physics motion

appleting · #1 $Old$ November 27th, 2008, 05:05 PM

i am really lost on what i am supposed to do here

'x = x(dot above) = dx/dt
"x = x(double dot above) = (d^2)*x/dt^2 for x=x(t)

second-order linear differential equation

"x + (w^2)*x = 0

w is a positive constant, initial conditions are x(0)=x0 and 'x(0)=0

a) show that this illustrates the motion of a simple harmonic oscillator with restoring force being linear in x and with spring contant k and attached mass m. (what is the equation of motion?) what does w correspond to in terms of k and m?
b) is the restoring force conservative?
c)given the intitial conditions, what are the solutions using trial solutions and the characteristic equation
d) proove that these solutions satisfy the initial conditions and the differential equations.

my mind is completely blank... help urgently needed please!!!

mr fantastic · #2 $Old$ November 27th, 2008, 11:57 PM

Quote:

Originally Posted by appleting $View Post$

i am really lost on what i am supposed to do here

'x = x(dot above) = dx/dt
"x = x(double dot above) = (d^2)*x/dt^2 for x=x(t)

second-order linear differential equation

"x + (w^2)*x = 0

w is a positive constant, initial conditions are x(0)=x0 and 'x(0)=0

a) show that this illustrates the motion of a simple harmonic oscillator with restoring force being linear in x and with spring contant k and attached mass m. (what is the equation of motion?) what does w correspond to in terms of k and m?
b) is the restoring force conservative?
c)given the intitial conditions, what are the solutions using trial solutions and the characteristic equation
d) proove that these solutions satisfy the initial conditions and the differential equations.

my mind is completely blank... help urgently needed please!!!

a) You're expected to know that simple harmonic motion satisfies the equation of motion $m \ddot{x} = - kx$ . So re-arrange your differential equation into this form.

b) Research for you to do. Do you know what conservative means in this context?

c) Trial solution is of the form $x = A \sin (\alpha t) + B \cos (\alpha t)$ .

Substitute this solution into the DE to get an expression for $\alpha$ in terms of w and hence in terms of k and m.

Use the initial conditions to solve for A and B.

d) I suppose you could substitute t = 0 into $x = x(t)$ and $\dot{x} = \dot{x}(t)$ .