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the shortest resonating wave in a tube with one end closed is $\lamba/4$ a knot at the closed end and a maximum at the open end, so the tubelenght is $\lamba/4$ while in the open tube it is $\lamba/2$ but both have to be 24cm longer for the next occurring resonance


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Based on structx.com formula here: https://structx.com/Beam_Formulas_017.html This is the formula: There is fomula for other types (like cantilever etc as well): https://structx.com/beams.html


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It's a nice idea. Sadly it won't work but there is still some interesting physics involved. As a general rule resonance is only an efficient way to transfer energy to an object if that object has a high Q factor. A high Q factor means the energy supplied builds up and increases the amplitude of the oscillation, and this may get large enough to damage the ...


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There is a difference between cancer and a virus infection. A virus invades a cell and uses it to produce more virus. These go on to invade more cells. Many viruses reproduce so much that the cell bursts. To many dead cells kills the patient, or at least makes him sick. Cancer is a problem where the patient's cells malfunction. They reproduce ...


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Coronavirus does not have cells. For that matter - no viruses have cells since they are sub-cellular organisms that hijack others' cells to reproduce. (Note the difference with bacteria, which are cellular organisms.) A viral particle typically consists of a protein coat, the viral genome hidden inside, and some auxiliary proteins to launch replication of ...


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The theoretical method is Draw the shear force and bending moment diagram for the beam with the point load Calculate the curvature at any point from the bending moment Integrate the curvature twice to get the deflection You should find this in any textbook on "strength of materials" with a section on beam bending. The practical method is find a beam ...


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The plot has several flaws. At small $\omega \equiv \epsilon$ $I$ should have the following behavior: ($\delta = \epsilon/\omega_0$) $$ I \sim \frac{\epsilon\tau}{|\epsilon\tau|} \frac{1}{\sqrt{1+ \frac{\delta^4}{((1-\delta^2)\tau\epsilon)^2 }}} \approx \pm 1\cdot C \quad \text{at} \quad |\epsilon|\ll \omega_0 $$ (C is some appropriate constant), so the ...


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The complex impedance $Z$ of the parallel arrangement of the inductor $L$ and capacitor $C$ is $\dfrac 1 Z = j\omega C + \dfrac {1}{j\omega L} \Rightarrow Z = \dfrac{j\omega L}{1-\omega^2LC}$. You will note that as $\omega \to \sqrt LC$ then $|Z| \to \infty$ so it is not unreasonable that your analysis shows that as resonance is approached the current in ...


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Yes, it makes sense to have the plot with a minimum. Why would you plot for negative frequencies, anyway? The reason for this is that you plot the total current, the current in the main branch. What you have there is a parallel LC circuit. At resonance the impedance of this circuit becomes very large (ideally infinite) so the current in the main circuit ...


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