Force transmitted to a block in viscous fluid through a spring, cord, and pulley

[zenterix]

Homework Statement

Consider a block of mass $m$ moving in a viscous fluid that provides frictional force.

The block is attached to a cord that runs over a pulley and is attached to a spring, as shown below. The spring has spring constant $K$.

You hold on to the other end of the spring and move it back and forth with displacement

$$d_0\cos{\omega_d t}\tag{1}$$

Relevant Equations

The change in length of the spring from its equilibrium length is the difference

$$x(t)-d_0\cos{\omega d t}\tag{2}$$

The equation of motions looks like

$$m\ddot{x}(t)+m\Gamma\dot{x}(t)=-K(x(t)-d_0\cos{\omega_d t})\tag{3}$$

Moving other end of the spring sinusoidally effectively produces a sinusoidally varying force on the mass.

Everything written above so far is as presented by the book "The Physics of Waves" by Howard Georgi.

I am a bit confused by the setup presented above.

The picture seems to show that we displace the end of the spring a distance $d_0\cos{\omega_d t}$ from the position the end occupies when it is in equilibrium.

It is not clear what $x(t)$ is measuring. It would seem that $x(t)=d_0\cos{\omega t}$.

How do we interpret (2)?

What does it mean to "move the other end of the spring" sinusoidally? Are they simply saying that we could alternatively have a different forcing function on the same end we are pulling and pushing on?

Finally, consider the situation in which the spring is stretched because we've pulled it and then the moment in which we start to push it back towards the equilibrium position. Would the cord connecting the spring to the mass not become slack?

 

[zenterix]

The text then goes on as follows.

Dividing the equation of motion by $m$ the equation becomes identical to a previously seen equation, namely

$$\ddot{x}(t)+\Gamma\dot{x}(t)+\omega_0^2x(t)=\frac{F_0}{m}\cos{\omega_d t}\tag{4}$$

but with $\frac{F_0}{m}=\frac{Kd_0}{m}=\omega_0^2d_0$.

The complexified equation is

$$\ddot{z}(t)+\Gamma\dot{z}(t)+\omega_0^2z(t)=\omega_0^2d_0e^{-i\omega_d t}\tag{5}$$

To me this seems to indicate that (3) is incorrect.

If it is incorrect, then my interpretation is that (2) is actually incorrect and (3) should be

$$m\ddot{x}(t)+m\Gamma\dot{x}(t)=Kd_0\cos{\omega_d t}$$

with the displacement from spring equilibrium being just $d_0\cos{\omega_d t}$.

 

[BvU]

Hi,

Can you agree that $x$ is the displacement of the left end of the spring (blue arrow)?

And then set up the force balance at that point ?
(I find the $\leftrightarrow$ with the $d_0\cos\omega_dt$ confusing -- I think it should be a $\leftarrow$ if $d_0>0$.
Start with $d_0=0$ if you want to set it up step by step)

$$x(t)-d_0\cos{\omega d t}\tag{2}$$

It would seem that $x(t)=d_0\cos\omega t$.

That would correspond to $K=\infty$: a very stiff spring

I agree with you that the description is awkward ($m\ddot x$ ignores Archimedes altogether !!)

$\$

 

[haruspex]

Can you agree that $x$ is the displacement of the left end of the spring (blue arrow)?

The author probably intends $x$ as the vertical displacement of the mass, but it comes to the same thing. Curiously, according to equations (2) and (3), $x$ is positive down and the displacement of the hand is positive to the left. But it is all consistent, which is what matters.

(I find the $\leftrightarrow$ with the $d_0\cos\omega_dt$ confusing -- I think it should be a $\leftarrow$ if $d_0>0$.

Surely $d_0$ is the amplitude, so a constant $>0$.

$m\ddot x$ ignores Archimedes altogether !!

How would that enter into it, so long as the block is fully immersed?

Moving other end of the spring sinusoidally effectively produces a sinusoidally varying force on the mass.

Interesting… that's not immediately obvious, but as the equations show, the identical motion could be achieved by a sinusoidal force applied to the mass instead of a sinusoidal displacement of the hand.

What does it mean to "move the other end of the spring" sinusoidally? Are they simply saying that we could alternatively have a different forcing function on the same end we are pulling and pushing on?

No, it means the right-hand end of the spring is moving sinusoidally.

Finally, consider the situation in which the spring is stretched because we've pulled it and then the moment in which we start to push it back towards the equilibrium position. Would the cord connecting the spring to the mass not become slack?

No, that would only happen if the spring becomes shorter than its relaxed length.

 

[zenterix]

Suppose we have a spring with spring constant $K$.

Let $O$ be any origin for our coordinate system (which is 1-dimensional).

Let $r_{O,R}$ denote the position of the right tip of the spring and $r_{O,L}$ the position of the left tip of the spring.

$$r_{O,R}=x_R$$

$$r_{O,L}=x_L$$

Let's also define the vectors

$$r_{L\to R}=r_{O,R}-r_{O,L}$$

$$r_{R\to L}=r_{O,L}-r_{O,R}=-r_{L\to R}$$

When the spring is in equilibrium we have

$$r_{L\to R,eq}=(r_{O,R}-r_{O,L})_{eq}=l_{eq}$$

$$r_{R\to L,eq}=-l_{eq}$$

The force on the right tip is

$$F_R=-K(r_{L\to R}-r_{L\to R,eq})$$

$$=-K((r_{O,R}-r_{O,L})-l_{eq})$$

Similarly, the force on the left tip is

$$F_L=-K((r_{R\to L}-r_{R\to L,eq})$$

$$=-K((r_{O,L}-r_{O,R})+l_{eq})$$

$$=K((r_{O,R}-r_{O,L})-l_{eq})$$

$$=-F_R$$

First let's apply these equations to the simple case of a spring attached to a wall.

Let the origin of the coordinate system be the position of the right tip when the spring is in equilibrium.

Since the left tip's position is fixed (to the wall), the origin of the coordinate system is also fixed.

$$r_{0,L}=-l_{eq}$$

$$r_{0,R}=0$$

$$F_R=-K((r_{O,R}-(-l_{eq})-l_{eq})=-Kr_{O,R}$$

The rightmost expression is the one that is usually expressed as Hooke's law in all the textbooks I have ever read.

In addition, we have

$$F_L=-F_R$$

Now let's consider the case of the OP. Here is a picture of that situation

In the leftmost picture the spring is in equilibrium.

If we consider the origin of our coordinate system to be the vertical light blue line then in the leftmost picture we have

$$r_{O,L}=0$$

$$r_{0,R}=l_{eq}$$

Then

$$F_R=-K((r_{O,R}-r_{O,L})-l_{eq})$$

$$=-K((l_{eq}-0)-l_{eq})$$

$$=0$$

$$F_L=-F_R=0$$

Now consider the two pictures on the right. That is, consider a situation in which the spring is out of equilibrium.

$$r_{O,L}=x(t)$$

$$r_{O,R}=l_{eq}+d_0\cos{\omega_d t}$$

$$F_R=-K((l_{eq}+d_0\cos{\omega_d t}-x(t))-l_{eq})$$

$$=-K(d_0\cos{\omega_d t}-x(t))$$

$$F_L=-F_R=-K(x(t)-d_0\cos{\omega_d t})$$

Thus, the book seems to be considering the force represented by this last expression. This is the force $F$ in the pictures above.

It is the force on the left tip of the spring.

Now, it seems that in case (3) of the pictures, ie the rightmost picture, the spring is compressed and so as noted by haruspex, this case would mean the cord tied to the mass is slack since the spring is pushing on the cord.

It is not clear to me why this case can or cannot happen for this problem.

The equation of motion used is for the block. It is assumed that the force $F_L$ derived above is always transmitted to the block.

Can the cord go slack?

I also just noticed that there is no gravitational force being considered.

 

[haruspex]

Now, it seems that in case (3) of the pictures, ie the rightmost picture, the spring is compressed and so as noted by haruspex, this case would mean the cord tied to the mass is slack since the spring is pushing on the cord.

It is not clear to me why this case can or cannot happen for this problem.

You are overlooking that $l_{eq}=l_{relaxed}+(mg-B)/k$, where B is the buoyant force. Only when $l_{eq}+x<l_{relaxed}$ will the cord go slack, i.e. $x<(B-mg)/k$, which must be negative or the mass would float.

 

[Chestermiller]

Let $x_L$ be the horizontal coordinate of the left side of the spring relative to the center of the pulley and let $x_R$ be the corresponding horizontal coordinate of the right side of the spring. Then the tension in the spring is $$T=k(x_R-x_L-l_0)$$where $l_0$ is the unextended length of the spring. Then the force balance on the mass, neglecting buoyancy is $$k(x_R-x_L-l_0)-mg-m\Gamma\frac{dx_L}{dt}=m\frac{d^2x_L}{dt^2}\tag{1}$$At equilibrium, this becomes: $$k(x_{Re}-x_{Le}-l_0)-mg=0\tag{2}$$where Le and Re refer to the equilibrium locations of the left end and the right end resprectively. If we subtract Eqn. 2 from Eqn.1, we obtain:$$-k(x-d_0\cos{\omega_d t})-m\Gamma\frac{dx}{dt}=m\frac{d^2x}{dt^2}\tag{3}$$where $x=x_L-x_{Le}$ and $d_0\cos{\omega_d t}=x_R-x_{Re}$