Question

A mass m sliding on a frictionless surface and connected to a spring with a positive, nonzero spring constant follows the DE mx'' + kx = 0. Note F = -kx. The energy of a solution to this equation is defined by E(t): = 1/2 (x')^2 + (1/2)kx^2. Show E(t) is constant.

Answer 1

Perhaps a nudge in the right direction is the best. I'm unsure of how to begin.

Answer 2

This is a physics question isn't it?

Answer 3

Eh, it's a Differential Equation question.

Answer 4

Is this your givens:

\[{\frac {d^2m} {dx^2} +kx=0}\]
\[E(t)={\frac 1 2}({\frac {d^2m}{dx^2})^2 + {\frac 1 2kx^2}}\]

Answer 5

I made a mistake. I'm sorry.
E(t) is supposed to be the first derivative of x, not second.

\[m \frac{ d ^{2x} }{ dt ^{2} } + kx = 0\]
\[\frac{ 1 }{ 2 } m (\frac{ dx }{ dt })^{2} + \frac{ 1 }{ 2 } kx^{2}\]

Answer 6

Sorry again, \[m \frac{ d ^{2}x }{ d t^{2} }\]

Answer 7

It also notes that the first equation is really F=ma since F exerted on the mass by the spring is -kx

Answer 8

Is the E(t) equation correct? I seems a bit odd to square a second derivative

Answer 9

Equations:
\[m \frac{ d^{2}x }{ dt ^{2} } + kx = 0\]
\[E(t) := \frac{ 1 }{ 2 } m (\frac{ dx }{ dt })^{2} + \frac{ 1 }{ 2 } kx^{2}\]

Answer 10

Would you have any idea on how to approach this problem?

Answer 11

It's been a while since i've done DE, so maybe not, but give me a sec

Answer 12

Not sure if this will help, but take a look at page 1.
http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-AppsOf2ndOrders_Stu.pdf

What I'm thinking is, if you obtain the arbitrary solution, take the first derivative, and substitute it into your E(t) equation, everything should cancel (i think) if E(t) is a constant. (Not too confident on that last part)

Answer 13

you have: \(m \ddot x + k x = 0 \)

which is same as: \(m \dot x \frac{d\dot x}{dx} + k x = 0\)...continuity assumed.

and from \(m \dot x \frac{d\dot x}{dx} + k x = 0\), we have \(\frac{1}{2} m \dot x ^ 2 + \frac{1}{2} k x ^ 2 = const \)

and you are choosing to equate that expression with E(t).......

if that is true, then \(\dot E = \dot {const} = 0\)

Answer 14

this is all very circular :-) but very interesting also.

Answer 15

Thanks you all!