X and Y coordinates of an oscillating object on a spring.

  • #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!
 
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  • #2
phantomvommand said:
##x(t) = A\cos(\omega t + \phi)##
##y(t) = A\sin(\omega t + \phi)## (from the above relevant equations).
except that, as indicated later, the amplitude and phase constants can be different. Only the frequency must be the same.
phantomvommand said:
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.
Easily proved. Write down the expression for ##x^2+y^2##.
Slight variation for the ellipse case.
phantomvommand said:
since the spring was never stretched in the y-direction to begin with!
You don’t know that, but it’s not just a question of initial displacement. In the general case, it may have been given some initial velocity.
 
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  • #3
phantomvommand said:
Is there any way to prove this, or explain this physically?
In first instance this is math. You have two seond order differential equations, so four initial conditions.
For example ##x, \dot x, y, \dot y## at ##t=0##.
Or ##A, \phi_1, B, \phi_2## at ##t=0##.
Setting
phantomvommand said:
ϕ1 = ϕ2 = 0 and A = B
leaves only one degree of freedom. And sure enough the "choice"
##x = A \cos(\omega _t),\ \ y=\sin(\omega t)## describes circular motion,
with very specific initial conditions ##x=A, \dot x=0, y=0, \dot y=\omega A## at ##t=0##

Ah, hello @haruspex !

##\ ##
 
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  • #4
phantomvommand said:
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!
So: no! Spring not stretched in the ##y## direction at ##t=0## doesn't mean never stretched. ##\dot y \ne 0## at t=0 means it does gets stretched in the ##y## direction.

##\ ##
 
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