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3

It's not that the final solution looks like that. Rather, you are looking for all the solutions of that form (normal modes) for two reasons: They are easy to find You can afterwards decompose any motion into a sum of normal modes. This comes from writing your equations of motion in the normal basis*. So first you work out the normal modes by assuming a ...


2

Let me answer your questions one by one. $\omega_0\gg 2\beta$ means the same thing with $\omega_0\gg\beta$. For example, we know that $10^4\gg 2$, so $10^4\gg2\times1$ and $10^4\gg1$. For $10^4$, $2$ has no differnece with $1$. Hence for $\omega_0$, $2\beta$ have no difference with $\beta$. Because of $\omega=\sqrt{\omega_0^2-2\beta^2}$ and $\omega_0\gg ...


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Gravity exerts a constant force downwards. If you apply a second constant force sideways you are in effect rotating the gravitational force: where the angle $\theta$ is given by: $$ \tan\theta = \frac{F_\text{ext}}{F_g} $$ The modulus of the net force is given by the usual Pythagoras rule: $$ F_\text{net}^2 = F_g^2 + F_\text{ext}^2 $$ So if you ...


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If the damping coefficient approaches zero, the differential equation we are looking for needs to approach the 2nd equation you wrote (the one for the physical pendulum). Therefore, we can conclude that the only thing we have to modify in the last equation to get the equation for a damped physical pendulum is to change g/L into mgL/(I_CM+mL^2). If we want ...


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A physical pendulum as described above behaves identically to a simple pendulum with a length $L^\prime=\frac{I_{\rm{CM}}+mL^2}{mL}$. So my inclination would be that you just replace both L's in your equation with the $L^\prime$ I just defined. That is: $\frac{\partial^2\theta}{\partial t^2}+\left(\frac{\xi m ...


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Hint : the block starts from rest 11 cm from equilibrium. You don't give it a push. That means that after one oscillation, it'll come back at the same place and never go further away than 11 cm. That makes the rest of the work become trivial.


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I'm sorry but it's a easy problem just using the ladder operator.You need to let x applied on the wavefunction one by one and the rest is easy. Applying the following property which can easily derived from the ladder operator (or just using the wavefunction,either way is ok,the specific procedure can be found in Griffiths' book): $\displaystyle ...


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Where in one dimension you would have $F=-kx$, you now have $\mathbf{F}=-k\mathbf{x}$ (assuming a simple Hookean spring) with $\mathbf{x}$ being the distance vector from the mass to the pivot of the spring.


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When dealing with operators it is often easier to see what is going on by applying it to the appropriate object. In this case you have the Hamiltonian operator that goes with a wave function that I will denote by $\psi(x,y,z,t)$. Now if you apply $\hat{H}$ to $\psi(x,y,z,t)$ you see that at some point you will have to take the partial derivative of $\psi$ ...



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