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)
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) )
@bw
could your please tell part c properly
I couldn't understand part c. In wolframalpha, what do we have to enter as the order of the taylor series and the point?
@phy
what other questions did you got?
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...
@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
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
what abt the rotating disc?
thnx mouse
@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..!
give me your a,b,lambda values....
Block on rotating disk is easy - I got option (a) :)
solve for v: mv^2/R=umg....
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 :(
@muse my lambda value is 3,a=5,b=35 and x_o is 6.0552
are you sure about your x0? I'm getting x0=3.651483717 for you...
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 :)
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
okay can u pls temme the value for F and K with ur crct values of x_o,lambda nd b values..? please..! mouse
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?
You should have done it by your own, son. I am very disappointed
@ 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.
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...
Yes i got x0, but i didn;t get k, can you please tell me how to calculate k for those values?
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...
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!
Please help me mouse
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!
@ KS, can you please tell me how you got the solution for Double Well potential? I am not getting the value for k.
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)
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 :)
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?
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! :)
mouse, can you explain how please?
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
@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
Thanks a lot mouse and KS,
Did anyone get the airliner and the pendulum problem?
anyone for the mass pushed by spring ?
ss01, can you please tell me the solution for the airliner problem?
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?
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?
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.
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
@ 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 :(
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? :/
@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 !
and yeah the u_k is the one used above.!!
Someone help with the t12? Please?
@ mouse, in the airliner's problem, what units do we have to keep? I converted kN to kg and tonnes to kg as well.
yes. convert everything to basic SI units....
HAS ANYBODY got t12? :/
I think i have got how to work for t12, let me try first.
mouse, t12 is very simple.
Just calculate a half period (T/2).
t = pi * (m/k)^0.5
Yes yes, even I got the same result, but solved it a little differently
so, t12 is just pi*sqrt(m/k) really? :/
Yes, try it yourself, I got a green tick!
Is 7e just K-W_engines or is there a trick to it?
Thanks Jack Package :) My job here is done :D
how do i get E_heat ?
PLease Help With emergency landine question :(
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
Please tell how to calculate value of " K " -- lambda = 3 , a = 9 , b = 223 .. :(
E_heat=0.5*M*v0^2-W
@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)
@ bw Thanks alot for that emergency plane question , but i am still confused and not able to get that value of " K " :(
@bw
The work and energy is not correct(emergency landing question). Is F0and M in kilograms? A huge number?
@Shashanoid: what is your U'(x) and your x0?
@Jack Package: whats wrong with W and E_heat it worked for me
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?
What are all these things
@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
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 ?
Block on rotating disk -
solve for v: mv^2/R=umg....what's u? and we don't know R..any help
@anonymous by u_s * mg/k you will get the x1 , | -2*u_k*mg/k + ( -x0 )|
@bw please help me with that.
@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 "
@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
@bw sorry can you please provide the value of " k " for x0 = 3.16 and u_x = 6x^3-60x=0 ..
@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
someone please help with the value for " k " please :(
@Shashanoid:
k=120