- #1

phantomvommand

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- Homework Statement
- Consider a mass m on a table attached to a spring at the origin with zero relaxed

length, which exerts the force

F = −kr

on the mass. We will find the general solution for r(t) = (x(t), y(t)) in two different ways.

(a) Directly write down the answer, using the fact that the x and y coordinates are independent.

(b) Sketch a representative sample of solutions. What kind of curve does the trajectory follow?

- Relevant Equations
- ##m\ddot x = kx## solution is ##x(t) = A\cos(\omega t + \phi)##

##r^2 = x^2 + y^2##

I get that:

##x(t) = A\cos(\omega t + \phi)##

##y(t) = A\sin(\omega t + \phi)## (from the above relevant equations). This agrees with the solution for part (a).

However, the solution manual claims in part (b) that:

In the case where ϕ1 = ϕ2 = 0 and A = B, the mass moves in a circle centered at the origin.More generally, when the angles ϕi are unequal, the mass can move in an ellipse with centerat the origin.

Is there any way to prove this, or explain this physically? I am especially dubious of the claim that when ϕ1 = ϕ2 = 0 and A = B, it is a circle, because this situation, physically, at time t = 0, should represent a simple 1D motion along the x-axis, since the spring was never stretched in the y-direction to begin with!

##x(t) = A\cos(\omega t + \phi)##

##y(t) = A\sin(\omega t + \phi)## (from the above relevant equations). This agrees with the solution for part (a).

However, the solution manual claims in part (b) that:

In the case where ϕ1 = ϕ2 = 0 and A = B, the mass moves in a circle centered at the origin.More generally, when the angles ϕi are unequal, the mass can move in an ellipse with centerat the origin.

Is there any way to prove this, or explain this physically? I am especially dubious of the claim that when ϕ1 = ϕ2 = 0 and A = B, it is a circle, because this situation, physically, at time t = 0, should represent a simple 1D motion along the x-axis, since the spring was never stretched in the y-direction to begin with!