Solving a Parabolic Bowl Oscillation Problem

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SUMMARY

The discussion focuses on calculating the period of oscillations for a particle moving in a frictionless parabolic bowl defined by the equation y = ax², with a = 1.460 m⁻¹ and an initial release point at b = 0.43 m. The user proposes a method involving the integration of velocity as a function of position, v(x), over the arc length from -b to +b to derive the period T. Although the approach yields a result close to that obtained through the small angle approximation, it is acknowledged as mathematically incorrect due to the non-constant acceleration and velocity in the system. The user seeks validation and comments on the applicability of their method.

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Homework Statement



Consider a particle moving back and forth on a frictionless parabolic bowl, y = ax2, where a = 1.460 m-1
If the particle is released from rest at the point on the
bowl at b = 0.43 m, find the period of the oscillations.

I have an equation for velocity(as a function of x). What i was thinking is that i could integrate this from -b to +b, and call this value 'v(x)*d', (it having units of m2/s.
since v=d/t, v(x)*d=d2/t.
thus t=d2/'v(x)*d'.
making the period, T=2*(d2/'v(x)*d'),
where d is the arc length of y(x), from -b to +b.

I realize this may be the least elegant and very possibly "physically incorrect", but it actually gave me a period very close to that obtained by the 'small angle approximation' (however, incorrect)

Can anybody comment on this method?
 
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i guess the crux of my argument is whether the area under the v(x) curve can be used in this manner, comments?
 
The acceleration and velocity are not at all constant, so the method is, mathematically speaking, completely wrong.
 

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