac{partial F}{partial x}=kLfrac{partial u^2}{partial x^2}=frac{m}{L}frac{partial ^2 u}{partial t^2}$$or, equivalently, $$frac{kL^2}{m}frac{partial u^2}{partial x^2}=frac{partial ^2 u}{partial t^2}$$This is the wave equation for the oscillating spring, with the wave speed c given by $$c^2=frac{kL^2}{m}$$
$endgroup$
For a spring that suffers a uniform stretching (i.e., without acceleration), the force-elongation equation is $$F=ku_L$$ where u is the displacement of the end of the spring at x = L relative to its undeformed length of L. This equation can also be expressed in the form $$F=kLfrac{u_L}{L}$$
For a spring that suffers a non-uniform stretching (such as is present with acceleration), the corresponding relationship is $$F=kLfrac{partial u}{partial x}$$ where u = u(x) is the displacement of the spring cross section that was at axial position x in the undeformed configuration of the spring.
If we apply a 2nd law force balance on the section of spring (with acceleration and oscillation) between x and $x+Delta x$, we obtain: $$F(x+Delta x)-F(x)=frac{m}{L}Delta xfrac{partial ^2 u}{partial t^2}$$where m is the total mass of the spring and the partial derivative on the right hand side is the acceleration of the cross section at location x. If we divide by $Delta x$ and take the limit as $Delta x$ approaches zero, we obtain:
$$frac{partial F}{partial x}=kLfrac{partial u^2}{partial x^2}=frac{m}{L}frac{partial ^2 u}{partial t^2}$$or, equivalently, $$frac{kL^2}{m}frac{partial u^2}{partial x^2}=frac{partial ^2 u}{partial t^2}$$This is the wave equation for the oscillating spring, with the wave speed c given by $$c^2=frac{kL^2}{m}$$
answered yesterday
Chet MillerChet Miller
16.1k2826
16.1k2826
add a comment |
add a comment |
$begingroup$
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.
$endgroup$
add a comment |
$begingroup$
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.
$endgroup$
add a comment |
$begingroup$
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.
$endgroup$
In principle, yes. Generally (i.e in problems) springs are indicated as massless because it makes solving easier in particular situations, but Hooke's Law ($F=-kx$) applies nonetheless.
When spring mass becomes non-negligible, such as determining the acceleration of an attached mass or in the case of spring oscillations, then the mass of the spring has to be considered, and the solution requires integrating over the length of the spring.
answered yesterday
MickMick
689413
689413
add a comment |
add a comment |
$begingroup$
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.
Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.
Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.
Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.
Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.
The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.
This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.
Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).
We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).
In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$
This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.
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$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
add a comment |
$begingroup$
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.
Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.
Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.
Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.
Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.
The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.
This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.
Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).
We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).
In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$
This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.
$endgroup$
$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
add a comment |
$begingroup$
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.
Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.
Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.
Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.
Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.
The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.
This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.
Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).
We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).
In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$
This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.
$endgroup$
Obviously as soon as we start taking into account realistic features of a spring we will lose the linear Hook law (the force is directly proportional to the extension), which is a good approximation for small extensions.
That said we can try to see if the presence of a mass alters the kinematics of the system: so let's think of adding a mass density to a classic ideal spring.
The conclusion is that when now you extend the spring by applying a force on it, you will have some inertia of the spring which will further oppose a rapid extension.
Let's see how this emerges from a simple model.
We can always write its energy like:$$E=K+V$$ where $E$ is the total energy of the system, $K$ its kinetic energy and $V$ its potential energy.
Focusing on the kinetic term, we know that we can write it as an integral over the mass distribution: $$K=frac{1}{2}int_0^L dl ;lambda; v^2(l,t)$$
where $lambda$ is the mass density per unit of length ( $l$ ) and $v(l)$ is the speed of each point of the spring (here $l$ follows the length of the spring, so indicates a point on the spring). Basically we have written the standard kinetic term $dK=frac{1}{2}dm ;v^2$ for every infinitesimal piece of the string of mass $dm$.
Now let's move to the potential. Does the mass influences the potential energy?
Let's neglect relativistic effects (which we neglected in the extimate of the new kinetic term and in the massless model to begin with); we could still have gravity: having a non-zero mass, the points of the string attract each other by gravitational interaction. This would favor configurations of the spring in which it is slightly compressed (so that all the points are slightly nearer to each other). However since this computation looks messy and since I assume your spring will not have a planetary mass while conserving the original elastic constant, we will neglect this effect too.
Note that I just said that if the spring was to be very heavy its elastic constant and therefore its potential energy would change. This is due to the fact that elastic energy is generated by electric bounds between molecules (or lattice ions in the case of a metallic spring). If you want a spring as heavy as a planet you will need its section to be hundreds of kilometer thick: therefore you will have a lot more microscopic bounds opposing the (giant)spring's torsion and compression.
The bottom line is that electric forces between the microscopic elements of the spring are always way more intense than the gravitational forces between them. So in any physically relevant spring, gravitational "self-interaction" will not play a role.
This was to say that the potential energy $V$ will remain that of the original massless spring: $V=frac{1}{2}k Delta x^2$, where $Delta x$ is the extension of the spring.
Now in order to understand how the Hook law is changed you should write down the equations of motion. If we add a constant external force $F_{ext}$ to the system it will act as a power source (modifying the energy with a term $+Delta x; F_{ext}$).
We can obtain the equations of motion just deriving the energy $E$ with respect to the time:
$$frac{dE}{dt}=dot{E}equiv0=int_0^L!!dl;[lambda;dot{v}(l,t);v(l,t)]+Deltadot{ x}left( kDelta x +F_{ext}right)$$
where we recognize that $Deltadot{x}=v(L,t)$ (if $v(0,t)=0$).
In conclusion, the extension resulting from the application of the external force will be given by:$$kDelta x=-F_{ext}-int_0^L!!dl;lambda;dot{v}(l,t);frac{v(l,t)}{v(L,t)}$$
This means that only in the static limit in which the spring has stopped accelerating the Hook law will be again valid.
answered yesterday
france95france95
37412
37412
$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
add a comment |
$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As we are discussing about the energy of a spring , I am having a question that like any rigid body when we apply work energy theorem we basically find out the work done by the external forces on the body and equals it with the change in the kinetic energy of the rigid body. But if we take a spring with some finite mass then can we apply the work energy theorem in the same way like that of a rigid body? Or int he equation we need to consider the work done by the internal forces inside the spring also?
$endgroup$
– Rifat Safin
13 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
$begingroup$
As you say, you need to take into account internal forces too: the "internal" energy of the spring changes ( the potential changes if you compress the spring). It is not a rigid body AND geometrically different configurations do not have the same energy
$endgroup$
– france95
5 hours ago
add a comment |
$begingroup$
The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.
Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.
We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.
Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.
However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.
$endgroup$
2
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
add a comment |
$begingroup$
The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.
Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.
We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.
Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.
However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.
$endgroup$
2
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
add a comment |
$begingroup$
The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.
Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.
We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.
Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.
However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.
$endgroup$
The law $F=-kx$ is a very good mathematical approximation of the real springs. But any real spring doesn't obey this law precisely. There are always some differences.
Massless spring is obviously a mathematical abstraction as well. In most circumstances we can treat real springs as massless. But not always, it depends on the problem to be solved.
We can use yet another mathematical abstraction: "ideal spring with mass". But we can't require that it always obeys the law $F=-kx$, because it would be not a spring but something weird.
Imagine that we have extended the spring and let it go. Even our mathematically ideal spring can't instantly get to it's original length parts of the spring have mass and can't move very fast. So there will be a period of time when the spring is still extended, but there is no force applied to it.
However this mathematical abstraction can obey the law $F=-kx$ when there are no movements in the system.
answered yesterday
lesniklesnik
2,1941612
2,1941612
2
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
add a comment |
2
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
2
2
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
In the context of the first paragraph, it might be worth adding that any real thing doesn't obey any law precisely. All physical laws are just approximations; we strive to make them so good that any deviations are unmeasurably small, but in practice that's not always necessary.
$endgroup$
– leftaroundabout
yesterday
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
$begingroup$
@leftaroundabout Agree. Sometimes I have a feeling that 2 + 2 does not equal 4 precisely. Because there are always some inaccuracy, always! But that's not correct of course. It's mathematics. It's only in physics that there are always some inaccuracies. But then charges of proton and positron are equal. Absolutely equal! Sum of charges of electron and proton is zero, and this zero is as good as zero element of group of natural numbers!
$endgroup$
– lesnik
18 hours ago
add a comment |
$begingroup$
Hooke's law differs from other physical laws in an important respect:
Springs closely obey Hooke's law because they are carefully manufactured to closely obey Hooke's law.
Hooke's law is not a fundamental law - this should be obvious when you consider that a spring is not a naturally occurring object, but a manufactured piece of engineering.
It is possible to buy springs which do not obey Hooke's law, which are manufactured for specific purposes. In this respect Hooke's law is more akin to an engineering specification than a physical law, though for the purposes of teaching physics it is treated as one.
So in truth the answer to "will a spring obey Hooke's law under some specific conditions", is that you should consult the manufacturer's data sheet. This will give information such as the maximum extension before the spring is permanently deformed and so on.
New contributor
$endgroup$
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
add a comment |
$begingroup$
Hooke's law differs from other physical laws in an important respect:
Springs closely obey Hooke's law because they are carefully manufactured to closely obey Hooke's law.
Hooke's law is not a fundamental law - this should be obvious when you consider that a spring is not a naturally occurring object, but a manufactured piece of engineering.
It is possible to buy springs which do not obey Hooke's law, which are manufactured for specific purposes. In this respect Hooke's law is more akin to an engineering specification than a physical law, though for the purposes of teaching physics it is treated as one.
So in truth the answer to "will a spring obey Hooke's law under some specific conditions", is that you should consult the manufacturer's data sheet. This will give information such as the maximum extension before the spring is permanently deformed and so on.
New contributor
$endgroup$
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
add a comment |
$begingroup$
Hooke's law differs from other physical laws in an important respect:
Springs closely obey Hooke's law because they are carefully manufactured to closely obey Hooke's law.
Hooke's law is not a fundamental law - this should be obvious when you consider that a spring is not a naturally occurring object, but a manufactured piece of engineering.
It is possible to buy springs which do not obey Hooke's law, which are manufactured for specific purposes. In this respect Hooke's law is more akin to an engineering specification than a physical law, though for the purposes of teaching physics it is treated as one.
So in truth the answer to "will a spring obey Hooke's law under some specific conditions", is that you should consult the manufacturer's data sheet. This will give information such as the maximum extension before the spring is permanently deformed and so on.
New contributor
$endgroup$
Hooke's law differs from other physical laws in an important respect:
Springs closely obey Hooke's law because they are carefully manufactured to closely obey Hooke's law.
Hooke's law is not a fundamental law - this should be obvious when you consider that a spring is not a naturally occurring object, but a manufactured piece of engineering.
It is possible to buy springs which do not obey Hooke's law, which are manufactured for specific purposes. In this respect Hooke's law is more akin to an engineering specification than a physical law, though for the purposes of teaching physics it is treated as one.
So in truth the answer to "will a spring obey Hooke's law under some specific conditions", is that you should consult the manufacturer's data sheet. This will give information such as the maximum extension before the spring is permanently deformed and so on.
New contributor
New contributor
answered yesterday
BenBen
1214
1214
New contributor
New contributor
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
add a comment |
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Why do I have a patronising "new contributor" flag? I've been here ages.
$endgroup$
– Ben
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
$begingroup$
Because it's your first answer.
$endgroup$
– BPP
yesterday
add a comment |
$begingroup$
Real springs have several properties that are not modeled by Hooke's linear law. You can find comparisons pretty easily. Experiments with springs might be one of the easiest to perform at home:
I don't see any experimental data on evaluating effect of spring's mass on force produced, probably because in most experiments mass of the spring is much smaller than the force applied to the spring to deform it. After all, mass of spring is linked with its spring constant k
. Thicker wire of a spring will increase its mass, but also will increase spring constant, i.e. it will require more force to deform it to the same length
$endgroup$
add a comment |
$begingroup$
Real springs have several properties that are not modeled by Hooke's linear law. You can find comparisons pretty easily. Experiments with springs might be one of the easiest to perform at home:
I don't see any experimental data on evaluating effect of spring's mass on force produced, probably because in most experiments mass of the spring is much smaller than the force applied to the spring to deform it. After all, mass of spring is linked with its spring constant k
. Thicker wire of a spring will increase its mass, but also will increase spring constant, i.e. it will require more force to deform it to the same length
$endgroup$
add a comment |
$begingroup$
Real springs have several properties that are not modeled by Hooke's linear law. You can find comparisons pretty easily. Experiments with springs might be one of the easiest to perform at home:
I don't see any experimental data on evaluating effect of spring's mass on force produced, probably because in most experiments mass of the spring is much smaller than the force applied to the spring to deform it. After all, mass of spring is linked with its spring constant k
. Thicker wire of a spring will increase its mass, but also will increase spring constant, i.e. it will require more force to deform it to the same length
$endgroup$
Real springs have several properties that are not modeled by Hooke's linear law. You can find comparisons pretty easily. Experiments with springs might be one of the easiest to perform at home:
I don't see any experimental data on evaluating effect of spring's mass on force produced, probably because in most experiments mass of the spring is much smaller than the force applied to the spring to deform it. After all, mass of spring is linked with its spring constant k
. Thicker wire of a spring will increase its mass, but also will increase spring constant, i.e. it will require more force to deform it to the same length
answered yesterday
aaaaaaaaaaaa
584213
584213
add a comment |
add a comment |
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If you are interested in Mathematica solutions, you can check out library.wolfram.com/infocenter/MathSource/7773
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– Bill Watts
yesterday