An advanced two spring weighing machine

[Rya_Sly]

Homework Statement

A more sophisticated weighing machine contains two springs. The effect of these is that the machine is more sensitive to masses less than some fixed value M0 than it is to greater masses. The equilibrium displacement is alpha1*M when M<M0, and (alpha2+c) when M≥M0 , where alpha1, alpha2 and c are constants.

An object of mass M0 is held in contact with the pan at x=0 and released at time t=0 .

Find an expression for the period of the subsequent motion, and find the period of the oscillations if the object mass M0=450g, and the values of the constants are: alpha1=0.800mkg−1 and alpha2=0.650mkg−1

Find an expression for the maximum reading of the pointer in terms of M0, alpha1 and alpha2. Calculate the maximum reading using M0=450g, alpha1=0.800mkg−1 and alpha2=0.650mkg−1

Homework Equations

F=-kx
SHM equation: mx''+kx=0

The Attempt at a Solution

The displacement has to be continuous when M=M0, so we can derive an expression for c:
alpha1*M0 = alpha2*M0+c
c= (alpha1-alpha2)*M0

I have been trying to figure out what the two spring system is - I tried two vertical springs in series with the pan and mass between them and I tried two springs in parallel, but neither reproduces the displacement behaviour of the problem. It's not clear to me if I should be trying to understand the setup of the system or they want me to solve a general case. Either way, I'm stumped.

 

[ehild]

Welcome to PF!
I think a picture of the set-up would be necessary. Is not there one somewhere around the problem text?

 

[haruspex]

You don't really need to understand the structure. You all the information about how it behaves. A possible design is that with no load only one spring is in contact with the pan. When the pan is depressed by alpha1*M0 it makes contact with the second spring.

 

[ehild]

You don't really need to understand the structure. You all the information about how it behaves. A possible design is that with no load only one spring is in contact with the pan. When the pan is depressed by alpha1*M0 it makes contact with the second spring.

Very wise! Can it be this?

 

[haruspex]

Very wise! Can it be this?

Yes, that's what I had in mind, thanks.

 

[Rya_Sly]

Thank you so much, that makes sense now! I think I was trying to make it more complicated than it needed to be.

 

[theoz]

Very wise! Can it be this?
View attachment 77494

How could this be used to give an answer for an expression for the period,

You don't really need to understand the structure. You all the information about how it behaves. A possible design is that with no load only one spring is in contact with the pan. When the pan is depressed by alpha1*M0 it makes contact with the second spring.

How does this help lead to an expression for the period of subsequent motion then?

 

[etotheipi]

How could this be used to give an answer for an expression for the period,
How does this help lead to an expression for the period of subsequent motion then?

At the critical mass, $M_0$, the equilibrium position of the pan has to be where it is just in contact with the shorter spring. This is so that any mass larger than $M_0$ causes both springs to be under compression, and any mass less than $M_0$ has only the larger spring under compression, at the respective equilibrium positions.

So for a mass of $M_0$, the '$x=0$' of the SHM will be at the level of the top of the shorter spring. So half of the cycle is SHM with both springs, the other half of the cycle is SHM with only the larger spring. You know the equations for the time period if the total spring constant is $k$, so just split the motion into two, half these, and add them together!

 

[theoz]

Thanks for the response. For the half of the period where both springs are in contact, how can K be determined as the spring constants of the two springs can't be assumed to be equal? The answer for the total time period has to be independent of the constant C. Many thanks

 

[etotheipi]

Thanks for the response. For the half of the period where both springs are in contact, how can K be determined as the spring constants of the two springs can't be assumed to be equal? The answer for the total time period has to be independent of the constant C. Many thanks

If in doubt, try and see what force results if you displace the springs by an amount $\delta x$, since by definition $\delta F = k\delta x$. Suppose the two springs have constants $k_1$ and $k_2$.

If you compress both springs by the same amount $\delta x$, the change in force is $\delta F = \delta F_1 + \delta F_2 = k_1 \delta x + k_2 \delta x = (k_1 + k_2)\delta x$. So in parallel, spring constants add.

Note that this is the same as what happens with capacitors, and the opposite of what happens with resistors. The analogy can sometimes be useful!

 

[theoz]

If in doubt, try and see what force results if you displace the springs by an amount $\delta x$, since by definition $\delta F = k\delta x$. Suppose the two springs have constants $k_1$ and $k_2$.

If you compress both springs by the same amount $\delta x$, the change in force is $\delta F = \delta F_1 + \delta F_2 = k_1 \delta x + k_2 \delta x = (k_1 + k_2)\delta x$. So in parallel, spring constants add.

Note that this is the same as what happens with capacitors, and the opposite of what happens with resistors. The analogy can sometimes be useful!

The issue I'm having is how can k be determined for the smaller spring independent of the constant c, so then k(effective) can be determined

 

[etotheipi]

The issue I'm having is how can k be determined for the smaller spring independent of the constant c, so then k(effective) can be determined

Ah, right, I see. Well the $\alpha$'s given in the problem statement have units $\text{m}\text{kg}^{-1}$, so something must be wrong with the definitions. I presume it really means that for $M < M_0$ the equilibrium extension is $M\alpha_1$ and for $M > M_0$ the equilibrium extension is $(M\alpha_2 + c)$, since that is at least dimensionally consistent. I presume also that all of the equilibrium extensions are measured from the top of the larger spring.

So now, in the case that $M<M_0$, $$Mg = k_1\alpha_1 M \implies k_1 = \frac{g}{\alpha_1}$$ Let's also define the difference in unstretched lengths of the springs to be $d$. Then for $M = M_0$, $$M_0 g = k_1 d \implies d = \frac{M_0 g}{k_1}$$ Finally, in the case $M>M_0$, $$Mg = k_1(\alpha_2 M + c) + k_2(\alpha_2 M + c -d)$$ Importantly this holds $\forall M>M_0$, so you can compare coefficients of $M$ and conclude that $g = k_1 \alpha_2 + k_2 \alpha_2$. This gives us $k_2 = g\left(\frac{1}{\alpha_2} - \frac{1}{\alpha_1}\right)$. We also conclude that $2c = d$, so the constant $c$ is half the difference between the unstretched lengths of the springs.

 

[theoz]

Ah, right, I see. Well the $\alpha$'s given in the problem statement have units $\text{m}\text{kg}^{-1}$, so something must be wrong with the definitions. I presume it really means that for $M < M_0$ the equilibrium extension is $M\alpha_1$ and for $M > M_0$ the equilibrium extension is $(M\alpha_2 + c)$, since that is at least dimensionally consistent. I presume also that all of the equilibrium extensions are measured from the top of the larger spring.

So now, in the case that $M<M_0$, $$Mg = k_1\alpha_1 M \implies k_1 = \frac{g}{\alpha_1}$$ Let's also define the difference in unstretched lengths of the springs to be $d$. Then for $M = M_0$, $$M_0 g = k_1 d \implies d = \frac{M_0 g}{k_1}$$ Finally, in the case $M>M_0$, $$Mg = k_1(\alpha_2 M + c) + k_2(\alpha_2 M + c -d)$$ Importantly this holds $\forall M>M_0$, so you can compare coefficients of $M$ and conclude that $g = k_1 \alpha_2 + k_2 \alpha_2$. This gives us $k_2 = g\left(\frac{1}{\alpha_2} - \frac{1}{\alpha_1}\right)$. We also conclude that $2c = d$, so the constant $c$ is half the difference between the unstretched lengths of the springs.

Very well explained, many thanks

 

[theoz]

In terms of finding a general expression for the maximum reading, at alpha1Mo, the reading will be Mo. However the amplitude of the motion when there are two springs touching is decreased by alpha2/alpha1 as spring constants have a linear relationship to extension. How is the maximum displacement from the top of the spring translated into the maximum reading from here?