- #1
Gregg
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This problem is about the Foucault pendulum. It is a mass, m, attached to a cable that does not restrict the motion. So it is sensitive to centrifugal and coriolis forces.
The first part states that it has a potential ## V = \frac{m \omega_0^2}{2}(x^2 + y^2)## and to find the force associated with that. To me this is a hint that you are going to use the equation
## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} ##
The Earth rotates at ## \omega## ccw. The pendulum is located at the North pole and the co-ordinate axes are such that x points 90 degrees longitude and y 180 degrees longitude. I don't really understand how this makes a difference.
I decided that ## \vec{\omega}=\omega \vec{\hat{k}} ## which I am unsure about.
So for the above equation, since ## \vec{F} = -\nabla V ## and after some computation/approximation, i.e. ## \dot{z} ## is small.
## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} \Rightarrow##
## \ddot{x} = \omega_0 x + \omega^2 x + 2\omega \dot{y} ##
## \ddot{y} = \omega_0 y + \omega^2 y - 2 \omega \dot{x} ##
Which seem OK for now.
Then we're given the hint to use ## \zeta(t) = x(t) + i y(t) ## to simplify things and hopefully turn it into a SHO problem with damping. Multiply the second equation by ##i##
## \ddot{x} + i\ddot{y} = \omega_0(x+i y) +\omega^2(x+i y) + 2\omega(\dot{y}-i\dot{x}) ##
## \ddot{\zeta} =(\omega_0^2+\omega^2) \zeta - 2\omega i \dot{\zeta} ##
Then plug in ## \zeta = \exp(\alpha t) ## get ##\alpha= \omega-\sqrt{\omega_0^2-\omega^2}## and from here it doesn't really get anywhere near the answer given.
First I assume that ##\omega^2 ## is really small.
## \zeta(t) = c_1 \exp(i (\omega + \omega_0) t) + c_2 \exp(i \omega - \omega_0)t) ##
But we are required to show that:
## \zeta(t) = e^{-i \omega t}\left[ \zeta(0) \left( \cos(\omega_0 t) + i\frac{\omega }{\omega_0}\sin(\omega_0 t)\right)+\frac{\dot{\zeta(0)}}{\omega} \sin (\omega_0 t) \right] ##
But I have no idea how to do this. I think that possibly I have made a mistake with the orientation of my axes or the angular velocity components. Any help would be appreciated!
The first part states that it has a potential ## V = \frac{m \omega_0^2}{2}(x^2 + y^2)## and to find the force associated with that. To me this is a hint that you are going to use the equation
## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} ##
The Earth rotates at ## \omega## ccw. The pendulum is located at the North pole and the co-ordinate axes are such that x points 90 degrees longitude and y 180 degrees longitude. I don't really understand how this makes a difference.
I decided that ## \vec{\omega}=\omega \vec{\hat{k}} ## which I am unsure about.
So for the above equation, since ## \vec{F} = -\nabla V ## and after some computation/approximation, i.e. ## \dot{z} ## is small.
## m \ddot{\vec{r}}= \vec{F} - 2 m \vec{\omega} \times (\vec{\omega} \times \vec{r})-m \vec{\omega}\times\vec{\dot{r}} \Rightarrow##
## \ddot{x} = \omega_0 x + \omega^2 x + 2\omega \dot{y} ##
## \ddot{y} = \omega_0 y + \omega^2 y - 2 \omega \dot{x} ##
Which seem OK for now.
Then we're given the hint to use ## \zeta(t) = x(t) + i y(t) ## to simplify things and hopefully turn it into a SHO problem with damping. Multiply the second equation by ##i##
## \ddot{x} + i\ddot{y} = \omega_0(x+i y) +\omega^2(x+i y) + 2\omega(\dot{y}-i\dot{x}) ##
## \ddot{\zeta} =(\omega_0^2+\omega^2) \zeta - 2\omega i \dot{\zeta} ##
Then plug in ## \zeta = \exp(\alpha t) ## get ##\alpha= \omega-\sqrt{\omega_0^2-\omega^2}## and from here it doesn't really get anywhere near the answer given.
First I assume that ##\omega^2 ## is really small.
## \zeta(t) = c_1 \exp(i (\omega + \omega_0) t) + c_2 \exp(i \omega - \omega_0)t) ##
But we are required to show that:
## \zeta(t) = e^{-i \omega t}\left[ \zeta(0) \left( \cos(\omega_0 t) + i\frac{\omega }{\omega_0}\sin(\omega_0 t)\right)+\frac{\dot{\zeta(0)}}{\omega} \sin (\omega_0 t) \right] ##
But I have no idea how to do this. I think that possibly I have made a mistake with the orientation of my axes or the angular velocity components. Any help would be appreciated!
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