Problem P. 4961. (October 2017)

P. 4961. An object executes simple harmonic motion of period \(\displaystyle T=0.2\) s. It takes \(\displaystyle \Delta t=0.01\) s to double its displacement of \(\displaystyle x=3\) cm. What is the amplitude of the motion?

(4 pont)

Deadline expired on November 10, 2017.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. 0,01 s alatt a rezgés fázisa \(\displaystyle \varphi=\frac{2\pi}{T}\Delta t=\tfrac1{10}\,\pi\) radiánt, vagyis \(\displaystyle 18^\circ\)-ot változik. A megadott feltétel szerint

\(\displaystyle A\sin(\omega t+ 18^\circ)=2A\sin(\omega t),\)

ahonnan

\(\displaystyle \cos18^\circ\sin(\omega t)+\sin18^\circ\cos(\omega t)=2\sin(\omega t),\)

azaz

\(\displaystyle \tg(\omega t)=\frac{\sin18^\circ}{(2-\cos18^\circ)}=0{,}295.\)

Innen

\(\displaystyle A=\frac{x}{\sin(\omega t)}=\frac{3~\rm cm}{0{,}283}\approx 10{,}6~\rm cm.\)

Statistics:

90 students sent a solution.
4 points:	71 students.
3 points:	4 students.
2 points:	5 students.
1 point:	7 students.
0 point:	3 students.

Problems in Physics of KöMaL, October 2017