- #1
naele
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Homework Statement
I have a 2D isotropic mass-spring system. The mass is pulled a distance A=1m, and then given an upwards kick with a velocity v_0. The k=1, m=1kg.
I need to find the furthest distance from the origin the mass will travel in its orbit.
Homework Equations
[tex]x(t)=A_x\cos(\omega t)[/tex]
[tex]y(t)=A_y\cos(\omega t-\delta)[/tex]
The Attempt at a Solution
The first thing I'm unsure of is how to handle the relative phase. I initially just solved for it using the initial conditions so when t=0
[tex]0=A_y\cos(-\delta)[/tex]
[tex] \delta=-\cos^{-1}(0)[/tex]
[tex]A_y=\frac{v_0}{\sin(\delta)\omega}[/tex]
I'm not 100% sure if this is acceptable.
Next, my initial approach was to set a function r(t)=sqrt(x(t)^2+y(t)^2), take the derivative set to zero and solve for t that way. In principle this would have given me the times at which the distance is at a max or a min. I only got one solution which I'm not sure makes sense.
So I have
[tex]r(t)=\sqrt{(\cos(\omega t))^2+(A_y\cos(\omega t-\delta))^2}[\tex]
where for simplicity I substituted [itex]A_x=1[/itex]. Taking the derivative gives a really messy result
[tex]r'(t)=\frac{-2\cos(\omega t)\sin(\omega t)\omega+2A_y^2\cos(-\omega t+\delta)\sin(-\omega t+\delta)\omega}{2\sqrt{(\cos(\omega t))^2+(A_y\cos(\omega t-\delta))^2}}[/tex]
Setting equal to 0 then
[tex]2\cos(\omega t)\sin(\omega t)=2A_y^2\sin(-\omega t+\delta)\sin(-\omega t+\delta)[/tex]
Using a trig identity
[tex]\sin(2\omega t)=A_y^2\sin(2(-\omega t+\delta))[/tex]
I should mention I used Maple to find the derivative, and then I used Maple again to solve for t. This is the expression it gave me
[tex]t=\frac{\arctan(\frac{A_y^2\sin(2\delta)}{1+A_y^2\cos(2\delta)})}{2\omega}[/tex]
This is the point where I am. I'm not at all sure about my work.
edit: it seems like I broke the latex interpreter. The equation it's not displaying is Ay=v_0/(sin(delta)w)
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