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This is an answer by an experimentalist who had been fitting data with mathematical models since 1968. When fitting data one goes to the simplest mathematical models. When the data display variations in time and space the Fourier expansion is extremely useful because it gives the frequencies and amplitudes that will fit a periodic data set. One gets as ...


4

Here is a solution for a spring force that varies directly with displacement. It thus varies with time implicitly, but has no explicit dependence on time or any other variable. Givens and Assumptions oscillator with mass $m$ amplitude of oscillation $A$ oscillator displacement, $x$, varies with time, but $x(t)$ is unknown spring applies force varying with ...


3

Observe that the number operator $N := a^\dagger a$ is positive-semidefinite because $$ \langle \psi \vert N \psi \rangle = \lvert a \lvert \psi \rangle \rvert^2 \ge 0$$ and hence has no negative eigenvalues. Since $a\lvert n \rangle = \sqrt{n}\lvert n - 1 \rangle$ for a normalized eigenstate of $N$ with eigenvalue $n$ (which one can derive from the ...


2

The FourierTransform.com is a website maintained by an enthusiast. The site is not peer reviewed, but it looks as though it might provide helpful explanations. Here's a link which provides some basic introduction to the Fourier transform. And here is another link to class notes provided by Prof. Carlton M. Caves for an introduction to the Fourier ...


2

Compare Hooke's law, $F=-kx$, with Newton's 2nd law, $F=ma$: $$ m\frac{d^2x}{dt^2}=-kx\tag{1} $$ where $[k]=\rm N/m=kg/s^2$ and $[m]=\rm kg$. (1) can also be written as, $$ \frac{d^2x}{dt^2}=-\frac{k}{m}x\tag{2} $$ So $[k/m]=\rm 1/s^2$. Why not make up a new variable, call it $\omega$, with units of $\rm1/s$, such that (2) becomes, $$ x''=-\omega^2x $$ i.e., ...


1

Hooke's law describes the force a spring exerts when its length is changed. Of course it makes intuitive sense that when you try to change the length of a spring (by pressing it, for example) the spring exerts a force on your hand, and resists the compression. For Hooke, this suggested that the force exerted by a spring is always in the opposite direction ...


1

You seem to have a hidden assumption in your work, that the springs are uncompressed before you displace them and take your measurement. It is easily possible that there is something in the mechanical setup of your device which keeps the springs compressed when your geophone is in its "resting" state.


1

To help you with section a: Your approach in writing the differential equation for the two positions (http://oi62.tinypic.com/dfx6yq.jpg) is correct. Since we know gravity won't affect the resonant frequency, I'll ignore gravity when writing the equations. $$\ddot{y_1} = -(k/m_1)(y_1-y_2)$$ $$\ddot{y_2} = -(k/m_2)(y_2-y_1) $$ Notice that the spring force ...


1

From Newton's second law we have (whether $k$ is constant or not) that: \begin{equation} m\ddot{x}+kx=0 \end{equation} The only difference is whether or not $k$ is a function of $t$ or not. If it is a function of $t$, the only general way to solve this differential equation is by using Taylor expansions. Let us take: \begin{equation} ...


1

The term 'amplitude' is often used somewhat ambiguously. The most rigorous definition is that amplitude is simply $|A|$ (the modulus of $A$). In your case (unmodulated wave or oscillation) there's only one amplitude. But others count the number of peaks and troughs as 'amplitudes' like you are doing. You count only three but that's because Fig.14.2 only ...



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