Finding maximum height of a string before it goes slack

[green-beans]

Homework Statement

A mass m is suspended by a light elastic string. When the mass remains at rest it is at a point 0, which is a distance a + b below the point from which the string is suspended from the ceiling, where a is the natural length of the string. The mass is pulled down a distance h below 0 and released from rest. The moving mass is subject to air resistance of magnitude cv per unit mass, where c is a positive constant and v is the speed of the mass.
Show that if bc² < 4g then the largest value of h such that the string never becomes slack is h = be^(πµ/ν), where µ = c/2 and ν = 1/2√{(4g/b) − c²}.

Homework Equations

I deduced the differential equation of motion in terms of the displacement x of the mass below 0 (until the string goes slack) to be x'' + cx' + (g/b)x = 0.

The Attempt at a Solution

bc² < 4g occurs when the characteristic equation has 2 complex roots λ = (-c/2) ± i{√|c² - 4g/b|}/2
So, the solution would be
x=e^(-c/2)t*(Asin{1/2*√|c²-4g/b|}t + Bcos{1/2*√|c²-4g/b|}t)
Then I tried differentiating the expression and setting first derivative to zero to find the maximum value of x but it did not work as I could not even solve the equation. Also I do not quite understand where π in the answer came from.
I would appreciate if someone could give me some direction on how to do this question.

Thanks!

 

[PeroK]

Homework Statement

A mass m is suspended by a light elastic string. When the mass remains at rest it is at a point 0, which is a distance a + b below the point from which the string is suspended from the ceiling, where a is the natural length of the string. The mass is pulled down a distance h below 0 and released from rest. The moving mass is subject to air resistance of magnitude cv per unit mass, where c is a positive constant and v is the speed of the mass.
Show that if bc² < 4g then the largest value of h such that the string never becomes slack is h = be^(πµ/ν), where µ = c/2 and ν = 1/2√{(4g/b) − c²}.

Homework Equations

I deduced the differential equation of motion in terms of the displacement x of the mass below 0 (until the string goes slack) to be x'' + cx' + (g/b)x = 0.

The Attempt at a Solution

bc² < 4g occurs when the characteristic equation has 2 complex roots λ = (-c/2) ± i{√|c² - 4g/b|}/2
So, the solution would be
x=e^(-c/2)t*(Asin{1/2*√|c²-4g/b|}t + Bcos{1/2*√|c²-4g/b|}t)
Then I tried differentiating the expression and setting first derivative to zero to find the maximum value of x but it did not work as I could not even solve the equation. Also I do not quite understand where π in the answer came from.
I would appreciate if someone could give me some direction on how to do this question.

Thanks!

I think what you have is correct. It must be simpler to set $\nu =$ the complicated expression you get from solving the quadratic. This keeps the algebra simpler.

The next step is to find the coefficients $A$ and $B$. Why not use the initial conditions $x(0) = \dot{x}(0) = 0$?

 

[green-beans]

I think what you have is correct. It must be simpler to set $\nu =$ the complicated expression you get from solving the quadratic. This keeps the algebra simpler.

The next step is to find the coefficients $A$ and $B$. Why not use the initial conditions $x(0) = \dot{x}(0) = 0$?

Thank you for your reply! However, I keep on getting that A and B are zero if I use that x(0)=x'(0)=0 and this reduces the whole equation to 0 which is impossible.

 

[BvU]

You have to define where $x=0$ is. I figured $x=0$ when length of string is $a+b$ ( Ah, the problem statement says so ). That way $x(0) = -h,\ \ \dot x(0)=0$ makes more sense.

 

[Delta2]

I think we need to define mathematically what the condition "the string never becomes slack" means. Does it mean (as I assume) that $x(t)\leq b$ because since the system will do oscillation around the point O, if the displacement becomes greater than b then the string will become(at a certain time point) less than its natural size which I suppose that's what slack means.

 

[PeroK]

Thank you for your reply! However, I keep on getting that A and B are zero if I use that x(0)=x'(0)=0 and this reduces the whole equation to 0 which is impossible.

Apologies, it should be, of course, $x(0) = h$ and $\dot{x}(0) = 0$.

The other point is that even if you had not been given the definitions of $\mu$ and $\nu$ you should have done this yourself. Once you have solved the characteristic equation, you should have set:

$\lambda = -\mu \pm i\nu$

Or, something similar. Then your general solution is:

$x(t) = e^{-\mu t}(A\cos(\nu t) + B\sin(\nu t))$

Which is much easier to work with, clearer to read and quicker to type into Latex.

 

[green-beans]

Apologies, it should be, of course, $x(0) = h$ and $\dot{x}(0) = 0$.

The other point is that even if you had not been given the definitions of $\mu$ and $\nu$ you should have done this yourself. Once you have solved the characteristic equation, you should have set:

$\lambda = -\mu \pm i\nu$

Or, something similar. Then your general solution is:

$x(t) = e^{-\mu t}(A\cos(\nu t) + B\sin(\nu t))$

Which is much easier to work with, clearer to read and quicker to type into Latex.

Should not $x(0) = - h$ since the mass starts below 0? Also I obtained that A = -h and B= -μh/nu but I don't quite understand how I can now find the greatest value of h. Also, since the string never becomes slack, does it mean that x(t) < 0?
Thanks in advance!

 

[PeroK]

Should not $x(0) = - h$ since the mass starts below 0? Also I obtained that A = -h and B= -μh/nu but I don't quite understand how I can now find the greatest value of h. Also, since the string never becomes slack, does it mean that x(t) < 0?
Thanks in advance!

I took downwards to be the positive direction. $x(0) = -h$ will work just as well, as it's the same general solution either way.

The solution will be damped SHM. The mass will move upwards. What is the key factor when it reaches its highest point?

PS The string will go slack if it reaches its natural length of $a$.

 

[green-beans]

I took downwards to be the positive direction. $x(0) = -h$ will work just as well, as it's the same general solution either way.

The solution will be damped SHM. The mass will move upwards. What is the key factor when it reaches its highest point?

PS The string will go slack if it reaches its natural length of $a$.

Thank you for your reply! At the highest point its velocity should become 0 so it will not go slack, is it right? If so I will need to solve v=0 as far as I understand. If yes, I then get that sin(nu)t *([μ²h/nu + h (nu)) = 0 but I cannot still see where the given expression for h comes from.

 

[PeroK]

Thank you for your reply! At the highest point its velocity should become 0 so it will not go slack, is it right? If so I will need to solve v=0 as far as I understand. If yes, I then get that sin(nu)t *([μ²h/nu + h (nu)) = 0 but I cannot still see where the given expression for h comes from.

Well, there's only one way that that expression is 0!

 

[green-beans]

Well, there's only one way that that expression is 0!

Yes, of course, it's when t=pi/nu. However, to obtain the answer from what I got it must be that (x(pi/nu)) = b which I don't quite understand why.

 

[PeroK]

Yes, of course, it's when t=pi/nu. However, to obtain the answer from what I got it must be that (x(pi/nu)) = b which I don't quite understand why.

The natural length of the string is $a$. It is stretched to a length $a+b$ by the mass $m$. This is the point $x = 0$. The string is then stretched to a length of $a+b+h$ (that's why I thought $x(0) = h$ was more natural than $x(0) = -h$, but no matter).

The mass rebounds to its highest point at $t = \frac{\pi}{\nu}$. If $x( \frac{\pi}{\nu}) > b$ (using your directions), then the mass has rebounded beyond the natural length of the string, which has gone slack (and, in fact, in this case the equations and solution are no longer valid). So, $x(\frac{\pi}{\nu}) = b$ is the maximum rebound.

That's the logic of the question.

 

[green-beans]

The natural length of the string is $a$. It is stretched to a length $a+b$ by the mass $m$. This is the point $x = 0$. The string is then stretched to a length of $a+b+h$ (that's why I thought $x(0) = h$ was more natural than $x(0) = -h$, but no matter).

The mass rebounds to its highest point at $t = \frac{\pi}{\nu}$. If $x( \frac{\pi}{\nu}) > b$ (using your directions), then the mass has rebounded beyond the natural length of the string, which has gone slack (and, in fact, in this case the equations and solution are no longer valid). So, $x(\frac{\pi}{\nu}) = b$ is the maximum rebound.

That's the logic of the question.

Oh right, this makes sense! Thank you so much - I now understand it :)