What Frequency Range Keeps the Spring Safe in Driven Oscillations?

[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 .

 

[Hootenanny]

I'm going to jump out here, two people helping in a thread isn't constructive.

Let me know if you need me to jump back in.

 

[gabbagabbahey]

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

x= -p^2Ae^ipt -B^2e^-ipt

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 .

You're missing a 1/m and cos(pt)≠e^ipt:

x''-1000x= (36/m)cos(pt) -10=18cos(pt)-10

=>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt)=18cos(pt)-10

But what is cos(pt) in terms of complex exponentials?

 

[Benzoate]

You're missing a 1/m and cos(pt)≠e^ipt:

x''-1000x= (36/m)cos(pt) -10=18cos(pt)-10

=>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt)=18cos(pt)-10

But what is cos(pt) in terms of complex exponentials?

Euler Formula:

e^i$$\theta$$= cos($$\theta$$)+sin($$\theta$$)
therefore

18*e^ipt-10= 18*(cos(pt) + sin (pt))-10

 

[gabbagabbahey]

But you don't have 18 e^ipt! You have 18 cos(pt).

What is cos(pt) in term of complex exponentials?

Hint: look under the section "Relationship to Trignometry" here

 

[Benzoate]

But you don't have 18 e^ipt! You have 18 cos(pt).

What is cos(pt) in term of complex exponentials?

Hint: look under the section "Relationship to Trignometry" here

sorry if you are becoming frustrated. I have a copy of Gregory Douglass's Classical mechanics books and there is an example like this in that book on p. 10 cos(t) and they say the comple counter part is 10e^it

18 cos(pt)= 18e^ipt - 10(e^0)

 

[gabbagabbahey]

cos(pt)≠e^ipt

Euler's formula gives cos(pt)=Real[e^ipt]=(e^ipt+e^-ipt)/2 (Real[z] is the real part of the complex number z)

so, 18cos(pt)-10=9e^ipt+9e^-ipt-10

=>-p^2Ae^ipt -Bp^2e^-ipt+1000(Ae^ipt +Be^-ipt+C)=9e^ipt+9e^-ipt-10

Compare the coefficients in front of each of the e^ipt,e^-ipt, and constant terms...what must A,B and C be (in terms of p)?

 

[Benzoate]

cos(pt)≠e^ipt

Euler's formula gives cos(pt)=Real[e^ipt]=(e^ipt+e^-ipt)/2 (Real[z] is the real part of the complex number z)

so, 18cos(pt)-10=9e^ipt+9e^-ipt-10

=>-p^2Ae^ipt -Bp^2e^-ipt-1000(Ae^ipt -B^2e^-ipt+C)=9e^ipt+9e^-ipt-10

Compare the coefficients in front of each of the e^ipt,e^-ipt, and constant terms...what must A,B and C be (in terms of p)?

A(p^2-1000)=9 ==> A=9/(p^2-1000)
B(1000-p^2)=9 ==>B=9/(1000-p^2)
C=-10/1000?

 

[gabbagabbahey]

Close, there were some typos in my last equation; you should get A=9/(1000-p^2), your B and C are correct though.

Now, since A and B are equal you have: x(t)=A(e^ipt+e^-ipt)+C=2Acos(pt)+C

Now, if the maximum extension of the spring is 4cm, what must the Value of A be?

 

[Benzoate]

Close, there were some typos in my last equation; you should get A=9/(1000-p^2), your B and C are correct though.

Now, since A and B are equal you have: x(t)=A(e^ipt+e^-ipt)+C=2Acos(pt)+C

Now, if the maximum extension of the spring is 4cm, what must the Value of A be?

4=9/(1000-p^2)*cos(pt)-1/100 am I trying to find p?

 

[gabbagabbahey]

Yes, since x(t)=Acos(pt)+C, it should be clear that the angular frequency of oscillation is p. So you want to find p.

You need to be careful of your units though; 4cm =0.04m so you should have:

0.04=9/(1000-p^2)*cos(pt)-1/100

since the rest of the quantities in the equation are in meters.

 

[Benzoate]

Yes, since x(t)=Acos(pt)+C, it should be clear that the angular frequency of oscillation is p. So you want to find p.

You need to be careful of your units though; 4cm =0.04m so you should have:

0.04=9/(1000-p^2)*cos(pt)-1/100

since the rest of the quantities in the equation are in meters.

won't there be two solutions to p? The p's I finding for the x_D are entirely different from the p's in the complementary function

 

[Benzoate]

Sorry to bumped this thread again, even though its been two days since its been active.
for my final solution I get c=36/(1000-p^2) and c=.04==> p1=10 and p2=-10; I'm not sure what my text means when it says 'spring is safe if p=<20 rad/s and p>=40. rad/s