Question

In lecture 13, how was the equation x(doubledot) + kx/m = 0 turned into a cosine function with t?

Answer 1

The full derivation is above the level of this course, and can be found here:
http://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Simple_Harmonic_Motion/Mathematical_Derivation
What professor Lewin does is just propose that a solution to this equation is in the form
\[x(t)=A \cos (\omega t+\phi)\]where A is the amplutide, w is the angular frequency and phi is the "phase angle" which depends on the initial conditions of the system. He then check that this proposed solution satisfies the differential "x double dot" equation. If you differentiate the cosine equation twice, you get your double dot. You get:\[-A \omega^2\cos (\omega t+\phi)+\frac{k}{m}A \cos (\omega t+\phi)=0\]This works if and only if \[\omega^2=\frac{k}{m}\]or,\[\omega=\sqrt{\frac{k}{m}}\]So, his proposed solution works. But yeah, to derive it in full takes a bit more than is needed for this course.

Answer 2

since the motion was periodic with time (if time period is 10 second means after every ten second it will be at the same place)
so he replaced the displacemet by a cosine function it could be sine function
atfter that it was simple
\[x=a \cos(wt+\phi)\]
\[x(\dot)=v=-a w \sin(wt+ph)\]
\[x(doubledot)=accelration=-a w ^{2} \cos(wt+\phi)\]
the on replacing x(doubledot) and x in your equation
\[-a w ^{2} \cos(wt+\phi)+k(a \cos(wt+\phi))/m=0\]
\[k/m=w ^{2}\]
basicaly it was the pridition