A particle of mass `m` moves due to the force `F=-alpha m r, ` where `alpha` is a positve constant ,` r` is the radius vector of the particle relative to the origin of coordinates. Find the trajectory of its motion if at the initial moment `r=r_(0)i` and the velocity `v=upsilon_(0)j`, whee `i` and `j` are the unit vectors of the `x` and `y` axes.

Explanation

In accordance with the problem
`vec(F)=-alpha m vec(r)`
So, `m(ddot(x)vec(I)+ddot(y)vec(j))=- alpha m ( x vec(i)+y vec(j))`
Thus `ddot(x) =-alpha x ` and `ddot (y)=-alphay`
Hence the solution of th edifferetnial equation
`ddot(x)=-alpha x` becomes `x=a cos ( omega_(0)t+delt)`, where `omega_(0)^(2)=alpha.... (1)`
So, `dot(x) =a omega_(0)sin ( omega_(o)t+alpha) .....(2) `
From the initial conditions of the problem, `v_(x)=0` and `x=r_(0)` at` t=0`
So from Eqn (2) `alpha=0`, and Eqn takes the form
`x=r_(0)cos omega_(0)t` so, ` cos omega_(0)t=x//r_(0) ....(3)`
One of the solution of other differential Eqn `ddo(y)=-alphay`, becomes
`y=a^(') sin (omega)t+delta^(')`, where `omega_(0)^(2)=alpha ...(4)` ,
From the initial condition, `y=0` at `t=0`, so `delta^(')=0` and Eqn `(4)` becomes `:`
`y=a^(') sin omega_(0)t(5)`
Differentiating w.r.t. time we get
`dot(y)=a^(')omega_(0)cos omega_(0)t .......(6)`
But from the initial condition of the problem, `dot(y)=v_(0)` at ` t=0` ,
So, from Eqn (6) `v_(0)=a^(')omega_(0)` or ,` a^(')=v_(0)//omega_(0)`
Using it in Eqn `(5)`, we get
`y=(v_(0))/(omega_(0))sin omega_(0)t` or `sin omega_(0)t=(omega_(0)y)/(v_(0)) ......(7)`
Squaring and adding Eqns`(3)` and (7) we get `:`
`sin ^(2)omega_(0)t+cos^(2)omega_(0)t=(omega_(0)^(2)y^(2))/(v_(0)^(2))+(x^(2))/(r_(0)^(2))`
or , `((x)/(r_(0)))^(2)+alpha((y)/(v_(0)))^(2)=1` (as `alpha,=omega_(0)^(2))`