Recent content by giulioo

7/12 like many of the others commenters i did guess almost all of them (which amazes me). well, it was pondered guessing :)

sorry my bad, at line "This is a special solution..." i meant ##\ddot x+ \omega ^2 x =f/m \cos(\gamma t+\beta)## instead of ##\ddot x+ \omega ^2 x =f/m \cos(\omega t+\beta)##.

I'm studying from landau lifšits "mechanics". I had some troubles in section small oscillations-->forced oscillations, especially from eq 22.4 to eq 22.5 i searched the web and came across this: https://www.physicsforums.com/threads/forced-oscillations-and-ressonance.488538/#post-3236442 this...

i forgot to say that ##\mathcal{D}_1## is the set of all smooth functions ##y:[a,b] \rightarrow \mathcal{R}##. Then obviously ##\mathcal{D}_1 \subset \mathcal{C}##. Let ##y_0 \in \mathcal{D}_1## (##\Rightarrow y_0 \in \mathcal{C}##). Then if ##J## has an extremum at ##y_0## in ##\mathcal{C}##...

well it is kind of late but yet.. I would say that extrema in "0-norm" are stronger. Consider a functional ##J## defined in ##\mathcal{C}[a,b]## (that is the space of all continuous functions from ##[a,b]## to ##\mathcal{R}##) then if ##J## has an extremum at ##y_0 \in \mathcal{C}## certanly...

well the wave is for sure not exactly 50/50 because of rise time. I haven't thought about it anyway. I will meditate upon it, i mean, rise time could add a dc component. Do you think input and output resistance could not affect signal in such way too? I already removed the dc component using an...

or probably it is due to internal impedance of the oscilloscope?

hi! when I send a square wave to a digital oscilloscope it detects the signal + dc signal. I wanted to know why. I was thinking about internal resistance of coaxial cable i was using. signal got out of an oscilloscope/wave generator and in channel A of the same oscilloscope/wg. It was +-1...