{"id":2423,"date":"2016-09-11T00:29:24","date_gmt":"2016-09-11T00:29:24","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2423"},"modified":"2024-08-23T15:06:53","modified_gmt":"2024-08-23T15:06:53","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-interferencia-de-ondas","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/ondulatoria\/ondas\/interferencia-de-ondas\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-interferencia-de-ondas\/","title":{"rendered":"Interfer\u00eancia de Ondas – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Interfer\u00eancia de Ondas<\/span><\/span><\/span><\/p>\n

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

\u00a001-<\/b><\/span><\/span><\/span> <\/b>R: E.\u00a0 Como o pulso A retornar\u00e1 invertido (extremidade fixa) ao se superpuserem os dois pulsos sofrer\u00e3o interfer\u00eancia destrutiva e em seguida, cada pulso seguir\u00e1 se caminho mantendo suas caracter\u00edsticas originais<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>Nulas; pois no instante da interfer\u00eancia destrutiva n\u00e3o h\u00e1 movimento dos pontos participantes.<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>B\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>(01+02+04) = 07<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

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

06-<\/b><\/span><\/span><\/span> <\/b>R – D<\/b><\/span><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 200=\u0394S\/0,02\u00a0 —\u00a0\u00a0\u0394S=4cm<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

07-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span><\/p>\n

a) A=2m;\u00a0l=4m e f=0,5Hz\u00a0 —\u00a0 V=lf\u00a0 —\u00a0 2=4f\u00a0 —\u00a0 f=0,5Hz<\/b><\/span><\/span><\/span><\/p>\n

b) Ap\u00f3s 4s as ondas se deslocaram de V=\u0394S\/\u0394t\u00a0 —\u00a0 2=\u0394S\/4\u00a0 —\u00a0\u00a0\u0394S=8m. Assim, a primeira onda que est\u00e1 voltando invertida e a segunda que est\u00e1 chegando, no instante 4s, se interferem construtivamente e a amplitude da onda resultante ser\u00e1 de 2+2=4m<\/b><\/span><\/span><\/span>\"\"<\/p>\n

08-<\/b><\/span><\/span><\/span> <\/b>Todos os pontos est\u00e3o em interfer\u00eancia construtiva cuja soma ser\u00e1 a altura do ret\u00e2ngulo. R: E<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>a) Os pontos A (7m) e B (9m) est\u00e3o no pulso I que se move para a direita e localizados conforme a figura e suas velocidades verticais indicadas na mesma.<\/b><\/span><\/span><\/span><\/p>\n

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

b) Enquanto o pulso I se desloca de 2m na dire\u00e7\u00e3o X com velocidade de 4m\/s, os pontos da corda deslocam-se de 3cm na dire\u00e7\u00e3o Y, no mesmo intervalo de tempo\u00a0\u0394t.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Pulso I\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>X\/<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t=<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>X\/V<\/b><\/span><\/span><\/span>I<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t=2\/4\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u0394<\/b><\/span><\/span><\/span><\/span>t=0,5s<\/b><\/span><\/span><\/span><\/p>\n

Nesse mesmo intervalo de tempo\u00a0\u0394t=0.5s, o ponto A, com velocidade V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>, percorreu \u0394Y=0,03m na dire\u00e7\u00e3o vertical\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=\u0394Y\/\u0394t\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,03\/0,5\u00a0 —\u00a0\u00a0V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=0,06m\/s ou 6cm\/s.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

c) Tanto o pulso I como o II, com velocidade de 4m\/s, em 1s, deslocam-se, na dire\u00e7\u00e3o Xde:\u00a0 V=\u0394X\/\u0394t\u00a0 —\u00a0 4=\u0394X\/1\u00a0 —\u00a0\u00a0\u0394X=4m<\/b><\/span><\/span><\/span><\/p>\n

Como cada pulso desloca-se 4m em sentidos contr\u00e1rios, no instante t=1s eles estar\u00e3o exatamente<\/b><\/span><\/span><\/span><\/p>\n

superpostos e ocorrer\u00e1 interfer\u00eancia destrutiva, com a corda exatamente na horizontal e com todos seus pontos tendo velocidade nula, inclusive C e D.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

10-<\/b><\/span><\/span><\/span> <\/b>A parte da frente de onda da direita (esquema1) percorreu 2cm at\u00e9 atingir o anteparo e mais 3cm de retorno (esquema 2) no intervalo de tempo\u00a0\u0394t. A mesma dist\u00e2ncia percorreu a outra onda.<\/b><\/span><\/span><\/span><\/p>\n

V=\u0394S\/\u0394t\u00a0 —\u00a0 2=5\/\u0394t\u00a0 —\u00a0\u00a0\u0394t = 2,5s<\/b><\/span><\/span><\/span><\/p>\n

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

12-<\/b><\/span><\/span><\/span> <\/b>Veja a figura abaixo:<\/b><\/span><\/span><\/span><\/p>\n

d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= (5\u03bb\u00a0+ X) \u2013 X = 5\u03bb\u00a0= 10.\u03bb\/2\u00a0 — portanto n \u00e9 par\u00a0 —\u00a0 interfer\u00eancia construtiva de amplitude 2 A<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

14-<\/b><\/span><\/span><\/span> <\/b>a) Como as duas fontes emitem ondas de mesma freq\u00fc\u00eancia e est\u00e3o lado a lado, estas ondas est\u00e3o em fase e Jos\u00e9 Guilherme ouve esses sons com refor\u00e7o (interfer\u00eancia construtiva). Mas, com o deslocamento de uma delas a interfer\u00eancia vai ficando destrutiva.<\/b><\/span><\/span><\/span><\/p>\n

b) V=\u03bbf\u00a0 —\u00a0 340=l680\u00a0 —\u00a0\u00a0\u03bb=0,5m.<\/b><\/span><\/span><\/span><\/p>\n

A primeira interfer\u00eancia destrutiva ocorre quando n=0. Assim, d=(2n+1).\u03bb\/2 —\u00a0 d=(2.0 +1).0,5\/2\u00a0 —-d=0,25m=25cm<\/b><\/span><\/span><\/span><\/p>\n

15-<\/b><\/span><\/span><\/span> <\/b>Como \u00e9 m\u00ednimo, trata-se de interfer\u00eancia destrutiva e \u00e9 v\u00e1lida a rela\u00e7\u00e3o d<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= (2n+1).\u03bb\/2, com n=0 (primeiro m\u00ednimo).<\/b><\/span><\/span><\/span><\/p>\n

Observe na figura abaixo que d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3m e d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=5m<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do comprimento de onda\u00a0\u03bb\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=5-3=2m\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sub><\/span>\u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0= (2n+1).\u03bb\/2\u00a0 —\u00a0 2=(2.0 + 1).\u03bb\/2\u00a0 —\u00a0\u00a0\u03bb=4m.<\/b><\/span><\/span><\/span><\/p>\n

V=\u03bbf\u00a0 —\u00a0 340=4f\u00a0 —\u00a0\u00a0f=85Hz<\/b><\/span><\/span><\/span><\/p>\n

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

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

A interfer\u00eancia \u00e9 destrutiva, pois o sinal captado \u00e9 mais fraco. \u00a0d(F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>P)=4m\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>P)=5m\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>P) –\u00a0 d(F<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>P)= (2n+1).\u03bb\/2\u00a0\u00a0\u00a0 5-4=(2.0 +1).\u03bb\/2\u00a0 —\u00a0 1=\u03bb\/2\u00a0 —\u00a0\u00a0\u03bb=2m<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>E<\/b><\/span><\/span><\/span><\/p>\n

Observe que o ponto P est\u00e1 exatamente no meio das duas fontes de onda e, ent\u00e3o como elas se movem com a mesma velocidade (mesmo meio), haver\u00e1 sempre uma interfer\u00eancia construtiva, pois as ondas est\u00e3o em fase.<\/b><\/span><\/span><\/span><\/p>\n

Assim, duas cristas e dois vales sempre chegar\u00e3o juntas ao ponto P e a b\u00f3ia sobe e desce com amplitude que \u00e9 o dobro da amplitude de cada fonte, mas com o mesmo per\u00edodo T das fontes originais.<\/b><\/span><\/span><\/span><\/p>\n

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

18-<\/b><\/span><\/span><\/span> <\/b>Observe na figura abaixo que d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=36m e d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=30m<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

a) Verifique, na express\u00e3o da interfer\u00eancia construtiva da onda, onde, d<\/b><\/span><\/span><\/span>1\u00ad<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2n.\u03bb\/2\u00a0 —\u00a0\u03bb=d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>-d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/n\u00a0 —\u00a0 que\u00a0\u03bb\u00a0n s\u00e3o inversamente proporcionais. Assim, o maior comprimento de onda (\u03bb) ocorre quando n=1.<\/b><\/span><\/span><\/span><\/p>\n

D<\/b><\/span><\/span><\/span>1\u00ad<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0= 2n.\u03bb\/2\u00a0 —\u00a0 36-30=2.1.\u03bb\/2\u00a0 —\u00a0\u00a0\u03bb=6m\u00a0<\/b><\/span><\/span><\/span><\/p>\n

b) Verifique na express\u00e3o V=\u03bbf\u00a0 —\u00a0 f=V\/l\u03bb —\u00a0 que a freq\u00fc\u00eancia (f) e o comprimento de onda (\u03bb) s\u00e3o inversamente proporcionais. Assim:<\/b><\/span><\/span><\/span><\/p>\n

1<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0menor freq\u00fc\u00eancia\u00a0 —\u00a0 V=\u03bbf\u00a0 —340=6f\u00a0 —\u00a0\u00a0f=56,7Hz<\/b><\/span><\/span><\/span><\/p>\n

2<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sup><\/span>\u00a0menor freq\u00fc\u00eancia\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>-d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=2n.\u00a0\u03bb\/2\u00a0 —\u00a0 36-30=2.2.\u00a0\u03bb\/2\u00a0 —\u00a0\u00a0\u03bb=3m\u00a0\u00a0\u00a0 —-\u00a0\u00a0\u00a0\u00a0 V=\u03bbf\u00a0 —\u00a0 340=3f\u00a0 —\u00a0\u00a0f=113,3Hz<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span> <\/b>R \u2013 B<\/b><\/span><\/span><\/span><\/p>\n

V=\u03bbf\u00a0 —\u00a0 340=\u03bb680\u00a0 —\u00a0\u00a0\u03bb=0,5m\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Interfer\u00eancia destrutiva \u2013 X-1,5=(2n+1).\u00a0\u03bb\/2\u00a0 —\u00a0 m\u00ednima (n=0)\u00a0 —\u00a0 X=1,5+ (2.0+1).\u00a0\u03bb\/2 — X=1,75m.<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span> <\/b>O comprimento de onda de cada uma das fontes vale\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 340=\u03bb.170\u00a0 —\u00a0\u00a0\u03bb=2m<\/b><\/span><\/span><\/span><\/p>\n

a) Ao atingirem A, as ondas provenientes de F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0devem sofrer a\u00ed, interfer\u00eancia destrutiva (m\u00ednimos de intensidade).<\/b><\/span><\/span><\/span><\/p>\n

Na situa\u00e7\u00e3o inicial\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>A)=10m e d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>A)=12,5m\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>A)- d(F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>A)=12,5 -10=2,5m =\u0394X<\/b><\/span><\/span><\/span>i<\/b><\/span><\/span><\/sub><\/span><\/p>\n

Na situa\u00e7\u00e3o de interfer\u00eancia destrutiva com a fonte F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0mais afastada, a dist\u00e2ncia entre as mesmas ser\u00e1\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span><\/p>\n

A equa\u00e7\u00e3o da interfer\u00eancia destrutiva \u00e9\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(2n+1).\u03bb\/2. Quando n=0, a solu\u00e7\u00e3o n\u00e3o satisfaz, pois nesse caso,\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=1m e como F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0deve ser afastada,\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0que \u00e9 a dist\u00e2ncia entre as fontes, deve ser maior que 2,5m.<\/b><\/span><\/span><\/span><\/p>\n

Para n=1\u00a0 —\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=(2n+1).l\/2.\u00a0 —\u00a0\u00a0\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0= (2.1+1).l\/2\u00a0 —\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>\u00a0= (2.1+1).2\/2\u00a0 —\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=3m (L<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

b) Ao atingirem B, as ondas provenientes de F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0e F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0devem sofrer a\u00ed, interfer\u00eancia construtiva (m\u00e1ximos de intensidade) e, nesse caso,\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2n.\u03bb\/2. A condi\u00e7\u00e3o n=0 n\u00e3o satisfaz, pois, nesse caso\u00a0DX<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=0 e as fontes estaria coincidentes e sabemos que elas est\u00e3o distanciadas de mais de 2,5m.<\/b><\/span><\/span><\/span><\/p>\n

d(F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>B)=10m\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>B)=?<\/b><\/span><\/span><\/span><\/p>\n

Para n=1\u00a0 —\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2n.l\/2\u00a0 —\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2.1.2\/2\u00a0 —\u00a0\u00a0\u0394X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>=2m<\/b><\/span><\/span><\/span><\/p>\n

\u0394<\/b><\/span><\/span><\/span><\/span>X<\/b><\/span><\/span><\/span>d<\/b><\/span><\/span><\/sub><\/span>= d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>B) – d(F<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>B)\u00a0 —\u00a0 2= d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>B) \u2013 10\u00a0 —\u00a0 d(F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>B)=12m<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pit\u00e1goras\u00a0 —\u00a0 (10)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ L<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>=(12)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0L<\/b><\/span><\/span><\/span>b<\/b><\/span><\/span><\/sub><\/span>=6,6m<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span> <\/b>R- D\u00a0\u00a0—\u00a0 veja teoria<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>a) N\u00e3o\u00a0 —\u00a0 a interfer\u00eancia \u00e9 um fen\u00f4meno que descreve a soma das amplitudes das onda e, ap\u00f3s a mesma, elas continuam seu movimento como se nada tivesse acontecido..<\/b><\/span><\/span><\/span><\/p>\n

b) N\u00e3o\u00a0 —\u00a0 a energia de uma onda est\u00e1 relacionada \u00e0 pot\u00eancia do gerador que a fez oscilar.<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span><\/p>\n

Observe a figura abaixo onde est\u00e1 representado o instante\u00a0 em que as ondas est\u00e3o totalmente sobrepostas\u00a0 —\u00a0 haver\u00e1 interfer\u00eancia construtiva na metade esquerda do intervalo e destrutiva na metade direita do mesmo intervalo\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

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

01- Como as ondas emitidas pelos dois alto-falantes est\u00e3o em fase elas possuem a mesma freq\u00fc\u00eancia (f), o mesmo comprimento de onda (\u03bb), e mesma amplitude (A) e em fase (ambas para cima ou ambas para baixo)\u00a0 —\u00a0 nos pontos O e M as ondas est\u00e3o em concord\u00e2ncia de fase (som de intensidade m\u00e1xima, refor\u00e7o)\u00a0 —\u00a0 para o ponto M, onde as ondas est\u00e3o em concord\u00e2ncia de fase (interfer\u00eancia construtiva),\u00a0 a dist\u00e2ncia de uma fonte at\u00e9 M (d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10m) menos a dist\u00e2ncia da outra\u00a0 fonte\u00a0at\u00e9 M (d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8m), deve valer um comprimento de onda 1\u03bb\u00a0 —\u00a0 d<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0\u2013 d<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1\u03bb\u00a0 —\u00a0 10 \u2013 8=1\u03bb\u00a0 —\u00a0\u00a0\u03bb=2,0m.<\/b><\/span><\/span><\/span><\/p>\n

02- Mais pr\u00f3xima\u00a0 —\u00a0 como V=\u03bb.f e como V \u00e9 constante, o comprimento de onda \u03bb e a frequ\u00eancia f s\u00e3o inversamente proporcionais\u00a0 —\u00a0 assim, como entre O e M existe um \u03bb, se a frequ\u00eancia aumenta, esse \u03bb diminui e o ponto M ficar\u00e1 mais perto do ponto O.<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span><\/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 Interfer\u00eancia de Ondas \u00a0 \u00a001- R: E.\u00a0 Como o pulso A retornar\u00e1 invertido (extremidade fixa) ao se superpuserem os dois pulsos sofrer\u00e3o interfer\u00eancia destrutiva e em seguida, cada pulso seguir\u00e1 se caminho mantendo suas caracter\u00edsticas originais 02- Nulas; pois no instante da interfer\u00eancia destrutiva n\u00e3o h\u00e1 movimento dos pontos participantes. 03- B\u00a0\u00a0\u00a0\u00a0 04-<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2419,"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-2423","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2423","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=2423"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2423\/revisions"}],"predecessor-version":[{"id":10917,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2423\/revisions\/10917"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2419"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2423"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}