Energy Integral Lemma-Differential Eq. HW

[calculator13]

Homework Statement

Use the energy integral lemma to show that motions of the free undamped mass-spring oscillator my"+ky=0 objey

m(y')^2 = ky^2=constant

Homework Equations

E(t)=1/2 y'(t)^2 - F (y(t))

The Attempt at a Solution

I am not sure how to even start this problem.I would like to know just what the problem is asking for.I do not have a problem doing the math to it.

Thank you

 

[gabbagabbahey]

If I remember correctly, the energy integral lemma says that the quantity E(t)=1/2 y'(t)^2 - F (y(t)) is a constant when y(t) satsifies the DE y''(t)=f(y) and f(y) has no explicit dependence on y' or t and F(y)=\int f(y)dy... So, I would start by placing your ODE in the form y''(t)=f(y)...what is f(y) in this case? Does it have any explicit dependence on y' or t? If not, then E(t)=1/2 y'(t)^2 - F (y(t)) is a constant.

 

[calculator13]

so is my f(y)=-ky/m then i get the F(y) of that expression and plug into E(t)=

 

[gabbagabbahey]

Yes, what does that give you?

 

[calculator13]

1st of all i apologize since i don't know how to write equation symbol and such..

here is what i get

y=-ky/m

-k/m integral (y)= -k/m (y^2)/2

(1/2)y'^2 - k/m(y^2)/2=k

then i am trying to solve for y'^2 so
i end up with

my'^2 =2km +ky^2

and what i am suppose to get is m (y'^2) +ky^2 =constant.

 

[calculator13]

so i realized i messed up my negative there since the original lemma has a negative and my f(y) has a negative so i end up with


my'^2 + ky^2 =2km

is it okay to re-write that expression as my'^2 + ky^2=constant since usually the values of k and m are #'s?

so it would be my'^2 + ky^2 = constant

??

 

[calculator13]

Okay,I figured it out my question out.I don't know why i was adding an extra K.Thanks for guiding me through it.

I have another question if you could just help me with setting it up.
it involves the energy lemma .

the question is
Use the energy integral lemma to show that pendulum motions obey

(theta prime)^2 /2 - (g/l)cos(theta) = C

what is my f(y) in this case?

 

[calculator13]

okay so i think i figured out this one too:

i did

(theta)"^2/2= [(g/l)cos(theta) ]'

(theta)"^2/2= -g/l sin(theta)

f(y)= -g/l sin(theta)

then after going through all the steps i ended up with

(theta)'^2 /2 - (g/l)cos(theta)=C which is the original DE.


let me know if this is okay.i am not sure if taking the 2nd derivative of the equation is allowed.