Question

An object of mass m=80 kg moves in one dimension subject to the potential energy

U(x)=λ/4*(x2−a2)2+(b/2*x2)

a) How many equilibrium points (stable and unstable ones) does this potential have?

b) Find a stable equilibrium point x0 such that x0 is positive. (in meters)

c) Do a Taylor expansion of the force F(x) for x close to the equilibrium point, x≃x0, that is F(x)=F0−k(x−x0)+… What are the values for F0 (in Newton) and k (in kg/s2)?

(d) What is the period T of small oscillations (in seconds) of this mass around the equilibrium point x0? (Note that the parameter k found in the previous question acts like a spring constant that wants to pull small deviations back to the equilibrium point)

Answer 1

a)3
b)take the derivative of U(x).. U'(x)
then take the zero-points U'(x)=0.. one zero is x=0, and it has two zeros symmetric to the y-axis.. take the positive one as answer for x0.
c)wolframalpha will do this for you. enter taylor expansion take U'(x) as the function and your x0..
d)2*pi/sqrt(k/m) (k from c) )

Answer 2

@bw
could your please tell part c properly

Answer 3

I couldn't understand part c. In wolframalpha, what do we have to enter as the order of the taylor series and the point?

Answer 4

@phy
what other questions did you got?

Answer 5

See... for x=x0... if I do my taylor serie expansion... my k is being accepted as correct but my F0 is not being accepted by edx... wonder why...

Answer 6

@anokneemouse what is d formula for part d?
what value of k did you got?
my x_o is 7.74
a=7 and b =-11 lamda=1

Answer 7

ss01

look.... its quite simple.... F0... comes to 0 for mine... and to get k, do the following:

Since I don't know the values of a,b,lambda you have...

F(x)=dU(x)/dx

now type in F(x) in wolfram alpha... you'll get 3 roots...
1)x=0
and 2 more equal roots with opposite signs.
the positive root must be chosen.

Now in wolfram alpha type in Talyor Expansion( and type in your F(x)) at x=the positive root.

the second term's coefficient is k.

d) T=2*pi/sqrt(k/m)

:) Cheers...

Mouse/Anokneemouse.

PS - I need help with the airliner and mass spring and pendulum sums

Answer 8

what abt the rotating disc?
thnx mouse

Answer 9

@anokneemouse nd @bw can u please woekout in the wolframalpha.iam not able to do it.! my value for x_0 is 6.0552. please tell the value of k and F .a=5 and b=35.please..!

Answer 10

give me your a,b,lambda values....

Answer 11

Block on rotating disk is easy - I got option (a) :)

solve for v: mv^2/R=umg....

Answer 12

Some help with mass spring??? Please? ANd Pendulum???

See, for the mass spring problem...
a)u_s*mg=kx1.... solve for x1....

but I'm not getting the other answers :(

Answer 13

@muse my lambda value is 3,a=5,b=35 and x_o is 6.0552

Answer 14

are you sure about your x0? I'm getting x0=3.651483717 for you...

Answer 15

the equation for x is x^2=(3a^2+b)/3. is it wrong?? x turns out to be sqrt(36.666)which is 6.0552...pls crct if i am wrong :)

Answer 16

no. Your equations are all messed up.

U(x)=λ/4*(x2−a2)2+(b/2*x2)
=>U'(x)=F(x)=λ/4*(4*x^3-4*a^2*x)-b*x
Now using your lambda,a,b values: U'(x)=3x^3-40x

hence: your roots are 0, 3.651483717, -3.651483717

Answer 17

okay can u pls temme the value for F and K with ur crct values of x_o,lambda nd b values..? please..! mouse

Answer 18

I'l give you the answers... but ask yourself, are you learning anything at all if you just fish for answers from strangers? Most of us come here if we get stuck!!


F0=0,k=80...do try to find out T on your own?

Answer 19

You should have done it by your own, son. I am very disappointed

Answer 20

@ mouse, my x_0 is 2, lambda is 5, a=4, b=60 and mass is 80kgs. Please tell me the value of K, i tried to solve it by your method, but i got it wrong.

Answer 21

λ/4*(4*x^3-4*a^2*x)-b*x just input your values and solve for x.... the positive x0 value is the one required phy...

Answer 22

Yes i got x0, but i didn;t get k, can you please tell me how to calculate k for those values?

Answer 23

now type in F(x) in wolfram alpha... you'll get 3 roots...
1)x=0
and 2 more equal roots with opposite signs.
the positive root must be chosen.

Now in wolfram alpha type in Talyor Expansion( and type in your F(x)) at x=the positive root.

the second term's coefficient is k.

d) T=2*pi/sqrt(k/m)

:) Cheers...

Answer 24

Thats what I am not getting! My x0 is 2, so F(x)= F₀-k(x-2), when i type this in taylor series calculator of wolframalpha, i get a huge equation!

Answer 25

Please help me mouse

Answer 26

Thanks alot mouse! after reading your hints I managed to learn how to use series in wolfram :) There should have been more directions provided with the question, the textbook description was insufficient! I havent got the spring one either but for the airliner question I have read a useful tip that you will have to differentiate the function given over t=0 to t=28s. This is for non constant force and acceleration so you will have to tweak accordingly the equations of kinematics, find out the heat produced due to friction (assuming it gets converted on 100% basis) and then integrate using indefinite and definite integration (whatever that means I have no clue, this course is too biased for maths students. wholly unfair to those who enjoy pure physios minus applied math) Hope that helped you somehow good luck!

Answer 27

@ KS, can you please tell me how you got the solution for Double Well potential? I am not getting the value for k.

Answer 28

Thank you mouse.
for those not getting the taylor stuff,do as mouse says, remembering
1- F(x) is the derivative of U(x)
2- for the taylor n wolfram you have to put
"taylor expansion (f(x)) at x=" and the value of x you found for b)

Answer 29

Phy... your k is 40... F0 is obviously 0.

I wasn't telling you because I thought it'd help you learn how to use wolfram better. Alas...

Somebody help me with the Mass Spring sum? Everything else i've finished off :)

Answer 30

And guys, those of you who're not interested in the math portions. Let's be honest, Lewin's lectures are the important bit for you. It's difficult to maintain the honour code and get certs if you're not happy with math. I think, just watching Lewin's lectures are fantastic if you don't want to do the math.

But, if you want the certs... the math is necessary... just as it is for the 8.02x course... the 2013 Spring edition I mean...

now somebody help me with the concept of mass-spring?

Answer 31

How does one get t_12 of the mass-spring system sum? I've got the other parts, for all other sums in case anybody needs help! :)

Answer 32

mouse, can you explain how please?

Answer 33

mouse how did you get q8.b) x2-x1? i know that for part d the answer is The block will stay at its resting position x2. for q6 d1) h~ is always smaller than h and for d2) t~up<t~dn and t~up<tup. both second options. need help with the other parts

Answer 34

@phy: input your derivated function (it should have an x^3 - x in it. In alphawolfram, write
series (then your function, then) point (then x_0 value) and the result will be of the form f=f_0 -k(x-x_0), check your value for f_0 and k in that equation (most likely your f_0 is a negative zero, so f_0=0 and k is either 40 or 80. if its 80 then your T will be 2*3.14 and if its 40 then t=8.885

Answer 35

Thanks a lot mouse and KS,

Did anyone get the airliner and the pendulum problem?

Answer 36

anyone for the mass pushed by spring ?

Answer 37

ss01, can you please tell me the solution for the airliner problem?

Answer 38

KS. The mod(x_2-x_1) is actually really simple. All you need to do is this...

1/2*k*(x_1)^2=(mu_k)*m*g*d
solve for d.... your d value will come equal to your x1...the magnitude I mean.

This means the body starts moving and comes to rest at x_0=0 itself. :D

So, your mod(x_2-x_1) value is simply
mod(0-x_1)=+x_1

My question is how do we figure our the time t12?

Answer 39

if you want to know how to solve the airliner problem....

since I can't input the link here... just type exactly this - "mit ocw 8.01 Airliner wheels locked landing" in google. Click on the first link... the link will be ending with something like ..../MIT8_01SC_problems08_soln.pdf.....

And for the pendulum... initial KE=final(PE+KE)...

as in treat your mean position as ground level...
so 1/2*m*v0^2=m*g*(l-l*cos(theta))+1/2*m*v1^2.... solve for v1.

Tension in the string=mgcos(theta)+m*v1^2/l

So after this you have a projectile motion with velocity v1... angle of projection = theta.... you should be able to do the rest from there. :)

Someone please give me the t12 from the mass-spring problem? How does one find out the time?

Answer 40

Thanks a lot mouse, I could solve the pendulum problem. But, in the mass spring problem, I got x1=0.5714 , so mod(x2-x1) should also be equal to x1, correct? But when i typed in the value, it showed that x1 was correct but mod(x2-x1) was wrong :(

Also, I had looked at the MIT page for the airliner problem, but I couldn't understand anything, its complicated and solutions for all the problems are not given there.

Answer 41

The soln is simple. Try and read and understand it? It's too large to explain! can you guys give me t12??

solve the equation I gave:

1/2*k*(x_1)^2=(mu_k)*m*g*d

Answer 42

@ mouse, I read the airliner prob properly, I solved it but got it wrong. I am confused in the equations. Their equation is F(t)=-F0+Bt , whereas ours is F(t)=-F0+(t/ts-1)*F1. What is B? I have only 1 try left now :(

Answer 43

Phy?
F(t)=m*a_x(t)=-F0+(t/ts-1)*F1
=>ax(t)=-F0/m+(t/ts-1)*F1/m
integrating from 0->t
vx(t)-vx(0)=-F0/m*t+(t^2/(2*ts)-t)*F1/m

you get Vo from here by putting t=0.

Now,

Vx(t)=Vx(0)-F0/m*t+(t^2/(2*ts)-t)*F1/m

Integrating between 0->t

We get

x(t)-x(0)=Vx(0)*t-F0/m*t^2/2+(t^3/(6*ts)-t^2/2)*F1/m

x(0)=0...therefore putting t=ts we get x(ts) from here... simple?

:)

please someone give me the mass-spring t12 ?? I've been helping you guys out as much as possible.... I scratch your back, you scratch mine? :/

Answer 44

@phy .
ΔU + ΔK = -fd = -μnd = -μmgd
(1/2k(x2)^2 - 1/2k(x1)^2) + (0 - 0) = -μmg(x2 - x1)
(x2)^2 - (x1)^2 = -2μmg(x2 - x1)/k
(x2 + x1)(x2 - x1) = -2μmg(x2 - x1)/k
x2 + x1 = -2μmg/k
x2 - x1 = -2μmg/k - 2x1 = -2(μmg/k + x1).
@mouse i was jus confused wid dat wolfram alpha and stuff..so pls don't misunderstand me .i also come here if i don't understand :). and yeah i solved for k too..! :) thnks mouse !

Answer 45

and yeah the u_k is the one used above.!!

Answer 46

Someone help with the t12? Please?

Answer 47

@ mouse, in the airliner's problem, what units do we have to keep? I converted kN to kg and tonnes to kg as well.

Answer 48

yes. convert everything to basic SI units....

HAS ANYBODY got t12? :/

Answer 49

I think i have got how to work for t12, let me try first.

Answer 50

mouse, t12 is very simple.

Just calculate a half period (T/2).
t = pi * (m/k)^0.5

Answer 51

Yes yes, even I got the same result, but solved it a little differently

Answer 52

so, t12 is just pi*sqrt(m/k) really? :/

Answer 53

Yes, try it yourself, I got a green tick!

Answer 54

Is 7e just K-W_engines or is there a trick to it?

Answer 55

Thanks Jack Package :) My job here is done :D

Answer 56

how do i get E_heat ?

Answer 57

PLease Help With emergency landine question :(

Answer 58

if you use the differential equations from mouse you get after simplification:

v0=(F0+F1/2)*t_s/M
|a(ts)|=F0/M
|a(0)|=(F0+F1)/M
s=t_s^2*(F0/2+F1/6)/M
W=F0*s

Answer 59

Please tell how to calculate value of " K " -- lambda = 3 , a = 9 , b = 223 .. :(

Answer 60

E_heat=0.5*M*v0^2-W

Answer 61

@Shashanoid: do you have x0?
make the taylor expansion of U'(x) with x0
K is the factor in the taylor expansion from (x-x0)

Answer 62

@ bw Thanks alot for that emergency plane question , but i am still confused and not able to get that value of " K " :(

Answer 63

@bw
The work and energy is not correct(emergency landing question). Is F0and M in kilograms? A huge number?

Answer 64

@Shashanoid: what is your U'(x) and your x0?

@Jack Package: whats wrong with W and E_heat it worked for me

Answer 65

The mod(x_2-x_1) is actually really simple. All you need to do is this...

1/2*k*(x_1)^2=(mu_k)*m*g*d
what's (mu_k)? mass *the kinetic friction coefficient?

Answer 66

What are all these things

Answer 67

@bw - please help me with the Double well potential .. My U'(x) is 3x^3-20x=0
X_0 = 2.58

help for the value of K

Answer 68

Phy
how did you get in the mass spring problem, x1=0.5714 ??

k*x1=u_s*m*g so x1=u_s*m*g/k ?

Answer 69

Block on rotating disk -
solve for v: mv^2/R=umg....what's u? and we don't know R..any help

Answer 70

@anonymous by u_s * mg/k you will get the x1 , | -2*u_k*mg/k + ( -x0 )|



@bw please help me with that.

Answer 71

@bw - please help me with the Double well potential .. My U'(x) is 3x^3-20x=0
X_0 = 2.58

please help fr the value of " K "

Answer 72

@Shashanoid:
the exact value for x0=2*sqrt(5/3)
taylor expansion:
3*(x-2*sqrt(5/3))^3+6sqrt(5)(x-2*sqrt(5/3))^2+40*(x-2*sqrt(5/3))

so K=40

Answer 73

@bw sorry can you please provide the value of " k " for x0 = 3.16 and u_x = 6x^3-60x=0 ..

Answer 74

@bw - And λ= 6 kg/(m2s2), a= 4 m, and b= 36 kg/s2. please provide the value of " k " i'm not getting it

Answer 75

someone please help with the value for " k " please :(

Answer 76

@Shashanoid:
k=120

Answer 77

@bw thanks alot bw you helped alot today :) very very thanks.

Answer 78

k can either be 40 or 80.Thanks mouse and to all that helped. We worked as a team and we nailed it :D

Answer 79

whoops, bw got here before me. I was wrong there. cheers sasha, hope we can help each other again in times of need.

Answer 80

Yes sure KS :)