Finding the period of pulley system

[songoku]

Homework Statement

Please see below

Relevant Equations

Simple harmonic motion

Newton's law

This is not homework. I found this question when browsing and there is also solution but I do not understand it. This is part of the solution:

My questions:
1)
From equation (i): ΔT = m.a, I think ΔT refers to W - T (weight of block - tension of string). I got this from free body diagram and taking $m$ moving downwards. Is it correct that ΔT = W - T?
If yes, why the solution states "net increment in the tension in the string will provide acceleration to the block" ? Shouldn't it be "decrement" in the tension, since the tension must be less than the weight for $m$ to move downwards? I assume initially the system is at rest.

2)
Why the tension on left string connected to pulley A is also ΔT? Shouldn't it be T?

3)
I don't understand equation (ii):
$$x_1=\frac{x-x_2}{2}$$

When $m$ moves $x$ meter downwards, both spring 1 and 2 will extend by $x_1$ and $x_2$ respectively. How to proceed to get that equation?

Thanks

 

[BvU]

Hello again,

If yes, why the solution states "net increment in the tension in the string

Check the signs: if $m$ is pulled down, $\Delta T$ is negative, as is $ma$ and the mass will accelerate upward as you would expect.

Why the tension on left string connected to pulley A is also ΔT? Shouldn't it be T?

With $\Delta T$ they mean the increment wrt equilibrium
As for the other variables in the picture.

The $x$ to the left of $K_1$ should be $x_1$.

When m moves x meter downwards, both spring 1 and 2 will extend by x1 and x2 respectively.

I don't agree with that.

How to proceed to get that equation?

By considering the length of the string !

$\$

 

[haruspex]

When m moves x meter downwards, both spring 1 and 2 will extend by x1 and x2 respectively.

I don't agree with that.

I believe that is how $x_1$ and $x_2$ are defined.
... unless you are objecting to the extraneous words "meter" and "both".

 

[haruspex]

... taking m moving downwards. Is it correct that ΔT = W - T?
If yes, why the solution states "net increment in the tension in the string will provide acceleration to the block" ? Shouldn't it be "decrement" in the tension, since the tension must be less than the weight for m to move downwards?

You are confusing the acceleration of the mass, its having moved downwards by some means, with the motion leading to that downward displacement.

 

[songoku]

Hello again,

With $\Delta T$ they mean the increment wrt equilibrium
As for the other variables in the picture.

Hello again BvU

What does it mean by increment with respect to equilibrium?

I don't agree with that.

When $m$ moves downwards, the string will pull spring 2 downwards so it extends by $x_2$ and to maintain equilibrium (maybe) spring 1 must be pulled upwards to provide downwards restoring force to balance upwards restoring force of spring 2 and upwards tension. That is what I think

By considering the length of the string !

Length of the string is constant but sorry I don't understand how to apply that. If $m$ moves downwards $x$ meters, spring 2 extends downwards by $x_2$ then I don't know how to continue

You are confusing the acceleration of the mass, its having moved downwards by some means, with the motion leading to that downward displacement.

I am really sorry, I don't understand this

This is how I will approach the problem.
(i) Draw free body diagram for $m$, which will be T upwards and W downwards

(ii) Assuming the system starts from rest and $m$ going down after being released, I set up Newton's 2nd law equation of motion:
$$W-T=m.a$$

(iii) Since the pulley is smooth, all the tension of string connected to pulley A and B will have same value T, so since spring 2 is extended by $x_2$, it means $T=k_2.x_2$

Then I stuck

Thanks

 

[Lnewqban]

The diagram should have identified the $x$ next to $k_1$ as $x_1$.

 

[haruspex]

Assuming the system starts from rest and m going down after being released

No, the mass m is hanging in equilibrium, then it is pulled down a small distance and released. Its acceleration will thereafter have opposite sign to its displacement from the equilibrium position.

 

[songoku]

No, the mass m is hanging in equilibrium, then it is pulled down a small distance and released. Its acceleration will thereafter have opposite sign to its displacement from the equilibrium position.

Oh I see, so initially ΔT should be T - W since after released, $m$ will go upwards

Two things I still do not understand:
1) Why the force in the left of pulley A is ΔT, not T

2) how to relate $x$ with $x_1$ and $x_2$

Thanks

 

[haruspex]

initially ΔT should be T - W

Yes, but it is because ΔT is defined to be the amount by which T exceeds the equilibrium value, which is W.

Why the force in the left of pulley A is ΔT, not T

The force is T, but, again, what we want is the amount by which the force exceeds the equilibrium value. That corresponds to the increase in extension beyond its equilibrium value.

2) how to relate $x$ with $x_1$ and $x_2$

If B moves up by $\delta$, what are the changes in $x, x_1, x_2$? So what combinations are constant?

 

[Lnewqban]

You could simplify this problem to behave like a system of one mass linked to two springs arranged in parallel, having one of them (#1), a mechanical advantage over the mass of 0.5 (class 3 lever).

 

[songoku]

The force is T, but, again, what we want is the amount by which the force exceeds the equilibrium value. That corresponds to the increase in extension beyond its equilibrium value.

Why the one being considered is the amount that exceeds the equilibrium value? The solution states that $\Delta T=k_2.x_2$ but from my own working I got $T=k_2.x_2$

If B moves up by $\delta$, what are the changes in $x, x_1, x_2$? So what combinations are constant?

I am not sure. I don't think I am able to visualize it correctly.

Maybe $x_1=\delta$, and the string connected to the left and right of pulley B will be shorter by the same amount (?) so $x_2=\frac{\delta}{2}$ and $x=\delta$?

You could simplify this problem to behave like a system of one mass linked to two springs arranged in parallel, having one of them (#1), a mechanical advantage over the mass of 0.5 (class 3 lever).

Sorry I don't understand this but I will go back to this later

Thanks

 

[BvU]

Why the one being considered is the amount that exceeds the equilibrium value? The solution states that $\Delta T=k_2.x_2$ but from my own working I got $T=k_2.x_2$

There should be no $T$ in your working. All variables appearing are zero at equilibrium. $T$ is not.
But $x_2$ is zero at equilibrium, so $T = k_2x_2$ does not hold. $\Delta T = k_2x_2$.

$\$

 

[haruspex]

Why the one being considered is the amount that exceeds the equilibrium value? The solution states that $\Delta T=k_2.x_2$ but from my own working I got $T=k_2.x_2$

$x_2$ is not the extension of spring 2. It is the change in extension as a result of the mass moving away from its equilibrium position.
Suppose, at equilibrium, the tension is T. Then the extension of that spring is $T/k_2$. A movement away from equilibrium increases the extension to $T/k_2+x_2$ and the tension to $T+\Delta T$. So
$T+\Delta T=k_2(T/k_2+x_2)=T+k_2x_2$.

Maybe $x_1=\delta$, and the string connected to the left and right of pulley B will be shorter by the same amount (?) so $x_2=\frac{\delta}{2}$ and $x=\delta$?

Yes, $x_1=\delta$. For the others, consider the total distance from the top of spring 2, round both pulleys and down to the mass. Since the string is constant, the increase in that total length is $x_2$.
Now break it down into
- the section from the top of spring 2, to pulley B
- the section from pulley B to pulley A, and
- the section from pulley A down to the mass.
Since the distance from ceiling to floor is constant, each of the first two decreases by $\delta$, while the third increases by x.

 

[songoku]

Since the distance from ceiling to floor is constant, each of the first two decreases by $\delta$, while the third increases by x.

You mean the third increases by $\delta$?

Yes, $x_1=\delta$. For the others, consider the total distance from the top of spring 2, round both pulleys and down to the mass. Since the string is constant, the increase in that total length is $x_2$.
Now break it down into
- the section from the top of spring 2, to pulley B
- the section from pulley B to pulley A, and
- the section from pulley A down to the mass.
Since the distance from ceiling to floor is constant, each of the first two decreases by $\delta$, while the third increases by x.

If for this case, the string on the left and right side of pulley connected to $m_2$ will each decrease by amount of $\frac x 2$ if $m_1$ moves downwards as far as $x$?

If yes, why the section from the top of spring 2, to pulley B and the section from pulley B to pulley A, each decreases by $\frac{\delta}{2}$?

Thanks

 

[haruspex]

You mean the third increases by δ?

No, x. Look at the diagram: x is defined to be how much further m gets from A, so is the distance m descends.

If for this case, the string on the left and right side of pulley connected to $m_2$ will each decrease by amount of $\frac x 2$ if $m_1$ moves downwards as far as $x$?

Yes.

If yes, why the section from the top of spring 2, to pulley B and the section from pulley B to pulley A, each decreases by $\frac{\delta}{2}$?

I wrote that each decreases by δ, not δ/2. And I gave the reason in post #13.

 

[songoku]

I wrote that each decreases by δ, not δ/2. And I gave the reason in post #13.

I am sorry, my question should be: why the section from the top of spring 2, to pulley B and the section from pulley B to pulley A, not each decreases by $\frac{\delta}{2}$

And sorry, I don't understand the hint about "the distance from ceiling to floor is constant". What it has to do with $\delta$ and $x$?

No, x. Look at the diagram: x is defined to be how much further m gets from A, so is the distance m descends.

I don't need to state $x$ in terms of $\delta$?

Thanks

 

[haruspex]

I am sorry, my question should be: why the section from the top of spring 2, to pulley B and the section from pulley B to pulley A, not each decreases by $\frac{\delta}{2}$

And sorry, I don't understand the hint about "the distance from ceiling to floor is constant". What it has to do with $\delta$ and $x$?I don't need to state $x$ in terms of $\delta$?

Thanks

If B moves up by δ then B is closer to the ceiling by that amount, so x1 is reduced by δ. Similarly, the vertical distance from ceiling to A is fixed, so the vertical distance from B to A is reduced by δ.

The steps I laid out in post #13 give you two ways of expressing the change in the total distance from the top of spring 2, round both pulleys and down to the mass. Equating these should give you the relationship between the three x variables.

 

[songoku]

If B moves up by δ then B is closer to the ceiling by that amount, so x1 is reduced by δ. Similarly, the vertical distance from ceiling to A is fixed, so the vertical distance from B to A is reduced by δ.

The steps I laid out in post #13 give you two ways of expressing the change in the total distance from the top of spring 2, round both pulleys and down to the mass. Equating these should give you the relationship between the three x variables.

Change in total distance from the top of spring 2, round both pulleys and down to the mass
= change in the section from the top of spring 2, to pulley B + change in the section from pulley B to pulley A + change in the section from pulley A down to the mass.
= $x_2 - \delta + x$ (where $x_2= - \delta$)

Is this correct?

Sorry, I don't know the other way to express the change in total distance

Thanks

 

[haruspex]

Change in total distance from the top of spring 2, round both pulleys and down to the mass
= change in the section from the top of spring 2, to pulley B + change in the section from pulley B to pulley A + change in the section from pulley A down to the mass.
= $x_2 - \delta + x$ (where $x_2= - \delta$)

Is this correct?

No. I may have misled you with a typo in post #13, corrected.
The change in distance from the top of spring 2 to pulley B is not x2. x2 is the change in length of spring 2. But yes, the change in the section from the top of spring 2, to pulley B is $-x_1=- \delta$.
Since spring 2 has extended by $x_2$, and the string length is fixed, what is another expression for the change in that total length?

 

[songoku]

Since spring 2 has extended by $x_2$, and the string length is fixed, what is another expression for the change in that total length?

Using your hint, I already get the equation relating $x,x_1$ and $x_2$.

I am just curious, is it possible to state $x_1$ in terms of $x_2$ only?

Thanks

 

[haruspex]

Using your hint, I already get the equation relating $x,x_1$ and $x_2$.

I am just curious, is it possible to state $x_1$ in terms of $x_2$ only?

Thanks

Since you have an equation relating x, x1 and x2, and you know x can vary, it is clear that the relationship between x1 and x2 varies.

 

[songoku]

You could simplify this problem to behave like a system of one mass linked to two springs arranged in parallel, having one of them (#1), a mechanical advantage over the mass of 0.5 (class 3 lever).

1) How to know that the system can be simplified to one mass linked to two springs arranged in parallel?

2) What I know about mechanical advantage is that it is the ratio between the weight of the load moved to force exerted to move the load, or the ratio of how far I need to pull a string to how far the load moves. In this case, what does mechanical advantage of 0.5 mean?
Two springs connected in parallel usually will have same extension, but not for this case. Is it because of the mechanical advantage?

Thanks

 

[haruspex]

You could simplify this problem to behave like a system of one mass linked to two springs arranged in parallel, having one of them (#1), a mechanical advantage over the mass of 0.5 (class 3 lever).

I think you mean in series.
For springs in parallel, forces add, displacements match.
For springs in series, forces match, displacements add.
The present case is like springs in series except that there is a factor to be applied to one displacement and to one force.