Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
 Already signed up? New to KöMaL?

# 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