{"id":1242,"date":"2015-08-26T00:38:24","date_gmt":"2015-08-26T00:38:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1242"},"modified":"2024-08-23T13:27:02","modified_gmt":"2024-08-23T13:27:02","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-funcao-horaria-da-elongacao-do-mhs","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/mhs\/movimento-harmonico-simples-mhs-funcao-horaria-da-elongacao\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-funcao-horaria-da-elongacao-do-mhs\/","title":{"rendered":"Fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o do MHS – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre<\/b><\/span><\/span><\/span><\/p>\n

fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o do MHS<\/b><\/span><\/span><\/span><\/p>\n

\u00a01-<\/b><\/span><\/span><\/p>\n

a)\u00a0 T=2s —\u00a0 f =1\/T — f= 1\/2Hz\u00a0 (percorre meia volta em cada 1s)<\/b><\/span><\/span><\/p>\n

b)\u00a0 w=2<\/b><\/span><\/span>p<\/b><\/span><\/span><\/span>\/T — w=2<\/b><\/span><\/span>p<\/b><\/span><\/span><\/span>\/2 — w=<\/b><\/span><\/span>p<\/b><\/span><\/span><\/span>rad\/s (varre um \u00e2ngulo de\u00a0prad em cada 1s)<\/b><\/span><\/span><\/p>\n

c) A=4m<\/b><\/span><\/span><\/p>\n

d) na posi\u00e7\u00e3o (elonga\u00e7\u00e3o) x=0 existem duas fases.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como ela est\u00e1 se deslocando em 0, para a esquerda, teremos quej<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=p\/2 rad\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

e)\u00a0j\u00a0=\u00a0j<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ w.t\u00a0 —\u00a0\u00a0j\u00a0=\u00a0p\/2 +\u00a0p.t<\/b><\/span><\/span><\/p>\n

x = A.cosj\u00a0 —\u00a0 x = 4.cos (p\/2 +\u00a0p.t)<\/b><\/span><\/span><\/p>\n

f)\u00a0<\/b><\/span><\/span><\/p>\n

t=0\u00a0 —\u00a0 x=4cos(p\/2 +\u00a0p.t)\u00a0 — x=4cos (p\/2 +\u00a0p.0) — x=4cos (p\/2)\u00a0 —\u00a0 x=4.0\u00a0 —\u00a0 x=0<\/b><\/span><\/span><\/p>\n

t=0,5s\u00a0 — x=4cos (p\/2 +\u00a0p.t)\u00a0 — x=4cos (p\/2 +\u00a0p.0,5) — x=4cos (p)\u00a0 —\u00a0 x=4.(-1)\u00a0 —\u00a0 x= -4m<\/b><\/span><\/span><\/p>\n

t=1s\u00a0 — x=4cos (p\/2 +\u00a0p.t)\u00a0 — x=4cos (p\/2 +\u00a0p.1) — x=4cos (3p\/2)\u00a0 —\u00a0 x=4.0\u00a0 —\u00a0 x=0<\/b><\/span><\/span><\/p>\n

t=1,5s\u00a0 — x=4cos (p\/2 +\u00a0p.t) — x=4cos (p\/2 +\u00a0p.1,5) — x=4cos (2p)\u00a0 —\u00a0 x=4.(+1)\u00a0 —\u00a0 x= +4m<\/b><\/span><\/span><\/p>\n

t=2s\u00a0 — x=4cos (p\/2 +\u00a0p.t) — x=4cos (p\/2 +\u00a0p.2) — x=4cos (5p\/2)\u00a0 —\u00a0 x=4.0\u00a0 —\u00a0 x=0<\/b><\/span><\/span><\/p>\n

t=4,5s\u00a0 — x=4cos (p\/2 +\u00a0p.t) — x=4cos (p\/2 +\u00a0p.4,5) — x=4cos (5p)\u00a0 —\u00a0 x=4.(-1)\u00a0 —\u00a0 x= -4m\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

g)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

02-<\/b><\/span><\/span><\/p>\n

a)\u00a0\u00a0\u00a0\u00a0 w=2p\/T = 2pf\u00a0\u00a0 —\u00a0\u00a0 2p\u00a0= 2pf\u00a0\u00a0 —\u00a0\u00a0 f = 1Hz
\nb) Para que DB e CD sejam pontos m\u00e9dios de AD e A\u2019D, os \u00e2ngulos est\u00e3o indicados na figura abaixo<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para se deslocar de B a C o ponto P deve varrer um \u00e2ngulo de 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0+ 30<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0= 60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>\u00a0=\u00a0p\/3 rad.<\/b><\/span><\/span><\/p>\n

Dj=\u00a0p\/3rad\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 w=Dj\/Dt\u00a0\u00a0 —\u00a0\u00a0 2p\u00a0=\u00a0\u00a0p\/3\/Dt\u00a0\u00a0 —\u00a0\u00a0\u00a0Dt = 1\/6s\u00a0<\/b><\/span><\/span><\/p>\n

w= 2prad\/s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

03-<\/b><\/span><\/span><\/span><\/p>\n

A = 2m<\/b><\/span><\/span><\/span><\/p>\n

T = 4s\u00a0\u00a0 —\u00a0\u00a0 w=2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/T\u00a0\u00a0 —\u00a0\u00a0 w=2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/4\u00a0\u00a0 —\u00a0\u00a0 w =\u00a0<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/2 rad\/s<\/b><\/span><\/span><\/span><\/p>\n

Quando x=0, W<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0pode ser\u00a0p\/2 rad ou 3p\/2. Observando o gr\u00e1fico verificamos que \u00e9\u00a0p\/2, pois, quando t=1s\u00a0 —\u00a0 A = -2m\u00a0<\/b><\/span><\/span><\/p>\n

 <\/p>\n

04- B<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

05-<\/b><\/span><\/span><\/span><\/p>\n

T<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0= 1,5s\u00a0\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0= 1\/1,5Hz\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 T<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0= 6s\u00a0\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0= 1\/6 Hz<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\u00a0\/ f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0= 1\/1,5 X1\/6\u00a0\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\/f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>\u00a0= 4<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

06-<\/b><\/span><\/span><\/span><\/p>\n

a) A=4m\u00a0\u00a0\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0\u00a0 w=2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/T\u00a0\u00a0\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 4<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\u00a0= 2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/T\u00a0\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 T = 1\/2s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 f=1\/T\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —- f=2Hz<\/b><\/span><\/span><\/span><\/p>\n

b) Como\u00a0j<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=0, ele partiu do ponto A=+4m<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

At\u00e9 chegar a 0, ele demorou um tempo t que \u00e9 igual a um quarto do per\u00edodo T=0,5s\u00a0\u00a0 —\u00a0\u00a0 t=0,5\/4\u00a0 —\u00a0\u00a0 t = 0,125s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

07-<\/b><\/span><\/span><\/p>\n

:C\u00e1lculo do per\u00edodo T\u00a0\u00a0 —\u00a0\u00a0 w=2p\/T\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0p\/2 = 2p\/T\u00a0\u00a0 —\u00a0\u00a0 T = 4s<\/b><\/span><\/span><\/p>\n

Para ir de +A at\u00e9 0 ela andou durante um quarto do per\u00edodo, ou seja, durante t=4\/4\u00a0\u00a0 —\u00a0\u00a0 t=1s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

08-<\/b><\/span><\/span><\/p>\n

O per\u00edodo \u00e9 T=4s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —–\u00a0\u00a0\u00a0 w=2p\/T\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 w= 2p\/4\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 w=p\/2rad\/s\u00a0\u00a0\u00a0 e A=2m<\/b><\/span><\/span><\/p>\n

Quando x=1m — t=0\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 x=Acos(wt+q)\u00a0\u00a0 —\u00a0\u00a0\u00a0\u00a0 1=2cos(w.0 +q)\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 1=2cosq\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 cosq=1\/2\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0\u00a0q=60<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>=p\/3<\/b><\/span><\/span><\/p>\n

x=Acos(wt+q)\u00a0\u00a0 —-\u00a0\u00a0\u00a0 x=2cos(p\/2+p\/3)<\/b><\/span><\/span><\/p>\n

 <\/p>\n

09-<\/b><\/span><\/span><\/span><\/p>\n

a) V=<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span>S\/<\/b><\/span><\/span><\/span>D<\/b><\/span><\/span>t\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 V=26\/13\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 V=2cm\/s<\/b><\/span><\/span><\/span><\/p>\n

b) f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>=1\/2Hz\u00a0\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=1\/8Hz\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\/f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=1\/2X8\/1\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub>\/f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub>=4<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

10-<\/b><\/span><\/span><\/span><\/p>\n

A=6m\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 T=8s\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 w=2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/T\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 w=2<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/8\u00a0\u00a0 \u00a0—-\u00a0\u00a0\u00a0 w=<\/b><\/span><\/span><\/span>p<\/b><\/span><\/span>\/4rad\/s<\/b><\/span><\/span><\/span><\/p>\n

Quando x=0, a part\u00edcula est\u00e1 na posi\u00e7\u00e3o angular inicial\u00a0p\/2 rad ou 3p\/2 rad. Se o MCU for no sentido anti-hor\u00e1rio, observando o gr\u00e1fico verificamos que\u00a0j<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=3p\/2 rad.<\/b><\/span><\/span><\/p>\n

X = A.cos(j<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>+ wt)\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0 x=6.cos(3p\/2 +\u00a0p\/4.t)<\/b><\/span><\/span><\/p>\n

11-\u00a0<\/b><\/span><\/span>Fun\u00e7\u00e3o hor\u00e1ria para o MHS\u00a0\u00a0 —\u00a0 x = a.cos(\u03a6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>+\u03c9.t)\u00a0 —\u00a0 \u00a0\u03c9 =2\u03c0\/T \u00a0\u00a0—\u00a0 T (per\u00edodo)\u00a0 —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=a.cos {(\u03c0\/12) + (3\u03c0\/4).t}\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

X<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=a.cos {(3\u03c0\/4).t}\u00a0 —\u00a0\u00a0 como oscilam de modo id\u00eantico a amplitude e o per\u00edodo s\u00e3o os mesmos para as duas part\u00edculas\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

Quando t=8\/9s\u00a0 —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=a.cos {(\u03c0\/12) + (3\u03c0\/4).(8\/9)} —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=a.cos(3\u03c0\/4)\u00a0 —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>=a.\u221a2\/2\u00a0 —\u00a0 X<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=a.cos {(3\u03c0\/4).(8\/9)}\u00a0 —\u00a0 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=a.cos(2\u03c0\/3)\u00a0 —\u00a0 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=a\/2\u00a0 —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=a\u221a2\/2 \u2013 a\/2\u00a0 —\u00a0 \u221a2\u22481,41\u00a0 —\u00a0 x<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0\u2013 x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>=0,21.a\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- D<\/b><\/span><\/span><\/p>\n

12-\u00a0<\/b><\/span><\/span>Fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o x de um MHS de amplitude x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0e pulsa\u00e7\u00e3o w\u00a0 —\u00a0 x=x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.cos(wt + \u0444<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 veja o esquema abaixo\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso em estudo, em t = 0 a part\u00edcula est\u00e1 no ponto de elonga\u00e7\u00e3o m\u00e1xima, portanto \u0444<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>= 0\u00a0 —\u00a0 quando t=1s\u00a0 —\u00a0 x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 a=x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>cos(w.1 + 0)\u00a0 —\u00a0 cosw=(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 a)\/x<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span><\/sub>(I)\u00a0 —\u00a0 quando t=2s\u00a0 —\u00a0 x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 (a + b)=x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>cos(w.2)\u00a0 —\u00a0 x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 a \u2013 b = x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.cos2w\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

cos2w=(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013a \u2013 b)\/x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0II)\u00a0 —\u00a0 cos2w=2.(cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>w) \u2013 1 (III)\u00a0 —\u00a0 (I) e (III) em (II)\u00a0 —\u00a0 2{(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u2013 a)\/x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>}<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 1 = (x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 a \u2013 b)\/x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

2{(x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 2.a.x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\/x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>} = (x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 a \u2013 b)\/x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ 1\u00a0 —\u00a0 2x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 ax<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0\u2013 bx<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 2x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>2<\/b><\/span><\/span><\/sup>\u00a0\u2013 4ax<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ 2.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 (4\u00aa \u2013 a \u2013 b).x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= 2.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —<\/b><\/span><\/span><\/p>\n

x<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2.a<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/(3.a \u2013 b)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

13-<\/b><\/span><\/span><\/p>\n

Num MHS a posi\u00e7\u00e3o angular\u00a0x\u00a0varia com o tempo conforme a fun\u00e7\u00e3o\u00a0\u00a0x = A.cos(\u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ wt)\u00a0\u00a0 que \u00e9 a fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o e onde\u00a0 x \u00e9 a elonga\u00e7\u00e3o; w, a pulsa\u00e7\u00e3o ou freq\u00fc\u00eancia angular ou ainda velocidade angular; \u00a0A, a amplitude (elonga\u00e7\u00e3o m\u00e1xima) e\u00a0\u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0a fase inicial\u00a0 da part\u00edcula em MHS.<\/b><\/span><\/span><\/p>\n

I. Correta\u00a0 —\u00a0 compare x(t) = 4.cos[(\u03c0\/2)t + \u03c0] com \u00a0x = A.cos(\u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ wt)\u00a0e verifique que a amplitude A=4m.<\/b><\/span><\/span><\/p>\n

II. Correta\u00a0 — observe que W=\u03c0\/2rad\/s\u00a0 —\u00a0 W=2\u03c0\/T\u00a0 —\u00a0 \u03c0\/2=2\u03c0\/T\u00a0 —\u00a0 T=4s.<\/b><\/span><\/span><\/p>\n

III. Correta\u00a0 —\u00a0 f=1\/T=1\/4=0,25Hz.<\/b><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/p>\n

I. Falsa\u00a0 — \u00a0Elonga\u00e7\u00e3o (x)\u00a0\u2013 posi\u00e7\u00e3o (localiza\u00e7\u00e3o) da part\u00edcula em MHS sobre o eixo x em rela\u00e7\u00e3o \u00e0 origem 0, ou seja,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

mostra a que dist\u00e2ncia de 0 a part\u00edcula se encontra em determinado instante.<\/b><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 Amplitude (A)\u00a0\u2013 em m\u00f3dulo \u00e9 a elonga\u00e7\u00e3o m\u00e1xima do MHS e corresponde ao raio da circunfer\u00eancia do<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0MCU (R=A).<\/b><\/span><\/span><\/p>\n

III. Falsa\u00a0 —\u00a0 \u00a0O MHS n\u00e3o \u00e9 qualquer movimento vibrat\u00f3rio\u00a0—\u00a0 \u00a0nele o m\u00f3vel se desloca sobre a mesma trajet\u00f3ria, indo<\/b><\/span><\/span><\/p>\n

\"\"\"\"\"\"\"\"<\/p>\n

 <\/p>\n

e vindo, em rela\u00e7\u00e3o a uma posi\u00e7\u00e3o m\u00e9dia de equil\u00edbrio (ponto O, onde a resultante das for\u00e7as que agem sobre ele \u00e9 nula)<\/b><\/span><\/span><\/p>\n

IV. Falsa\u00a0 — ela varia de acordo com a fun\u00e7\u00e3o\u00a0 —\u00a0 a= -w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.A.cos(\u03c6<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ w.t)\u00a0\u00a0\u00a0 fun\u00e7\u00e3o hor\u00e1ria da acelera\u00e7\u00e3o do MHS<\/b><\/span><\/span><\/p>\n

Observe que:<\/b><\/span><\/span><\/p>\n

* quando x=0\u00a0 —\u00a0\u00a0 a=0<\/b><\/span><\/span><\/p>\n

* quando x= +A\u00a0 —\u00a0\u00a0 a= -w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>A\u00a0 — valor m\u00ednimo de a, pois\u00a0\u03c6=\u03c0\u00a0rad e cos\u00a0\u03c0= -1 —\u00a0a<\/b><\/span><\/span>m\u00ednimo<\/b><\/span><\/span><\/sub>= – w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.A<\/b><\/span><\/span><\/p>\n

* quando x= -A\u00a0 —\u00a0\u00a0 a= w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>A\u00a0\u00a0\u00a0 — valor m\u00e1ximo de a, pois\u00a0\u03c6=2\u03c0\u00a0\u00a0rad\u00a0 e cos 2\u03c0= 1\u00a0 —\u00a0a<\/b><\/span><\/span>m\u00e1ximo<\/b><\/span><\/span><\/sub>\u00a0=\u00a0w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.A<\/b><\/span><\/span><\/p>\n

V. Correta\u00a0 —\u00a0 Per\u00edodo (T)\u00a0\u2013 corresponde ao tempo que o MCU demora \u00a0para efetuar uma volta completa ou ao tempo que o MHS demora para efetuar um \u201cvai e vem\u201d completo sobre a reta x\u00a0 —\u00a0 observe, pela defini\u00e7\u00e3o, que ele independe da amplitude A do MHS.<\/b><\/span><\/span><\/p>\n

R- E<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

Voltar para os Exerc\u00edcios<\/a><\/span><\/h3>\n","protected":false},"excerpt":{"rendered":"

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre fun\u00e7\u00e3o hor\u00e1ria da elonga\u00e7\u00e3o do MHS \u00a01- a)\u00a0 T=2s —\u00a0 f =1\/T — f= 1\/2Hz\u00a0 (percorre meia volta em cada 1s) b)\u00a0 w=2p\/T — w=2p\/2 — w=prad\/s (varre um \u00e2ngulo de\u00a0prad em cada 1s) c) A=4m d) na posi\u00e7\u00e3o (elonga\u00e7\u00e3o) x=0 existem duas fases. Como ela est\u00e1 se deslocando em 0, para a<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1238,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-1242","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1242","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/comments?post=1242"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1242\/revisions"}],"predecessor-version":[{"id":10845,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1242\/revisions\/10845"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1238"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1242"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}