Water tower/spring system Diff Eq

SithsNGiggles
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Homework Statement



Suppose a water tower in an earthquake acts as a mass-spring system. Assume that the container on top is full and the water does not move around. The container then acts as a mass and the support acts as the spring, where the induced vibrations are horizontal. Suppose that the container with water has a mass of 10,000 kg. It takes a force of 1000 N to displace the container 1 m. For simplicity, assume no friction. When the earthquake hits the water tower is at rest.

Suppose that an earthquake induces an external force ##F(t)=mA\omega^2\cos(\omega t)##.

What is the natural frequency of the water tower?

Find a formula for the maximal amplitude of the resulting oscillations of the water container (the maximal deviation from the rest position). The motion will be a high frequency wave modulated by a low frequency wave, so simply find the constant in front of the sines.

Homework Equations



The Attempt at a Solution



Here's the differential equation I set up:

##10,000x''+1,000x=mA\omega^2\cos(\omega t)##

For the natural frequency, I used the formula ##\omega_0=\sqrt{\frac{k}{m}}##, which gives me ##\omega_0=\sqrt{\frac{1}{10}}\text{ rad/s}=\frac{1}{2\pi}\sqrt{\frac{1}{10}}\text{ Hz}##. Is this right?

And for the second part, do I just solve this equation? I'm not sure what it means to find the "constant in front of the sines."
 
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What you have is that y(t)= cos(\sqrt{1/10}t) and y(t)= sin(\sqrt{1/10}t) are solutions to the associated homogeneous equation, 10000x''+ 1000x= 0. Can you find the general solution to the entire equation?
 
HallsofIvy said:
What you have is that y(t)= cos(\sqrt{1/10}t) and y(t)= sin(\sqrt{1/10}t) are solutions to the associated homogeneous equation, 10000x''+ 1000x= 0. Can you find the general solution to the entire equation?

Yup, I've found that the general solution is
##\displaystyle x(t)=C_1\cos\left(\sqrt{\frac{1}{10}}t\right)+C_2 \sin \left(\sqrt{\frac{1}{10}}t\right)+\frac{mA \omega ^2}{1000-10000\omega^2}\cos(\omega t)##

We're also assuming ##\omega\not=\omega_0## for the second part. I forgot to put that in my first post.

By the way, is the ##m## in the solution the same as the mass of the water tower?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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