Cum, asupra corpului nu va mai actiona nicio forta externa, distanta de 3cm de la care a fost lasat liber este defapt amplitudinea(elongatia maxima)
k = 20 N/m
m = 0,2 kg
A = 3 cm = 0,03 m
a) T = ?
[tex]\omega=\frac{2\pi}{T}\\
\omega=\sqrt{\frac{k}{m}}\rightarrow\frac{2\pi}{T}=\sqrt{\frac{k}{m}}\rightarrow T=2\pi\sqrt{\frac{m}{k}}\\
T=2\pi\sqrt{\frac{0,2}{20}}s=0,2\pi s\approx 0,628 s[/tex]
b) Ep = ?
In momentul echilibrarii, elongatia y(distanta de la corp la punctul de echilibru) va fi 0 m
[tex]E_p=\frac{ky^2}{2}\\
y=0\ m\rightarrow \boxed{E_p=0\ J}[/tex]
c) Am aflat ca in pozitia de echilibru, energia potentiala este 0. Mai stim ca la pornire(cand elongatia este maxima), viteza este 0, asadar si energia cinetica in acel moment va fi 0.
Stim ca suma dintre energia cinetica si energia potentiala este constanta, atata timp cat nu exista frecare.
Noi trebuie sa aflam energia cinetica in momentul echilibrarii (Ec2):
[tex]E_{p1},\ E_{c1}\ - \ \text{Energiile in punctul de elongatie maxima}\\
E_{p2},\ E_{c2}\ -\ \text{Energiile in momentul echilibrarii}\\\\
E_{c1}=E_{p2}=0\ J\\
E_{p1}+E_{c1}=E_{p2}+E_{c2}=\ \text{const.}\\
E_{p1}+0=0+E_{c2}\rightarrow E_{p1}=E_{c2}\\\\
E_{c2}=E_{p1}=\frac{ky^2}{2}=\frac{kA^2}{2}\\\\
E_{c2}=\frac{20\cdot (0,03)^2}{2}\ J=\boxed{0,009\ J}\\\\
E_{c2}=\frac{mv^2}{2}\rightarrow v = \sqrt{\frac{2E_{c2}}{m}}=\boxed{0,3\ m/s}[/tex]
d)Aici ne vom folosi de formula energiei totale. Pozitia oscilatorului inseamna elongatia y.
[tex]E_t=\frac{kA^2}{2}\\
E_c+E_p=E_t\\
E_c=E_p\rightarrow 2E_p=E_t\\
2\cdot\frac{ky^2}{2}=\frac{kA^2}{2}\\
y^2=\frac{A^2}{2}\rightarrow y=\pm \frac{A\sqrt{2}}{2}\rightarrow y=\pm \frac{0,03\sqrt{2}}{2}\ m\approx \pm \boxed{0,0212\ m}\\\\
\text{Ecuatia elongatiei in functie de timp:}\\
y=Asin(\omega t)\\
\pm\frac{A\sqrt{2}}{2}=Asin(\omega t)\rightarrow sin(\omega t)=\pm \frac{\sqrt{2}}{2}\\
arcsin(\pm\frac{\sqrt{2}}{2})=\pm\frac{\pi}{4}\rightarrow
\boxed{\omega t\in \{ \frac{\pi}{4}+n\pi\ |\ n \in Z\}}\\\\[/tex]
Am folosit formula ecuatiei trigonometrice pentru sinus:
sin x = a ==> x = (-1)^k * arcsin(a) + kπ, k ∈ Z
Noi am aflat faza ωt, dar trebuie sa-l aflam pe t:
[tex]\omega=\sqrt{\frac{k}{m}} \rightarrow t\sqrt{\frac{k}{m}}\in \{\frac{\pi}{4}+n\pi\ |\ n \in Z\}}\\
t\in \{\pi\frac{\frac{1}{4}+n}{\sqrt{\frac{k}{m}}}\ |\ n \in Z\}\\
t\in\{\frac{\pi(n+1)}{3}\sqrt{\frac{m}{k}}\ |\ n \in Z\}\\\\
\sqrt{\frac{m}{k}}=0,1\ m/s^2\\
t\in \{\frac{\pi(n+1)}{30}\ |\ n\in Z\}[/tex]
