What Frequency Range Keeps the Spring Safe in Driven Oscillations?

[Benzoate]

1. The problem statement, alml variables and given/known data

A block of mass 2 kg is suspended from a fixed support by a spring of strength 2000 N m^-1. The block is subject to the vertical driving force 36 cos pt N. Given that the spring will yield if its extension exceeds 4 cm, find the range of frequencies that can safely be applied. Take g = 10 m s^-2

Homework Equations

m*x'' + b*x'+kx= F(t)
x''+2*K*x' + $$\Omega$$₂=F₀cos(pt)

The Attempt at a Solution

2*x''+ b*x'+ 2000*x=36*cos(pt) N

what is the extension, and how would I find the range of frequencies.

 

[Hootenanny]

m*x'' + b*x'+kx= F(t)
x''+2*K*x' + $$\Omega$$₂=F₀cos(pt)

Are you sure that you have set up your ODE correctly? There are no velocity dependent forces specified in the question and you don't seem to have any term to account for the weight of the mass.

 

[Benzoate]

Are you sure that you have set up your ODE correctly? There are no velocity dependent forces specified in the question and you don't seem to have any term to account for the weight of the mass.

I had the last equation written wrong.

m*x'' + b*x'+kx= F(t)
x''+2*K*x'+x*$$\Omega$$²=F₀*cos(pt)
I didn't include the weight into the equation because I wasn't sure how to incorporate weight into the oscillation driven equation since there is no weight term.

 

[Hootenanny]

Let's start again from first principles. Newton's second law states that

m*x'' = F_net

Where F_net is the sum of all forces acting on the body. So what is the sum of all the forces acting on you block?

 

[Benzoate]

Let's start again from first principles. Newton's second law states that

m*x'' = F_net

Where F_net is the sum of all forces acting on the body. So what is the sum of all the forces acting on you block?

F_net= F_spring+F_grav

F_spring= m*x'' + b*x'+kx
F_grav= m*g

can ignore drag forces for now.

 

[gabbagabbahey]

Are you sure about your F_spring? Is there really a damping force specified in this question? And why is there an mx'' in there?!

 

[Benzoate]

Are you sure about your F_spring? Is there really a damping force specified in this question? And why is there an mx'' in there?!

I got the equation
Fspring= m*x'' + b*x'+kx , from the textbook Classical Mechanics written by John R. Taylor.

The problem doesn't specify that there is a damping constant present, but I thought that a damping constant could not be present only when the oscillator is under ideal conditions, i.e. no dissipative force presents and the oscillator oscillates forever.

 

[Hootenanny]

F_net= F_spring+F_grav

F_spring= m*x'' + b*x'+kx
F_grav= m*g

can ignore drag forces for now.

You say that we can ignore drag forces, yet you include a damping term (b*x') in your spring equation. You also pull a second order differential term out of thin air.

One other thing to be careful of is your signs. The weight of the mass will always act down and would there usually be negative is a standard coordinate system.

 

[Hootenanny]

I got the equation
Fspring= m*x'' + b*x'+kx , from the textbook Classical Mechanics written by John R. Taylor.

You can't just blindly apply equations from texts without knowing what they mean! Close that textbook and read the question. Now, what are the forces acting on the mass?

 

[Benzoate]

You can't just blindly apply equations from texts without knowing what they mean! Close that textbook and read the question. Now, what are the forces acting on the mass?

Fnet= Fspring-Fgrav

You are right. drag forces would be present only when a damping constant is present, well at least for a simple harmonic.
x''+2Kx'+(x*(omega)^2)=X''

X=36 cos (pt)
X'= -36 p sin (pt)
X''= -36 p² cos(pt)

Therefore

x''+2Kx'+(x*(omega)^2)=-36 p² cos(pt)

 

[Hootenanny]

Fnet= Fspring-Fgrav

You are right. drag forces would be present only when a damping constant is present, well at least for a simple harmonic.
x''+2Kx'+(x*(omega)^2)=X''

X=36 cos (pt)
X'= -36 p sin (pt)
X''= -36 p² cos(pt)

Therefore

x''+2Kx'+(x*(omega)^2)=-36 p² cos(pt)

I still have no idea where you're pulling these equations from. You still have a damping here in there (x'), there is no damping force. As I said above, the best way to start this question is to simply write down the sum of the forces acting on the mass. In other words, write the following expression explicitly.

Fnet= Fspring-Fgrav

F_net = ...

 

[Benzoate]

I still have no idea where you're pulling these equations from. You still have a damping here in there (x'), there is no damping force. As I said above, the best way to start this question is to simply write down the sum of the forces acting on the mass. In other words, write the following expression explicitly.

F_net = ...

x'' + 2Kx' + $$\Omega$$²=F(t) is the spring equation for driven harmonic motion. I am not pulling these equations out of thin air. How can there not be a damping present when the only case where a damping constant is not present in the equation for an oscillator is when the linear oscillator is a classical linear oscillator, where external conditions allow oscillator to bounce up and down forever without stopping. And the problem doesn't say that this external driving force will act upon the oscillator forever , so I don't think I can just assume no damping constant is present.

 

[Benzoate]

F_net = ...

F_external=F_spring -F_grav ; what other forces could there possibly be?

 

[gabbagabbahey]

F_external=F_spring -F_grav ; what other forces could there possibly be?

There is also a driving force isn't there? ;0)

What are the expressions for each of these three forces,

F_spring=?
F_grav=?
F_driving=?

 

[Hootenanny]

x'' + 2Kx' + $$\Omega$$²=F(t) is the spring equation for driven harmonic motion. I am not pulling these equations out of thin air. How can there not be a damping present when the only case where a damping constant is not present in the equation for an oscillator is when the linear oscillator is a classical linear oscillator, where external conditions allow oscillator to bounce up and down forever without stopping. And the problem doesn't say that this external driving force will act upon the oscillator forever , so I don't think I can just assume no damping constant is present.

If you assume that there is a damping force, then you can not find a numerical solution (the solution will be a function for the damping parameter). You are not told that the system is damped, nor are you given any information regarding the damping constant, nor can you work out the damping constant with the information provided.

 

[Benzoate]

There is also a driving force isn't there? ;0)

What are the expressions for each of these three forces,

F_spring=?
F_grav=?
F_driving=?

My external force is my driving force

F_grav=-mg
F_spring=-kx
F_external= 36*cos(pt)
36*cos(pt)= -mg - kx

 

[gabbagabbahey]

Shouldn't F_net=F_grav+F_spring+F_driving? Also since F_spring is presumably the restoring force that the spring imparts onto the mass, shouldn't it be +kx to signify that it is pulling the mass upward (while gravity is pushing it downward)?

 

[Benzoate]

Shouldn't F_net=F_grav+F_spring+F_driving? Also since F_spring is presumably the restoring force that the spring imparts onto the mass, shouldn't it be +kx to signify that it is pulling the mass upward (while gravity is pushing it downward)?

For an damped undriven oscillator

x''+2*K*x' +omega^2*x=0

for a driven harmonic oscillator

x''+2*K*x'+omega^2*x=F_driven

and F_grav i thought was always pointing down so shouldn't F_grav have a negative sign?

 

[gabbagabbahey]

For an damped undriven oscillator

x''+2*K*x' +omega^2*x=0

for a driven harmonic oscillator

x''+2*K*x'+omega^2*x=F_driven

and F_grav i thought was always pointing down so shouldn't F_grav have a negative sign?

Forget about those two equations for a second and continue on with F_net (the sum of all forces acting on m).

F_net=?

 

[Benzoate]

Forget about those two equations for a second and continue on with F_net (the sum of all forces acting on m).

F_net=?

Okay.

F_net= F_spring + F_driven-F_gravWhy should I be concerned with the net force?

 

[gabbagabbahey]

Okay.

F_net= F_spring + F_driven-F_gravWhy should I be concerned with the net force?

Because from Newton's second law, the net force is also equal to ma=mx'', so you get a differential equation for x:

mx''=F_net=F_spring + F_driven+F_grav

(this is how the equations of motion for the oscillator in your book are derived!)

Substitute your expressions for F_spring,F_driven,F_grav into this; what do you get?

 

[Benzoate]

Because from Newton's second law, the net force is also equal to ma=mx'', so you get a differential equation for x:

mx''=F_net=F_spring + F_driven+F_grav

(this is how the equations of motion for the oscillator in your book are derived!)

Substitute your expressions for F_spring,F_driven,F_grav into this; what do you get?

F_spring= -kx

F_grav= -mg

F_driven = 36 cos(pt)

mx''= 36 cos(pt)-mg-kx

 

[gabbagabbahey]

Close, F_spring should be positive since it is pulling the mass upwards. So,

mx''= 36 cos(pt)-mg+kx

=> mx''-kx=36cos(pt)-mg

Which is an inhomogeneous ODE. Have you tackled one of these before?

 

[Benzoate]

Close, F_spring should be positive since it is pulling the spring upwards. So,

mx''= 36 cos(pt)-mg+kx

=> mx''-kx=36cos(pt)-mg

Which is an inhomogeneous ODE. Have you tackled one of these before?

I think so. I would treat this as an auxillary equation:

mx''-kx= 36 cos(pt)-mg ==>What should I assume about pt
==> x''-(k/m)*x = 36/m*cos(pt) - g
Should I let x= ce^ipt?

 

[Benzoate]

should I also take the amplitude into account:

a= F₀/(($$\Omega$$²-p²)²-4K²p²)^1/2

r^2-1000= 18 cos pt -10

let F= 18e^(ipt)

x'' -1000x = 18 e^(ipt)
x= ce^(ipt)
x'= cip*e^(ipt)
x''= -cp^2*e^(ipt)

e^(ipt) cancel and I am left with :

(cp^2*e^(ipt))-1000(ce^(ipt))= 18e^(ipt) -10

Normally e^(ipt) would cancel , but since I have an extra term where the e^(ipt) is not a coefficient therefore , I can't cancel out e^ipt easily

I don think I will have any qualms with obtaining the complementary solution since:

r^2-1000= 0

therefore

x=Ae^(-31.6277t) + B*t*e^(-31.62777t)

what role will the extension of the spring play in my general equation ?

 

[Hootenanny]

Close, F_spring should be positive since it is pulling the mass upwards. So,

mx''= 36 cos(pt)-mg+kx

No, Benzoate had it correct first time. F_spring is a restoring force so it is going to be positive sometimes, but negative other times. In the ODE, and Hooke's law for that matter, x represents the displacement from the equilibrium position; in other words when the mass is simply hanging stationary from the spring, it is at position x=0. Now, when the mass drops below this point (negative x) the restoring [spring] force is going to be directed upwards, toward the origin (equilibrium position). However, when the mass is above this point (positive x), the restoring [spring] force will be directed downwards. Therefore, the correct expression is

F_spring = -kx

As Benzoate had it originally. Hence, the correct form of the ODE is

mx'' = 36cos(pt) - kx - mg

Do you follow Benzoate? This is an important point to note. I'll help you with solving the ODE next.

 

[Benzoate]

No, Benzoate had it correct first time. F_spring is a restoring force so it is going to be positive sometimes, but negative other times. In the ODE, and Hooke's law for that matter, x represents the displacement from the equilibrium position; in other words when the mass is simply hanging stationary from the spring, it is at position x=0. Now, when the mass drops below this point (negative x) the restoring [spring] force is going to be directed upwards, toward the origin (equilibrium position). However, when the mass is above this point (positive x), the restoring [spring] force will be directed downwards. Therefore, the correct expression is

F_spring = -kx

As Benzoate had it originally. Hence, the correct form of the ODE is

mx'' = 36cos(pt) - kx - mg

Do you follow Benzoate? This is an important point to note. I'll help you with solving the ODE next.

yes, I understand now why spring force is in the negative direction. However , my x=e^ipt fails to cancel out nicely .

 

[gabbagabbahey]

Start by finding the complimentary solution; that is the solution to mx''+kx=0 (sorry about the +-kx confusion BTW)

 

[Benzoate]

Start by finding the complimentary solution; that is the solution to mx''+kx=0 (sorry about the +-kx confusion BTW)

x=e^ipt
x'=ipe^ipt
x''=-p^2*e^ipt

e^ipt cancels and I am left with

mp^2 - k=0

dividing out the mass I now have:

p^2-k/m=0

plugging in k and m I get:

p^2=1000

p1= $$\sqrt{1000}$$

p2=- ($$\sqrt{1000}$$)the equation for complementary function is :

x= c1e^p1*t + c2e^p2*t

now I think what I only have left is to calculate the amplitude and the equation for the driven response x^D

Not sure how to calculate to x^D easily since the e^ipt factor doesn't dissipear so easily .

what role will the extension play in the problem

 

[Hootenanny]

x=e^ipt
x'=ipe^ipt
x''=-p^2*e^ipt

e^ipt cancels and I am left with

mp^2 - k=0

dividing out the mass I now have:

p^2-k/m=0

plugging in k and m I get:

p^2=1000

p1= $$\sqrt{1000}$$

p2=- ($$\sqrt{1000}$$)the equation for complementary function is :

x= c1e^p1*t + c2e^p2*t

now I think what I only have left is to calculate the amplitude and the equation for the driven response x^D

Not sure how to calculate to x^D easily since the e^ipt factor doesn't dissipear so easily .

what role will the extension play in the problem

Your solution looks okay, except that you're missing a factor i from your complementary function. In the first line of your solution you say assume that

x=e^^ipt

And then you determine the values of p,

$$p_{1,2}=\pm10\sqrt{10}$$

Hence, your complementary function should be

x= c₁e^ip1*t + c₂e^ip2*t

Do you follow? It is the fact that these exponentials are complex, that we allow us to write the complimentary function in a form that is little easier to deal with.

To answer your question regarding extension, x is the extension of the spring beyond it's equilibrium position.

 

[Benzoate]

Your solution looks okay, except that you're missing a factor i from your complementary function. In the first line of your solution you say assume that

x=e^^ipt

And then you determine the values of p,

$$p_{1,2}=\pm10\sqrt{10}$$

Hence, your complementary function should be

x= c₁e^ip1*t + c₂e^ip2*t

Do you follow? It is the fact that these exponentials are complex, that we allow us to write the complimentary function in a form that is little easier to deal with.

yes but what about the function for the driven response ? I still have p^2 in my equation; I will right down what I mean:

x''-1000x= 36e^ipt -10
let x= ce^ipt. then x'= cipe^ipt and x''= -cp^2e^ipt;
Therefore ,-cp^2e^ipt-1000( ce^ipt)=36e^ipt -10 ; canceling out e^ipt , I am left with:

-cp^2-1000=36-10e^-ipt; I still have that pesky e^-iptterm in my equation , and therefore cannot have an amplitude composed of only real and imaginary parts and therefore cannot calculate x^D.

To answer your question regarding extension, x is the extension of the spring beyond it's equilibrium position.

not sure what you mean. Should I subtract 4 cm from the equilibrium point?

 

[gabbagabbahey]

Why are you choosing e^ipt as a particular solution?

The nonhomogeneous terms in your ODE (mx''+kx=36cos(pt)-mg) are the 36cos(pt) and -mg terms. Are you familiar with the method of undetermined coefficients? If so, what are the forms of the suggested trial particular solutions for each of these two terms?

 

[Benzoate]

Why are you choosing e^ipt as a particular solution?

The nonhomogeneous terms in your ODE (mx''+kx=36cos(pt)-mg) are the 36cos(pt) and -mg terms. Are you familiar with the method of undetermined coefficients? If so, what are the forms of the suggested trial particular solutions for each of these two terms?

There is a problem like this in my textbook where F(t) has a cosine term and Euler's formula is applied.

the solutions to this problem say p is safe if p < 20 rd/s and p>40 rd/s; I don't know what that means since I found that p₁= $$+10$$$$\sqrt{10}$$ and p₂= $$-10$$$$\sqrt{10}$$

 

[gabbagabbahey]

There is a problem like this in my textbook where F(t) has a cosine term and Euler's formula is applied.

the solutions to this problem say p is safe if p < 20 rd/s and p>40 rd/s; I don't know what that means since I found that p₁= $$+10$$$$\sqrt{10}$$ and p₂= $$-10$$$$\sqrt{10}$$

But p doesn't need to be one of those two values (p1,p2). p1 and p2 are the values needed to obey the homogeneous part of the ODE, not the inhomogeneous part.

The general form of the particular solution should be Ae^ipt +Be^-ipt +C not just e^ipt. The first two terms are necessary to account for the 36cos(pt) term on the RHS, while the constant term is needed to account for the -mg constant term.

What do you get when you plug this particular solution into the ODE?

 

[Benzoate]

But p doesn't need to be one of those two values (p1,p2). p1 and p2 are the values needed to obey the homogeneous part of the ODE, not the inhomogeneous part.

The general form of the particular solution should be Ae^ipt +Be^-ipt +C not just e^ipt. The first two terms are necessary to account for the 36cos(pt) term on the RHS, while the constant term is needed to account for the -mg constant term.

What do you get when you plug this particular solution into the ODE?

So I wouldn't write out use Euler term because of the extra constant term?

x= Ae^ipt +Be^-ipt +C

x'= ipAe^ipt -ipBe^-ipt + 0

x''=-p^2Ae^ipt -B^2e^-ipt+0


x''-1000x= 36eipt -10 , but F(t)=36 cos(t), so trig terms will not completely go away .