Spring with mass hangs from the ceiling

  • #1
MatinSAR
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Homework Statement
A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2
meter. A mass m is now hung on the lower end of the spring. Imagine supporting the mass
with your hand so that the spring remains relaxed (that is, not expanded or compressed),
then suddenly remove your supporting hand. The mass and spring oscillate. The lowest
position of the mass during the oscillations is 0.1 m below the place it was resting when
you supported it. What is the spring constant?
Relevant Equations
Conservation of energy.
I know that we can answer it using conservation of energy or using N's 2nd law.
1700228585000.png


Using N's 2nd Law:
##F = mv \frac {dv}{dx}##
##Fdx = mvdv##
For spring we have : ##F=-kx##
##(mg-kx)dx=mvdv##
We'll get same result using above equation.

My question:
Average spring force from 0 to x is ##-\frac {1}{2}kx## ... Can't we say that at lowest point spring force could be equal to ##-\frac {1}{2}kx## instead of ##-kx##? Then at lowest point we have ##mg = -\frac {1}{2}kx## ...
 
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  • #2
MatinSAR said:
Homework Statement: A massless spring with no mass attached to it hangs from the ceiling. Its length is 0.2
meter. A mass m is now hung on the lower end of the spring. Imagine supporting the mass
with your hand so that the spring remains relaxed (that is, not expanded or compressed),
then suddenly remove your supporting hand. The mass and spring oscillate. The lowest
position of the mass during the oscillations is 0.1 m below the place it was resting when
you supported it. What is the spring constant?
Relevant Equations: Conservation of energy.

I know that we can answer it using conservation of energy or using N's 2nd law.
View attachment 335630

Using N's 2nd Law:
##F = mv \frac {dv}{dx}##
##Fdx = mvdv##
For spring we have : ##F=-kx##
##(mg-kx)dx=mvdv##
We'll get same result using above equation.

My question:
Average spring force from 0 to x is ##-\frac {1}{2}kx## ... Can't we say that at lowest point spring force could be equal to ##-\frac {1}{2}kx## instead of ##-kx##? Then at lowest point we have ##mg = -\frac {1}{2}kx## ...
At the lowest point, it's not at equilibrium.
 
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  • #3
Chestermiller said:
At the lowest point, it's not at equilibrium.
I certainly agree. If it was at equilibrium then It wouldn't come back upward after reaching the lowest point.
I was trying to answer using average spring force.
 
  • #4
MatinSAR said:
I certainly agree. If it was at equilibrium then It wouldn't come back upward after reaching the lowest point.
I was trying to answer using average spring force.
Why? The gravitational force is constant therefore its average over any length is equal to its value at any length. As you have found, the average spring force depends on the length over which you average it. What is the motivation for setting them equal when they are not?
 
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  • #5
kuruman said:
Why?
Because my friend is trying to solve it using N's law withhout integrals and ...
I wanted to know If it is possible.
kuruman said:
The gravitational force is constant therefore its average over any length is equal to its value at any length. As you have found, the average spring force depends on the length over which you average it. What is the motivation for setting them equal when they are not?
At lowest point, Aren't they equal?! (##mg=\frac 1 2 kx##)

He doesn't know why average spring force at lowest point is ##\frac 1 2 kx## but he believes he can use this to solve ...
 
  • #6
The maximum spring force is 2mg and the maximum displacement is ##\frac{2mg}{k}##. The average spring tension is mg.
 
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  • #7
Chestermiller said:
The maximum spring force is 2mg and the maximum displacement is ##\frac{2mg}{k}##. The average spring tension is mg.
I have used:
##F_{av}=\frac {1} {x} \int_0^x -kxdx = \frac {1} {x}( -\frac {1} {2} kx^2)=-\frac {1} {2} kx##

According to this, Can't I say that at lowest point ##mg## should be equal to ##-\frac {1} {2} kx##?
 
  • #8
I think I have found a reasonable way to answer the way my friend suggests.

kuruman said:
As you have found, the average spring force depends on the length over which you average it.
According to this post I know at distance x below the point that mass M released average spring force is :
##F_{av}=\frac {1} {x} \int_0^x -kxdx = \frac {1} {x}( -\frac {1} {2} kx^2)=-\frac {1} {2} kx##
Chestermiller said:
The maximum spring force is 2mg and the maximum displacement is ##\frac{2mg}{k}##. The average spring tension is mg.
And according to this post average spring force at lowest point at distance x below the point that mass M released is equal to ##mg##.

So at lowest point ##mg## should be equal to ##\frac {1} {2} kx## be cause we've just calculated average force in 2 different ways. It is not related to equilibirium. Am I right?!
 
  • #9
Chestermiller said:
The maximum spring force is 2mg and the maximum displacement is ##\frac{2mg}{k}##. The average spring tension is mg.
Average over time. Not average over position as op is doing.
MatinSAR said:
I have used:
##F_{av}=\frac {1} {x} \int_0^x -kxdx = \frac {1} {x}( -\frac {1} {2} kx^2)=-\frac {1} {2} kx##

According to this, Can't I say that at lowest point ##mg## should be equal to ##-\frac {1} {2} kx##?
No. You are averaging over length. Force is change of momentum per time.
 
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  • #10
Orodruin said:
Average over time. Not average over position as op is doing.
No. You are averaging over length. Force is change of momentum per time.
Good point. So there is no way to use average spring force over lenght (## -\frac {1} {2} kx##) to answer the question.
 
  • #11
Well, the argument using the integral of force over length is just the energy argument. Since the gravitational side has a constant force the energy argument actually becomes average force over length equal to the gravitational force (with opposite sign).
 
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  • #12
Orodruin said:
Well, the argument using the integral of force over length is just the energy argument. Since the gravitational side has a constant force the energy argument actually becomes average force over length equal to the gravitational force (with opposite sign).
Yes! If ##mg## is constant, According to what I said, at any distance x below the released point ##\frac {1} {2} kx## should be equal to ##mg##. But this is wrong.
 
  • #13
MatinSAR said:
Yes! If ##mg## is constant, According to what I said, at any distance x below the released point ##\frac {1} {2} kx## should be equal to ##mg##. But this is wrong.
No, this is wrong. The energy argument is based on zero kinetic energy and the mass does not have zero kinetic energy everywhere, only at the turning points.
 
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  • #14
Orodruin said:
No, this is wrong. The energy argument is based on zero kinetic energy and the mass does not have zero kinetic energy everywhere, only at the turning points.
Things are getting complicated. Now I prefer to use conservation of energy Or N's law instead of trying to use average force ...

Thanks everyone for their help and time.
 
  • #15
MatinSAR said:
Things are getting complicated. Now I prefer to use conservation of energy Or N's law instead of trying to use average force ...
It can still be done with time averages if you start with an equation that you know is valid at all times and take its time average. What is true at all times should be true for the averages. Newton's law is true at all times but ##mg=-\frac{1}{2}kx## is not.

Here is how to do it.

We want to find the equilibrium point which is distance ##x_0## below the starting point. We also know that the mass oscillates with amplitude ##A## about the equilibrium point with frequency ##\omega =\sqrt{k/m}.## So we write the position of the mass relative to the starting point as $$x(t)=x_0-A\sin(\omega t).$$ Since ##x_0## is constant, the time-average of this over interval ##\Delta t## is $$x_{\text{avg}}=x_0-\frac{\int_0^{\Delta t}A\sin(\omega t)~dt}{\int_0^{\Delta t} dt}.$$ Over a complete oscillation ##\Delta t = T=~##one period, the average of the sinusoidal is zero and ##x_{\text{avg}}=x_0.##

Now we write Newton's second law $$\begin{align} m\ddot x & =-kx-mg \nonumber \\
-m\omega^2 A\sin(\omega t) & =-k[x_0-A\sin(\omega t)]-mg \nonumber
\end{align} $$ and average both sides of the equation to get $$ 0 =-k[x_0-0]-mg \implies x_0=-\frac{mg}{k}.$$ Where there's a will, there's a way.
 
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  • #16
kuruman said:
Where there's a will, there's a way.
Yes! I was doing it in a wrong way.
Thank you for your time @kuruman .
 
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  • #17
kuruman said:
We want to find the equilibrium point
That's not what the question asks for. Perhaps you mean that as a step towards finding the lowest point.
The equilibrium point is easy: tension=weight. The challenge is to relate that to maximum displacement.
Consider a position x above equilibrium. The net upward force there is ##k(x_0-x)-mg=-kx##. Since this is symmetric about x=0, the whole motion has that symmetry. It was at rest at ##x=x_0##, so must also be at rest at ##x=-x_0##, so the lowest position is ##2x_0## below the release position.
 
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  • #18
Yes it's a step. I mixed it up with another similar thread in which I posted recently.
haruspex said:
The challenge is to relate that to maximum displacement.
Not much of a challenge if one considers that the mass is started at a turning point. We already found that the distance to the equilibrium point, a.k.a. amplitude, is ##x_0=\dfrac{mg}{k}.## The distance between turning points is twice the amplitude. Same argument as the last sentence in post #17.
 
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  • #19
kuruman said:
The distance between turning points is twice the amplitude.
That fact rests on, e.g., solving the SHM equation. If the target is to avoid integration, that's cheating. The aim of my symmetry argument is to avoid that.
 
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  • #20
haruspex said:
That fact rests on, e.g., solving the SHM equation.
Not necessarily. Consider a plot of the parabola representing the potential energy. Draw a line parallel to the x-axis representing the constant mechanical energy. The points of intersection of line and parabola are the turning points, symmetrically situated about the origin.

Besides, why is the target to avoid integration? In post #1 the OP presents the solution using energy conservation from the link in post #18 and then wonders if the answer can be obtained by some sort of averaging procedure. There is no mention of avoiding integration.
 
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  • #21
kuruman said:
Not necessarily. Consider a plot of the parabola representing the potential energy. Draw a line parallel to the x-axis representing the constant mechanical energy. The points of intersection of line and parabola are the turning points, symmetrically situated about the origin.

Besides, why is the target to avoid integration? In post #1 the OP presents the solution using energy conservation from the link in post #18 and then wonders if the answer can be obtained by some sort of averaging procedure. There is no mention of avoiding integration.
According to post #5, the wish is to solve using neither energy conservation nor integrals. Why bother?, you ask. To that I have no answer.
 
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  • #22
haruspex said:
According to post #5, the wish is to solve using neither energy conservation nor integrals. Why bother?, you ask. To that I have no answer.
I have no answer either and the "friend" who wishes to see this solved without integrals is not available to clarify the point. I think I'm done here.
 
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  • #23
haruspex said:
The equilibrium point is easy: tension=weight. The challenge is to relate that to maximum displacement.
I'm not sure why i didn't notice this last night. Thank you.
kuruman said:
Yes it's a step. I mixed it up with another similar thread in which I posted recently.

Not much of a challenge if one considers that the mass is started at a turning point. We already found that the distance to the equilibrium point, a.k.a. amplitude, is ##x_0=\dfrac{mg}{k}.## The distance between turning points is twice the amplitude. Same argument as the last sentence in post #17.
Now it makes sense. Thanks.

@haruspex @kuruman Thank you for your time. I'm sorry for the poor question I asked and I'm sorry that I've wasted your time. I hope I won't ask such questions again.
 

FAQ: Spring with mass hangs from the ceiling

What is Hooke's Law and how does it apply to a spring hanging from the ceiling?

Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position, which can be mathematically expressed as F = -kx. In the case of a spring hanging from the ceiling, the weight of the mass attached to the spring causes it to stretch until the restoring force of the spring balances the gravitational force, resulting in a new equilibrium position.

How do you calculate the spring constant (k) for a spring hanging from the ceiling?

The spring constant (k) can be calculated by measuring the displacement (x) of the spring from its equilibrium position when a known mass (m) is attached. Using Hooke's Law, F = kx, and knowing that the force (F) is equal to the weight of the mass (mg), we get k = mg/x.

What is the potential energy stored in a spring hanging from the ceiling?

The potential energy (U) stored in a stretched or compressed spring is given by the formula U = (1/2)kx^2, where k is the spring constant and x is the displacement from the equilibrium position. This energy is the result of the work done to stretch or compress the spring.

How does the mass affect the oscillation period of a spring hanging from the ceiling?

The oscillation period (T) of a mass-spring system is determined by the formula T = 2π√(m/k), where m is the mass and k is the spring constant. This means that the period of oscillation increases with the mass and decreases with a stiffer spring (higher k).

What factors can affect the behavior of a spring-mass system hanging from the ceiling?

Several factors can affect the behavior of a spring-mass system, including the spring constant (k), the mass (m) attached to the spring, gravitational acceleration (g), and any damping forces such as air resistance. Additionally, the material properties of the spring and any limitations in its elastic range can also influence its behavior.

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