Recent content by SianRR

Ah yeah, ##V_x=\frac{E_y}{B_z}-[\frac{E_y}{B_z}-u]cos(wt)## Thanks again for all your help!

Ah okay but are my solutions; ##V_x(t)=\frac{E_y}{B_z}[1-ucos(wt)]## ##V_y(t)=[\frac{E-y}{B_z}-u]sin(wt)## correct? or what we'd expect?

Okay I'm getting that ##V_y = e^{\pm iwt} ## So I want to write this as ##V_y(t)=Ce^{+iwt} + De^{-iwt}## where C and D are some constants. That from the initial conditon that ##V_y(0)=0## that C=-D ##V_y(t)=Ce^{+iwt} -Ce^{-iwt}## , using Euler's expansion...

Ah okay so fixing that I have a negative in front of the ##w^2 V_y## Which looks a lot like a form of simple harmonic motion. I don't want to come across as silly for asking this, but I'm still not too sure what my actual final equations of ##V_y(t) ## and ##V_x(t)## should look like. But...

Okay using the same method as above and assuming that ##E_y## is constant (because it is a uniform field?) so that differentiating means that term goes to zero. Subbing into the ##F_x## equation I get ##\frac{d^2V_y}{dt^2} = w^2V_y ##

I have done ##\frac {d^2V_x}{dt^2} = w\frac{V_y}{dt} ## Then rearranging for ## \frac{V_y}{dt} ## I subbed that into the equation I have for ##F_y## and that leaves me with ##\frac{m}{w} \frac{d^2V_x}{dt^2} -qB_zV_x =qE_y ## Can this then be solved by setting ##V_x=## to some exponential or a...

Ah thank you for replying, I’ll know next time to type my work thank you. I have a question though, I’m not sure about taking the time derivative of ##\large \frac{d v_x}{dt} = \omega v_y##. Because I’d end up with accelerations, unless you mean I’d be able to sub in ##\large \frac{d...

I've attached my attempt at a solution below, I thought integrating it would be the best way to go but I'm just getting so confused and could use some help. This isn't my first attempt at a solution either I've been working on this for just under two hours now.