Spring with mass hangs from the ceiling

[MatinSAR]

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.

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 $mg = -\frac {1}{2}kx$ ...

 

[Chestermiller]

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.

 

[MatinSAR]

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.

 

[kuruman]

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?

 

[MatinSAR]

Why?

Because my friend is trying to solve it using N's law withhout integrals and ...
I wanted to know If it is possible.

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 ...

 

[Chestermiller]

The maximum spring force is 2mg and the maximum displacement is $\frac{2mg}{k}$. The average spring tension is mg.

 

[MatinSAR]

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 $-\frac {1} {2} kx$?

 

[MatinSAR]

I think I have found a reasonable way to answer the way my friend suggests.

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 :
$F_{av}=\frac {1} {x} \int_0^x -kxdx = \frac {1} {x}( -\frac {1} {2} kx^2)=-\frac {1} {2} kx$

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?!

 

[Orodruin]

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.

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.

 

[MatinSAR]

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.

 

[Orodruin]

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).

 

[MatinSAR]

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.

 

[Orodruin]

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.

 

[MatinSAR]

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.

 

[kuruman]

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.

 

[MatinSAR]

Where there's a will, there's a way.

Yes! I was doing it in a wrong way.
Thank you for your time @kuruman .

 

[haruspex]

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.

 

[kuruman]

Yes it's a step. I mixed it up with another similar thread in which I posted recently.

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.

 

[haruspex]

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.

 

[kuruman]

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.

 

[haruspex]

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.

 

[kuruman]

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.

 

[MatinSAR]

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.

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.