Question

Un corp de masa m,legat de un resort ideal,oscileaza armonic cu amplitudinea A, energia totala a oscilatorului fiind E.Calculati: a) pulsatia w(omega doar ca nu o am pe tel) a miscarii oscilatorii b)elongatia x si viteza v ale corpului in momentele in care energia sa cinetica este egala cu cea potentiala;c) forta elastica F datorata resortului in conditiile de la punctul B.
m=1kg
A=Radical din 2 pe 4.(m) metri
E=4 J

Answer 1

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Answer 2

m = 1 kg

$\displaystyle{ A = \frac{\sqrt{2}}{4} \ m }$

E = 4 J

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a) $\displaystyle{ \omega }$ = ?

b) x = ? v = ? (Ec = Ep)

c) F = ?

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a)

$\displaystyle{ E = \frac{K \cdot A^{2}}{2} \rightarrow K = \frac{2E}{A^{2} }$

$\displaystyle{ K = \frac{2 \cdot 4}{\frac{2}{16}}= \frac{8}{1} \cdot \frac{16}{2} = 64 \ N/m }$

$\displaystyle{ K = m \cdot \omega ^{2} \rightarrow \omega = \sqrt{\frac{K}{m}} }$

$\displaystyle{ \omega = \sqrt{64} = 8 \ rad/s }$

b)

Ec = Ep          (1)

E = Ec + Ep    (2)

(1) + (2) ⇒ E = 2 × Ec

$\displaystyle{ Ec = \frac{mv^{2}}{2} }$

• se inmulteste cu 2

$\displaystyle{ E = mv^{2} }$

$\displaystyle{ v = \sqrt{\frac{E}{m}} = \sqrt{4} = 2 \ m/s }$

$\displaystyle{ x = A \cdot sin (\omega t + \phi _{0}) }$

$\displaystyle{ x = \frac{\sqrt{2}}{4} \cdot sin (8t + \phi _{0}) }$

c)

$\displaystyle{ Ep = \frac{E}{2} }$               (3)

$\displaystyle{ Ep = \frac{K \cdot x^{2}}{2} }$        (4)

$\displaystyle{ (3) + (4) \rightarrow \frac{K \cdot x^{2}}{2} = \frac{E}{2} }$

$\displaystyle{ E = x^{2} \cdot K \rightarrow x = \sqrt{\frac{E}{K}} }$

$\displaystyle{ x = \sqrt{\frac{4}{64}} = \sqrt{\frac{1}{16}} }$

$\displaystyle{ x = \frac{1}{4} = 0,25 \ m }$

$\displaystyle{ F = -K \cdot x = -64 \cdot \frac{1}{4} }$

$\displaystyle{ F = -\frac{64}{4} = -16 \ N }$