Conservation of Energy with springs

[heroslayer99]

Homework Statement

A particle of mass m is suspended from a fixed point O by a light spring of natural length a and modulus of elasticity 4mg, the particle is pulled vertically down a distance d from its equilibrium position and released, find d if the particle just reaches O.

Relevant Equations

T = modulus of elasticity (extension) / natural length
EPE = modulus of elasticity (extension)^2 / 2 x natural length

Start by finding the equilibrium position, so we have {4mgx}/{a} = mg giving us x = a/4, therefore the spring's length is 5a/4. Now the loss in EPE (and therefore gain in energy of the particle) between the bottom and the equilibrium position is clearly 4mg((a/4 + d)^2 , and then from the equilibrium position up to O, the particle loses 4mg(a^2)/2a (as this is the gain in EPE), therefore in total the particle's energy has changed by 4mg((a/4 + d)^2 - 4mg(a^2)/2a, and this should equal the change in GPE (- the work done by gravity), which is clearly mg(5a/4 + d), this gives us d=5/4(a), whereas my textbook does something else, could someone let me know if either mine or the textbook's working is wrong? Thanks

 

[PeroK]

The textbook answer is correct. Although, this is for an elastic string, which requires no force/energy to compress.

The total EPE stored in the string is $\frac{4mg}{2a}(x - a)^2$, where $x$ is the total extension. In this case $x = a + \frac a 4 + d$.

Alternatively, the total EPE in the string is $\frac{4mg}{2a}(\Delta x)^2$, where $\Delta x$ is the extension beyond the natural length. In this case $\Delta x = \frac a 4 + d$.

I'm struggling a little to understand what you have done.

 

[heroslayer99]

The textbook says that the energy gained from the spring's extension is enough to reach the top if the distance is 3/4a, I said that we need a little more energy to squish the spring down from its natural length to 0.

Textbook also says this, does that not imply that it costs energy to squish the spring?

 

[PeroK]

The textbook says that the energy gained from the spring's extension is enough to reach the top if the distance is 3/4a, I said that we need a little more energy to squish the spring down from its natural length to 0.
View attachment 339387Textbook also says this, does that not imply that it costs energy to squish the spring?

The textbook solution is for a string, not a spring. A spring makes little physical sense in this case - although that would be nothing new!

But, you still have a mistake in using the intermediate equilibrium point. I don't understand why you subtracted that term.

 

[heroslayer99]

As the particle moves a distance of d + a/4 up, it gains energy, we know it gains an equal amount to the EPE of the string. Then as the particle moves a distance of a up, it loses energy (As it costs energy to squish the spring down to a length of 0, from its natural length). Is my misunderstanding that the spring does not need any force to compress? I have edited my post aswell, check if that makes more sense now.

 

[PeroK]

Is my misunderstanding that the spring does not need any force to compress?

The textbook solution is for a sTring (with a T) not a sPring (with a P). A sTring requires no energy to compress from its natural length.

 

[heroslayer99]

The textbook solution is for a sTring (with a T) not a sPring (with a P). A sTring requires no energy to compress from its natural length.

Ah I see. This is not mentioned explicitly in my book, is there a reason for this?

 

[PeroK]

I think I understand your approach. There is EPE stored in the initial extension from $a$ to $\frac{5a}{4}$ to reach the equilibrium position. The question is whether the mass is pulled far enough to utilise that energy. If you don't pull the mass at all, then that energy cannot be released. There is an assumption that the mass is pulled far enough to be accelerated up beyond the natural length and hence release all the EPE.

In this case, as $d = \frac{3a}{4}$, that is much greater than $\frac a 4$, so the string will definitely release all its EPE as the mass will go up and beyond the natural length.

 

[heroslayer99]

I am happy with all the EPE being released, the issue was that I thought the particle needed some energy to squish the string down from a length of a to 0, but apparently you don't need any extra energy for that.

 

[PeroK]

I am happy with all the EPE being released, the issue was that I thought the particle needed some energy to squish the string down from a length of a to 0, but apparently you don't need any extra energy for that.

In reality (think of a bungee jump), the string has some mass that needs to be accelerated upward. In this case, the assumption is that this is negligible.

 

[heroslayer99]

Alright, so we use the word light to mean little to no mass (effectively 0 mass), and this means that the string doesn't need any force to compress. However it still needs force to extend right? And if we use the word light to describe a spring, as my textbook does, do the same things still apply?

 

[PeroK]

Alright, so we use the word light to mean little to no mass (effectively 0 mass), and this means that the string doesn't need any force to compress. However it still needs force to extend right?

Yes. Think of a bungee rope as a good example.

And if we use the word light to describe a spring, as my textbook does, do the same things still apply?

A spring definitely needs force to compress. That would be a different problem - and EPE would again be stored in the spring as the mass moves upwards beyond the natural length.

My main complaint about even considering a spring is that it's unrealistic to assume that a spring (however light) is compressed to zero length.

A string simply gets out of the way and you can imagine the mass touching the ceiling at point O.

That was partly why I assumed that the question must be talking about a string in th first place!

 

[heroslayer99]

So here we would say that the epe at the start is equal to the epe at the end + the change in gpe.

 

[PeroK]

View attachment 339388
So here we would say that the epe at the start is equal to the epe at the end + the change in gpe.

Best to start a new thread. I'm offline now!