{"id":1263,"date":"2015-08-26T01:12:01","date_gmt":"2015-08-26T01:12:01","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1263"},"modified":"2024-08-23T13:22:11","modified_gmt":"2024-08-23T13:22:11","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-mhs-sistema-massa-mola","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/mhs\/mhs-sistema-massa-mola\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-mhs-sistema-massa-mola\/","title":{"rendered":"MHS Sistema massa-mola – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

MHS Sistema massa-mola<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/p>\n

01.a) F(t) = ma\u00a0 —\u00a0 F(t) = mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>Acos(wt +\u00a0j)<\/b><\/span><\/span><\/p>\n

b) mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>Acos(wt =\u00a0j) = mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>x\u00a0 —\u00a0 x(t) = A cos(wt +\u00a0j)<\/b><\/span><\/span><\/p>\n

c) Usando as equa\u00e7\u00f5es para a energia cin\u00e9tica e potencial, juntamente com as equa\u00e7\u00f5es hor\u00e1rias da posi\u00e7\u00e3o e velocidade, temos que\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\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>k<\/b><\/span><\/span><\/p>\n

Ec(t) = 1\/2mv2(t) = 1\/2 m(wAsen(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>))<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0=\u00a0 1\/2<\/b><\/span><\/span>mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>sen<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= 1\/2 kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0sen<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\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>Ep(t) = 1\/2kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(t) = 1\/2 k(Acos(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>) )<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>(t)=1\/2kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>)<\/b><\/span><\/span>\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\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

A energia mec\u00e2nica \u00e9 a soma da energia cin\u00e9tica com a energia potencial. Logo,<\/b><\/span><\/span><\/p>\n

Emec = 1\/2 kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0sen<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>) + 1\/2kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>)\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1\/2kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(sen<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>) + cos<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(wt +\u00a0<\/b><\/span><\/span>j<\/b><\/span><\/span><\/span>))\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1\/2kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>(1)<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1\/2kA2,\u00a0<\/b><\/span><\/span>que \u00e9 uma constante<\/b><\/span><\/span><\/p>\n

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

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

E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>-1J\u00a0\u00a0\u00a0\u00a0\u00a0 A=0,5m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 V<\/b><\/span><\/span>m\u00e1xima<\/b><\/span><\/span><\/sub>=2m\/s<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1\/2.kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 1=1\/2.k.(0,5)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>k=8N\/m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=1\/2.mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m\u00e1xima<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1=1\/2.m.(2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>m=0,5kg<\/b><\/span><\/span><\/p>\n

T=2p\u00d6m\/k\u00a0 —\u00a0 T=2p\u00d60,5\/8\u00a0 —\u00a0 T=2p.1\/4\u00a0 —\u00a0 T=p\/2 s\u00a0 —\u00a0 f=1\/T\u00a0 —\u00a0 f=1\/p\/2\u00a0 —<\/b><\/span><\/span><\/span>\u00a0f=2\/pHz<\/b><\/span><\/span><\/span><\/p>\n

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

01- Verdadera \u2013 a for\u00e7a el\u00e1stica iguala a for\u00e7a peso no ponto m\u00e9dio onde a velocidade \u00e9 m\u00e1xima e a acelera\u00e7\u00e3o nula<\/b><\/span><\/span><\/p>\n

02- Falsa \u2013 a velocidade da pessoa aumenta at\u00e9 o ponto m\u00e9dio e a partir da\u00ed come\u00e7a a diminuir.<\/b><\/span><\/span><\/p>\n

04- Falsa \u2013 pois, T=2p\u00d6m\/k<\/b><\/span><\/span><\/p>\n

08- Verdadeira\u2014veja equa\u00e7\u00e3o acima<\/b><\/span><\/span><\/p>\n

16- Falsa- a acelera\u00e7\u00e3o \u00e9 nula no ponto m\u00e9dio, a partir do qual ela inverte seu sentido, retardando a pessoa.<\/b><\/span><\/span><\/p>\n

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

E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 (constante) e E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=m.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m\u00e1xima<\/b><\/span><\/span><\/sub>\/2\u00a0 —k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m\u00e1xima<\/b><\/span><\/span><\/sub>\/2\u00a0 —\u00a0 k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= m.V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m\u00e1xima\u00a0<\/b><\/span><\/span><\/sub>= constante, ou seja, k \u00e9 inversamente proporcional a A, e E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0\u00e9 sempre constante\u00a0 —\u00a0 alternativa a<\/b><\/span><\/span><\/p>\n

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

a)\u00a0 Como ela est\u00e1 sujeita a apenas uma for\u00e7a, o movimento \u00e9 horizontal e essa for\u00e7a \u00e9 a for\u00e7a el\u00e1stica.Quando x=1m\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=1J\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 1=k.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 k=2N\/m. A amplitude A vale 2m, pois \u00e9 a\u00ed que v=0.<\/b><\/span><\/span><\/p>\n

E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 — E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=2.2<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=4J<\/b><\/span><\/span><\/p>\n

b) Quando x=0\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=0\u00a0 — E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>p\u00a0 —\u00a0<\/b><\/span><\/span><\/sub>\u00a0\u00a04=mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 0\u00a0 —\u00a0 4=0,5V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 V=\u00d616\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V=4m\/s<\/b><\/span><\/span><\/p>\n

c) E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ Ep\u00a0 —\u00a0 4=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 4=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ 2.1<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=3J<\/b><\/span><\/span><\/p>\n

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

Vamos calcular o per\u00edodo T, que \u00e9 o tempo que ele demora para efetuar um vai e vem completo.<\/b><\/span><\/span><\/p>\n

T=2p\u00d6m\/k\u00a0 —\u00a0 T=2.3.\u00d60,35\/100\u00a0 —\u00a0 T=6.5.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 T@0,3s<\/b><\/span><\/span><\/p>\n

0,3s —— 1 vai e vem completo<\/b><\/span><\/span><\/p>\n

1,0s ——\u00a0\u00a0\u00a0 n<\/b><\/span><\/span><\/p>\n

n@3,3\u00a0\u00a0\u00a0\u00a0 —–\u00a0 alternativa C<\/b><\/span><\/span><\/p>\n

08-
\na) E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 \u00a0\u00a0—\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=7\/9.k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2
\nE<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= E\u00ad<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = 7\/9.k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 = (7.k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0+ 9.k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>)\/18\u00a0 —\u00a0 9.k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= 16.k.x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 x =\u00a0\u00d69\/16.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/p>\n

X =\u00a0\u00b1\u00a03\/4.A\u00a0 — Nas posi\u00e7\u00f5es x = + 3\/4.A e X = – 3\/4.A\u00a0<\/b><\/span><\/span><\/p>\n

b) Sim. Por exemplo, no ponto O quando toda a energia mec\u00e2nica estar\u00e1 na forma de energia cin\u00e9tica.<\/b><\/span><\/span><\/p>\n

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

(01) Falsa. A for\u00e7a el\u00e1stica n\u00e3o \u00e9 constante, pois varia de acordo com a deforma\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

(02) Correta. Desprezando-se as for\u00e7as externas dissipativas o sistema oscilar\u00e1 sempre.<\/b><\/span><\/span><\/p>\n

(04) Correta\u00a0 —\u00a0 w =\u00d6k\/m\u00a0 —\u00a0 w =\u00a0\u00d6200\/2\u00a0 —\u00a0 w = 10rad\/s<\/b><\/span><\/span><\/p>\n

(08) Falsa. A velocidade m\u00e1xima do corpo vale\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1xima<\/b><\/span><\/span><\/sub>\u00a0= w.A = 10.0,1 = 1,0m\/s, mas n\u00e3o \u00e9 no ponto de m\u00e1ximo deslocamento, mas sim na posi\u00e7\u00e3o central 0.<\/b><\/span><\/span><\/p>\n

(16) Correta. O per\u00edodo T \u00e9 dado por T = 2p\u00d6m\/k\u00a0 —\u00a0 T = 2p\u00d62\/200\u00a0 —\u00a0 T = 2p.1\/10\u00a0 —\u00a0 T =\u00a0p\/5 s.<\/b><\/span><\/span><\/p>\n

Soma=(02 + 04 + 16) = 22<\/b><\/span><\/span><\/p>\n

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

Em X<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0\u00e9 m\u00e1xima e E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0\u00e9 nula. Em X<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0= a\u00a0 —\u00a0\u00a0 E = E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E = a + b\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ a = a + b\u00a0 —\u00a0 E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0= +b<\/b><\/span><\/span><\/p>\n

R: C<\/b><\/span><\/span><\/p>\n

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

a) No equil\u00edbrio\u00a0 —\u00a0 F<\/b><\/span><\/span>e<\/b><\/span><\/span><\/sub>\u00a0= P\u00a0 —\u00a0 k.x = m.g\u00a0 —\u00a0 k = m.g\/x\u00a0 —\u00a0 k = 0,4.10\/0,05\u00a0 —\u00a0 k=4\/0,05\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>k =80N\/m<\/b><\/span><\/span><\/p>\n

b) O movimento \u00e9 um MHS e o seu per\u00edodo n\u00e3o depende da amplitude A e \u00e9 fornecido pela express\u00e3o\u00a0 —\u00a0 T = 2.p\u00d6m\/k\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\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

T = 2p\u00d60,4\/80\u00a0 —\u00a0 T = 2p.2,24\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T = 4,48p\u00a0s<\/b><\/span><\/span><\/p>\n

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

a) Escolhendo qualquer ponto por exemplo, quando E\u00a0 —\u00a0 F = 0,75nN, x=-15nm.\u00a0F =- k.x\u00a0 —\u00a0 0,75.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\u00a0= -(-15).10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>.k\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

k = 0,75\/15—\u00a0\u00a0<\/b><\/span><\/span>k = 0,05N\/m<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

b) T = 2p\u00d6m\/k\u00a0 —\u00a0 T = 2p\u00d6180.10<\/b><\/span><\/span>-26<\/b><\/span><\/span><\/sup>\/0,05\u00a0 —\u00a0 T = 12p10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>\u00a0s\u00a0 —\u00a0 w =2p\/T\u00a0 —\u00a0 w = 2p\/12p10<\/b><\/span><\/span>-12<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 w = 1\/6.10<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sup>rad\/s<\/b><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>m\u00e1aima<\/b><\/span><\/span><\/sub>\u00a0= w.A\u00a0 —\u00a0 V<\/b><\/span><\/span>m\u00e1xima\u00a0<\/b><\/span><\/span><\/sub>= 1\/6.10<\/b><\/span><\/span>12<\/b><\/span><\/span><\/sup>.30.10<\/b><\/span><\/span>-9<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V<\/b><\/span><\/span>m\u00e1xima\u00a0<\/b><\/span><\/span><\/sub>= 5.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0m\/s<\/b><\/span><\/span><\/p>\n

13-\u00a0<\/b><\/span><\/span>Com o cubo em equil\u00edbrio, a for\u00e7a resultante sobre ele \u00e9 nula\u00a0 —\u00a0 empuxo = peso\u00a0 —\u00a0 \u03c1.g.v = m.g\u00a0 —\u00a0 1.000v=81\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

V=0,081m<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 como a \u00e1rea da base \u00e9 1 m<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>, isto significa que o ponto de equil\u00edbrio fica a 0,081 m ou 8,1 cm abaixo da linha da superf\u00edcie, pois V=s.h\u00a0 —\u00a0 0, 081=1.h\u00a0 —\u00a0 h=0,081m\u00a0 —\u00a0 o cubo \u00e9 for\u00e7ado para baixo, digamos uma profundidade x al\u00e9m de 0,081 m\u00a0 —\u00a0 a for\u00e7a resultante sobre o cubo funcionar\u00e1 como a for\u00e7a restauradora do MHS\u00a0 —\u00a0 F<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>\u00a0= Empuxo \u2013 Peso\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>\u00a0= \u03c1.g.(0,081 + x) \u2013 m.g\u00a0 —\u00a0 F<\/b><\/span><\/span>resultante<\/b><\/span><\/span><\/sub>\u00a0= \u03c1.g.0,081 + \u03c1.g.x \u2013 m.g = \u03c1.g.x\u00a0 —\u00a0 se esta for\u00e7a \u00e9 a restauradora do MHS ent\u00e3o \u03c1.g.x = k.x \u00a0—\u00a0 \u00a0k = \u03c1.g\u00a0 — frequ\u00eancia angular de um sistema oscilante\u00a0 — w=\u221a(k\/m)=\u221a (\u03c1.g\/m)=\u221a(1.000×10\/81)\u00a0 —\u00a0 w=\u221a(10.000\/81)\u00a0 —\u00a0 w=100\/9 rad\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- A<\/b><\/span><\/span><\/p>\n

14-\u00a0<\/b><\/span><\/span>No MHS (movimento harm\u00f4nico simples) o sistema apresenta energia potencial el\u00e1stica m\u00e1xima nas extremidades (A e \u2013A) e energia cin\u00e9tica m\u00e1xima no centro (0). Desta forma a velocidade da part\u00edcula no centro do sistema \u00e9 dada por\u00a0 —\u00a0 mv<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2=kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

0,05v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=20.(0,2)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=0,8\/0,05\u00a0 —\u00a0 v=\u221a16\u00a0 —\u00a0 v=4m\/s\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

15-\u00a0<\/b><\/span><\/span>a) Express\u00e3o da frequ\u00eancia\u00a0 —\u00a0 f = (1\/2\u03c0).\u221a(k\/m)\u00a0 —\u00a0 30.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= (1\/2.3,14).\u221a(k\/5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 30.6,28.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>\u00a0= \u221a(k\/5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>)\u00a0 —\u00a0 (188,4.10<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0= k\/5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 k=35.495.10<\/b><\/span><\/span>6<\/b><\/span><\/span><\/sup>.5.10<\/b><\/span><\/span>-3<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 k=1,77.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>\u00a0N\/m\u00a0 — \u00a0energia potencial el\u00e1stica\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=(1\/2).k.\u2206x<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 — E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=0,5.1,77.10<\/b><\/span><\/span>8<\/b><\/span><\/span><\/sup>.(0,02.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>= 3.54.10<\/b><\/span><\/span>-8<\/b><\/span><\/span><\/sup>\u00a0J\u00a0<\/b><\/span><\/span>\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

b) Express\u00e3o fornecida\u00a0 —\u00a0 (\u2206t\/t)=(1\/2).(\u2206L\/L)\u00a0 —\u00a0 (\u2206t\/1.800)=(1\/2).(- 0,2)\/90)\u00a0 —\u00a0 \u2206t\/1.800= – 1\/900\u00a0 —\u00a0 \u2206t=-1.800\/900\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u2206t= – 2s\u00a0 — \u00a0novo intervalo de tempo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u2206t\u2019= 1800 \u2013 2 = 1798 s<\/b><\/span><\/span><\/p>\n

16-\u00a0<\/b><\/span><\/span>Nas figuras abaixo\u00a0 —\u00a0 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>: dist\u00e2ncia da extremidade fixa da mola at\u00e9 o centro de oscila\u00e7\u00e3o para o sistema n\u00e3o em rota\u00e7\u00e3o\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

R: dist\u00e2ncia da extremidade fixa da mola at\u00e9 o centro de oscila\u00e7\u00e3o para o sistema em rota\u00e7\u00e3o\u00a0 —\u00a0 R\u2019: dist\u00e2ncia da extremidade fixa da mola at\u00e9 um ponto qualquer da trajet\u00f3ria\u00a0 —\u00a0 x: deforma\u00e7\u00e3o da mola\u00a0 —\u00a0 \u2206x: varia\u00e7\u00e3o da deforma\u00e7\u00e3o entre o centro de oscila\u00e7\u00e3o em rota\u00e7\u00e3o e um ponto qualquer da trajet\u00f3ria\u00a0 —\u00a0 se o sistema apenas girasse sem oscilar, o movimento circular uniforme teria raio R\u00a0 —\u00a0 a for\u00e7a resultante sobre a part\u00edcula seria apenas a for\u00e7a el\u00e1stica agindo como resultante centr\u00edpeta\u00a0 —\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=F<\/b><\/span><\/span>el\u00e9trica<\/b><\/span><\/span><\/sub>=mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>R=kx\u00a0 —\u00a0 mw<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>R=k(R \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) (I)\u00a0 —\u00a0 para o sistema girando e oscilando vamos considerar um referencial fixo ao oscilador (referencial n\u00e3o-inercial)\u00a0 —\u00a0 para esse referencial h\u00e1 um movimento oscilat\u00f3rio, com uma deforma\u00e7\u00e3o aparente da mola igual a \u2206x, quando a part\u00edcula est\u00e1 numa posi\u00e7\u00e3o de raio R\u2019\"\"R\u00a0 —\u00a0 para esse referencial, temos que introduzir a \u201cfor\u00e7a de in\u00e9rcia\u201d ou for\u00e7a centr\u00edfuga (F<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>), dirigida para fora, oposta \u00e0 for\u00e7a el\u00e1stica, como mostrado na Fig 2\u00a0 —\u00a0 nesse referencial, obedecendo ao sentido de orienta\u00e7\u00e3o, a for\u00e7a resultante vale\u00a0 — F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>= – F<\/b><\/span><\/span>el\u00e9trica<\/b><\/span><\/span><\/sub>\u00a0+ F<\/b><\/span><\/span>i<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 m.a = -k.x + m\u22062R\u2019 (II)\u00a0 —\u00a0 figura 2\u00a0 —\u00a0 x= (R\u2019 \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

R\u2019=\u2206x + R\"\"\u00a0 — substituindo-os em (II)\u00a0 —\u00a0\u00a0 ma = -k(\u2206x + R \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m\u22062(\u2206x + R)\u00a0 —\u00a0 ma = -k\u2206x \u2013 k(R \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + m\u22062\u2206x + m\u22062R (IV)\u00a0 —\u00a0 substituindo (I) em (IV)\u00a0 —\u00a0 ma = -k\u2206x \u2013 k(R \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) + 2\u2206x + k(R \u2013 R<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 fazendo os cancelamentos e isolando a\u00a0 —\u00a0 a= ( – k\/m + w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>).\u2206x\u00a0 —\u00a0 a= – (k\/m \u2013w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>).\u2206x (V)\u00a0 —\u00a0 a propriedade fundamental de um MHS diz que a acelera\u00e7\u00e3o \u00e9 diretamente proporcional \u00e0 elonga\u00e7\u00e3o (\u2206x)\u00a0 —\u00a0 a constante de proporcionalidade \u00e9 o oposto do quadrado da pulsa\u00e7\u00e3o do movimento oscilat\u00f3rio (\u2206<\/b><\/span><\/span>osc<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 a= – w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>osc<\/b><\/span><\/span><\/sub>.\u2206x\u00a0 —\u00a0 a= – 2(\u2206f)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.\u2206x (VI)\u00a0 —\u00a0 igualando (V) com (VI)\u00a0 —\u00a0 4\u22062f<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=k\/m \u2013 w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 f<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=k\/4\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>m \u2013 1\/4\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 assim, o quadrado da frequ\u00eancia do MHS depende linearmente do quadrado da velocidade angular\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- D<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>Rela\u00e7\u00e3o fundamental do MHS\u00a0 —\u00a0 a= – w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.x\u00a0 —\u00a0 quando t=1\/3s\u00a0 —\u00a0 x=0,3cos\u03c0t=0,3cos\u03c0.1\/3\u00a0 —\u00a0 x=0,3cos\u03c0\/3=0,3.1\/2\u00a0 —\u00a0 x=0,15m\u00a0 —\u00a0 x=0,3c0swt\u00a0 —\u00a0 w=\u03c0\u00a0 —\u00a0 a=-w<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.x= – \u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.0,15\u00a0 —\u00a0 \u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=10\u00a0 —\u00a0 a=1,5m\/s<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 lei fundamental da din\u00e2mica\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=ma\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,5.1,5\u00a0 —\u00a0 F<\/b><\/span><\/span>R<\/b><\/span><\/span><\/sub>=0,75N\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/p>\n

18-\u00a0<\/b><\/span><\/span>a) Observe na figura abaixo, as posi\u00e7\u00f5es ocupadas\u00a0 pela crian\u00e7a na roda gigante nos instantes t=0, t=60s, t=120s, t=180s e t=240s,<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0e o gr\u00e1fico da altura da crian\u00e7a em fun\u00e7\u00e3o do tempo equivale a um MHS de amplitude a=20m, com a origem no ponto h=20m..<\/b><\/span><\/span><\/p>\n

b) Supondo que o carrossel efetue uma volta completa em 20s, ou seja, T=20s\u00a0 —\u00a0 velocidade escalar\u00a0 —\u00a0 v=2\u03c0r\/T=2.(3).4\/20\u00a0 —<\/b><\/span><\/span><\/p>\n

v=1,2m\/s\u00a0 —\u00a0 a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/R=1,44\/4=0,36ms<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>\u00a0—\u00a0\u00a0<\/b><\/span><\/span>a<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>=0,36m\/s<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>—\u00a0 o gr\u00e1fico pedido est\u00e1 representado abaixo:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

 <\/p>\n

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

Como as for\u00e7as dissipativas s\u00e3o desprez\u00edveis, a energia mec\u00e2nica \u00e9 sempre constante no MHS e vale\u00a0E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>= kA<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 ou\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 ou\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=kx<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + m.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=k.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 0,4=20.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>=8,8\/20\u00a0 —\u00a0 A=\u221a(4.10<\/b><\/span><\/span>-2<\/b><\/span><\/span><\/sup>)=2.10<\/b><\/span><\/span>-1<\/b><\/span><\/span><\/sup>m\u00a0 — A=0,2m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre MHS Sistema massa-mola \u00a0 01.a) F(t) = ma\u00a0 —\u00a0 F(t) = mw2Acos(wt +\u00a0j) b) mw2Acos(wt =\u00a0j) = mw2x\u00a0 —\u00a0 x(t) = A cos(wt +\u00a0j) c) Usando as equa\u00e7\u00f5es para a energia cin\u00e9tica e potencial, juntamente com as equa\u00e7\u00f5es hor\u00e1rias da posi\u00e7\u00e3o e velocidade, temos que\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0k Ec(t) = 1\/2mv2(t) = 1\/2 m(wAsen(wt +\u00a0j))2\u00a0=\u00a0 1\/2mw2A2sen2(wt<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1259,"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-1263","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1263","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=1263"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1263\/revisions"}],"predecessor-version":[{"id":10840,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1263\/revisions\/10840"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1259"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1263"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}