springs question

I need to find out the distance each of the springs has stretched from its equilibrium position.

It needs to be in terms of d (the max displacement of each ball from its initial position).

Any ideas? I have tried but to no avail.

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Reply 1

physicsgirlie

is v a force? if it is, you could consider juse one spring and the forces on it... not entirely sure how i woulod do this!

Reply 2

Sidhe

I'm a little confused as well, there are no variables here? We don't know anything about the system, are you sure there isn't more to this question?

Reply 3

Sidhe

Worzo

If that's just a static picture, then each string just stretches according to Hooke's law:

F = kx \rightarrow x = \frac{F}{k} = \frac{2v}{k}\text{cos}45

Hmm interesting I'm lost then? We have no numbers here? So how do you derive an answer.

Is it just an algebraic answer without an experimental result? I guess that's it, you just relate the answer to an equation.

d=

Reply 4

Theorist

Worzo

If that's just a static picture with force

v

, then each string just stretches according to Hooke's law:

F = kx \Rightarrow x = \frac{F}{k} = \frac{2v}{k}\text{cos}45

Unless it's dynamic, and it means velocity

v

, in which case it becomes a SHM problem.

1. But we need to give the answer in terms of

d,

where

d

is the maximum displacement of each ball from its initial position.
2. How would you solve the problem if it was 'dynamic'? How would the system operate then?

Reply 5

Morbo

Right, ignoring my earlier static approach (because who calls their force v, and it wants something to do with maximum displacement,

d

, which implies vibration), let's assume it's dynamic and it means velocity

v

, in which case it becomes a SHM problem.

The problem is identical for all the springs, so let

v' = v\text{cos}45 = v/\sqrt{2}

, the component of velocity of a mass in the direction of a spring with spring constant

k

.

Consider one such spring with a mass at either end:
..|->x₁...............|->x₂
(m) oooooooooo (m)

Hooke's law for each mass, with

w'^2 = \frac{k}{m}

\ddot x_2 = -w'^2(x_2 - x_1)

\ddot x_1 = w'^2(x_2 - x_1)

\ddot x_2 - \ddot x_1 = -2w'^2(x_2 - x_1)

So the frequency for the oscillation we are interested in is:

w^2 = 2w'^2 = \frac{2k}{m}

Now know from SHM for the i^th mass that:

\dot x_i^2 = w^2(a^2 - x_i^2)

Which in our problem translates to:

v'^2 = w^2((\frac{d}{2})^2 - (\frac{x}{2})^2)

The

\frac{x}{2}

and

\frac{d}{2}

are because the displacement of each mass only contributes to half the extension of the spring, so as to keep our notation consistent.

x^2 = d^2 - \frac{4v'^2}{w^2}

x^2 = d^2 - \frac{4m(v^2/2)}{2k}

x = \pm \sqrt{d^2 - \frac{mv^2}{k}}

I think!

Reply 6

bally

This is some more additional info about the question:

Four balls, each of mass m, are connected by four identical relaxed springs with spring constant k. The balls are simultaneously given equal initial speeds v directed away from the center of symmetry of the system.

Worzo: That is wayyyyy toooo complicated for this question.

It says to use 'geometry' ?

Reply 7

plmokn

Generally it helps if you tell people the whole question...
Anyway since the springs are initially relaxed equate initial kinetic energy and elastic potential energy when the balls are at their maximum displacement. Then it should be straightforward to find the maximum displacement of the balls in terms of the extension of the springs.

Reply 8

bally

sorry, i missed that part, was late at night :frown:

Reply 9

Theorist

I think that this is the right answer, but I'm really not sure. However, maybe it might guide you to the solution, if it isn't correct.

* * * * *

Say that the maximum displacement of each ball from its equilibrium position is

d

. (This is due to the kinetic energy of the balls and energy transfer - the loss of each ball's kinetic energy is converted to increased elastic potential energy in the spring. However, as the question asks us to give the answer in terms of

d

, we simply mask all of the energy transfer between the ball and the spring underneath this one variable).

Each of the balls will move diagonally, but its displacement will be used to extend two springs horizontally - the diagonal displacement of any ball can be resolved into horizontal and vertical components, which is responsible for the extension of two springs.

Therefore, the extension of one spring caused by the displacement of one ball is given by

d\cos45

. But, as is shown in the diagram, each spring is connected to two balls and so its total magnitude of displacement is

2d\cos45

.

We can equate this displacement to the increased elastic potential energy of the spring, which is given by

\Delta E_{P} = \frac{1}{2}kx^{2}

where

\Delta E_{P}

is the change in elastic potential energy of the springs,

k

is the spring constant and

x

is the extension of each spring.

Thus, we equate to give:

[indent]

2d\cos45 = \frac{1}{2}kx^{2}

[/indent]
[indent]

4d\cos45 = kx^{2}

[/indent]
[indent]

\frac{4d\cos45}{k} = x^{2}

[/indent]
[indent]

\therefore x = \sqrt{\frac{4d\cos45}{k}}.

[/indent]

Reply 10

Christophicus

Dharma

I think that this is the right answer, but I'm really not sure. However, maybe it might guide you to the solution, if it isn't correct.

* * * * *

Say that the maximum displacement of each ball from its equilibrium position is

d

d

d\cos45

. But, as is shown in the diagram, each spring is connected to two balls and so its total magnitude of displacement is

2d\cos45

.

We can equate this displacement to the increased elastic potential energy of the spring, which is given by

\Delta E_{P} = \frac{1}{2}kx^{2}

where

\Delta E_{P}

is the change in elastic potential energy of the springs,

k

is the spring constant and

x

is the extension of each spring.

Thus, we equate to give:

[indent]

2d\cos45 = \frac{1}{2}kx^{2}

[/indent]
[indent]

4d\cos45 = kx^{2}

[/indent]
[indent]

\frac{4d\cos45}{k} = x^{2}

[/indent]
[indent]

\therefore x = \sqrt{\frac{4d\cos45}{k}}.

[/indent]

S.I. units of 2dcos45 are m and S.I. units of 0.5kx² are kgm²s^-2 so they can't be equal.

Reply 11

Robob

Dharma

I think that this is the right answer, but I'm really not sure. However, maybe it might guide you to the solution, if it isn't correct.

* * * * *

Say that the maximum displacement of each ball from its equilibrium position is

d

d

d\cos45

. But, as is shown in the diagram, each spring is connected to two balls and so its total magnitude of displacement is

2d\cos45

.

We can equate this displacement to the increased elastic potential energy of the spring, which is given by

\Delta E_{P} = \frac{1}{2}kx^{2}

where

\Delta E_{P}

is the change in elastic potential energy of the springs,

k

is the spring constant and

x

is the extension of each spring.

Thus, we equate to give:

[indent]

2d\cos45 = \frac{1}{2}kx^{2}

[/indent]
[indent]

4d\cos45 = kx^{2}

[/indent]
[indent]

\frac{4d\cos45}{k} = x^{2}

[/indent]
[indent]

\therefore x = \sqrt{\frac{4d\cos45}{k}}.

[/indent]

wtf?

Where have you used the initial kinetic energy?

Reply 12

Theorist

Widowmaker

S.I. units of 2dcos45 are m and S.I. units of 0.5kx² are kgm²s^-2 so they can't be equal.

robob

wtf?

Where have you used the initial kinetic energy?

Right, so I've messed this up completely haven't I?! :redface:

I mean, if the SI units don't even match up then....

OK. What do you guys make of this? :smile:

For one ball acting on one spring, the horizontal component of the work done by the ball on the spring (due to the ball's kinetic energy) is transferred to the spring, increasing its elastic potential energy, to the point that at the maximum extension, the kinetic energy of the ball is zero.

However, the diagram shows that for each spring, two balls are acting on it. Therefore, the total work done on a spring will be double that stated in the previous paragraph.

Therefore:

[indent]

2\times\frac{1}{2}mv^{2} = Fd\cos 45

[/indent]
[indent]

mv^{2} = kxd\cos 45

[/indent]
[indent]

\therefore x = \frac{\sqrt{2}mv^{2}}{kd}.

[/indent]

Is that any better, if not correct?

Reply 13

physicsgirlie

well, your units work out at least!
looks ok to me, but then i'm not brilliant at mechanics.

edit: so a pretty pointless post then.

Reply 14

Morbo

I think this question is simpler than has been made out.

It wants to know the maximum extension of each spring,

x_0

, in terms of the maximum displacement of a mass from its equilibirum position,

d

. It doesn't ask for an expression for

d

.

It is just geometry. We know that a ball which moves a distance

d

in the direction of

v

will have a component of distance

d' = d\text{cos}45 = \frac{d}{\sqrt{2}}

along the direction of the spring.

Hence,

x_0 = 2 \times d' = \frac{2d}{\sqrt{2}} = \sqrt{2}d

.

That does exactly what the question asks.

Reply 15

Theorist

Worzo

I think this question is simpler than has been made out.

It wants to know the maximum extension of each spring,

x_0

, in terms of the maximum displacement of a mass from its equilibirum position,

d

. It doesn't ask for an expression for

d

.

It is just geometry. We know that a ball which moves a distance

d

in the direction of

v

will have a component of distance

d' = d\text{cos}45 = \frac{d}{\sqrt{2}}

along the direction of the spring.

Hence,

x_0 = 2 \times d' = \frac{2d}{\sqrt{2}} = \sqrt{2}d

.

That does exactly what the question asks.

Man points for that. :cool:

Typical of me (if you knew the sort of person I am) to overcomplicate things!

Incidentally though Worzo, if you wanted to find the extension of the spring in terms of all the variables given there, how would you do it?

You see I think that my method is right but I'm starting to wonder:

To find the magnitude of a ball's kinetic energy which does work in increasing the elastic potential energy of the spring, would you say it is equal to

\frac{mv^{2}}{2\cos 45}

\frac{m(v\cos 45)^{2}}{2}?

Basically what I'm trying to say is: to find the work done on the spring, do you resolve the velocity of the ball to find its horizontal component and then work out the work done by the ball; or do you work out the work done by the ball and then resolve it horizontally in the direction of the spring's motion? I ask because both ways yield different answers.

Reply 16

Morbo

Dharma

You resolve the velocity, because that is the vector. Kinetic energy has no direction and cannot be "resolved". If you want to split the kinetic energy into that of each component of motion, the energy just adds like:

KE_{total} = KE_x + KE_y + KE_z

because

\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2

See how because of the v², KE from each component just adds like a scalar? It's a very important result. (In thermodynamics, it leads to the Equipartition of Energy.) Anyway, I hope that helps you understand, because I'll use that in the next part...

You see I think that my method is right but I'm starting to wonder

No, your method is wrong. You use Work Done = Force x Distance, which is not right in this case. Elastic potential energy stored in a spring is given by

\text{EPE }= \frac{1}{2}kx^2

.

As I showed before, the maximum extension of the spring is given by

x_0 = \sqrt{2}d

. If you want to express

d

in terms of the variables given,

v

and

k

, then you have to solve the dynamics.

At maximum extension of the spring, all the initial kinetic energy of two masses along the direction of the spring goes into the elastic potential energy stored in the stretched spring:

2 \times \frac{1}{2}mv'^2 = \frac{1}{2}kx_0^2

Now, we know that

v' = v\text{cos}45^o = \frac{v}{\sqrt{2}}

, and we have an expression for

x_0

, hence:

2m(\frac{v}{\sqrt{2}})^2 = k(2d^2)

mv^2 = 2kd^2

Unparseable latex formula:

\Rightarrow d = \sqrt{\frac{mv^2}{2k}

Now using

Unparseable latex formula:

x_0 = \sqrt{2}d \Rightarrow x_0 = \sqrt{\frac{mv^2}{k}

Reply 17

bally

Thanks guys it was sqrt2 * d.

Reply 18

Theorist

Worzo

KE_{total} = KE_x + KE_y + KE_z

because

\frac{1}{2}mv^2 = \frac{1}{2}mv_x^2 + \frac{1}{2}mv_y^2 + \frac{1}{2}mv_z^2

That's exactly what I was thinking, after I posted the wrong method. This is because, like you said, if you resolve the velocity into its horizontal and vertical components, you'll get the kinetic energy of one ball in each direction is

\frac{1}{4}mv^{2}

, which if you double to take into account the velocity component in the other direction will give you

\frac{1}{2}mv^{2}

which is the total kinetic energy of the spring. I did it just now, and Pythagoras confirms it also.

However, one thing which I would like to clarify. I can see how the kinetic energies just add up like a scalar. But could you explain why this is because of the

v^{2}

(is it because squaring the velocity gets rid of the negative sign?), and why this is so significant?

Worzo

You use Work Done = Force x Distance, which is not right in this case.

Why? I'm not doubting you, it's just I can't get my head around physically why this is not the case.

Reply 19

Morbo

Dharma

However, one thing which I would like to clarify. I can see how the kinetic energies just add up like a scalar. But could you explain why this is because of the

v^{2}

(is it because squaring the velocity gets rid of the negative sign?), and why this is so significant?

Because Pythagoras says that the magnitude of a vector - for example the velocity - is given by

v^2 = v_x^2 + v_y^2 + v_z^2

. This is the geometry. The physics is that it just so happens that kinetic energy depends on

v^2

. Multiply the Pythagoras equation by

\frac{1}{2}m

and voila, you have the formula for adding kinetic energy contributions from each component.

Besides, kinetic energy is a scalar and scalars add like normal addition. The squaring is significant becase kinetic energy must always be positive, even if your velocity is negative.

Why? I'm not doubting you, it's just I can't get my head around physically why this is not the case.

Because

W = F \times d

only applies when a constant force is moved through a distance. In the case of stretching a spring, the force that must be applied to extend the spring changes with the extension. From physical experience, you know that it becomes harder to stretch a spring the more extended it is.

I'm not sure what level of calculus you're familar with, so I'll explain it simply. Imagine stretching the spring just a little bit, a distance