Recent content by Gil

The square root of 1 is ±√1 = ±1, isn't it?

You have (1)1/2, and it can be +1 or -1. The mistake you do is that you sould consider ((-1)2)1/2=|-1| = 1

For point a, I don't understand exactly what is your question. For point c think the following. Where are the charges? Are there charges INSIDE the box? The electrical field inside the box is related to the electric charge that is inside the box. Another way to see this is the following. In...

The ##\dot x## in your expression is linear when it should be squared, that's what I mean.

If you know the 3 angles, then you can apply the sines law and the cosines law, and because you know A and B you have a the length of the side "b". From this you get "a" and "c". So, you have 3 angles, the size of the 3 sides and 2 points, getting the last one is easy from here.

The time derivative of ##x^2## is ##2x\dot x##. The second derivative is ##2 (\dot x^2 + x \ddot x)##.

You can use an energy approach to solve the velocity. You know the potential of both springs (##U=\frac{1}{2}kx^2##) and the mass. Just set ##E= K + U##. When ##V=0## then ##K=0## and ##E=U## and when ##V=max## then ##U=0## and ##E=K##.