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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 m^{2}/s.

since v=d/t, v(x)*d=d^{2}/t.

thus t=d^{2}/'v(x)*d'.

making the period, T=2*(d^{2}/'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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# Homework Help: Parabolic Bowl

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