{"id":2360,"date":"2016-08-01T23:50:36","date_gmt":"2016-08-01T23:50:36","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2360"},"modified":"2024-08-23T14:48:07","modified_gmt":"2024-08-23T14:48:07","slug":"resolucao-comentada-das-questoes-de-vestibulares-sobre-efeito-doppler","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/ondulatoria\/acustica\/efeito-doppler\/resolucao-comentada-das-questoes-de-vestibulares-sobre-efeito-doppler\/","title":{"rendered":"Efeito Doppler – Resolu\u00e7\u00e3o"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de vestibulares sobre <\/b><\/span><\/span><\/span><\/p>\n

Efeito Doppler<\/b><\/span><\/span><\/span><\/p>\n

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

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

I \u2013 Errada\u00a0 —\u00a0 mais agudo\u00a0 —\u00a0 ambul\u00e2ncia se aproximando\u00a0 —\u00a0 recebe mais frentes de onda.<\/b><\/span><\/span><\/span><\/p>\n

II \u2013 Correta\u00a0 —\u00a0 a dist\u00e2ncia entre o observador e a fonte \u00e9 sempre a mesma (raio da circunfer\u00eancia).<\/b><\/span><\/span><\/span><\/p>\n

III \u2013 Errada\u00a0 —\u00a0 mais grave\u00a0 —\u00a0 ambul\u00e2ncia se afastando\u00a0 —\u00a0 recebe menos frentes de onda.<\/b><\/span><\/span><\/span><\/p>\n

IV \u2013 Errada\u00a0 —\u00a0 vide I, II e III acima.<\/b><\/span><\/span><\/span><\/p>\n

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

02-<\/b><\/span><\/span><\/span> <\/b>(1) Falsa\u00a0 —\u00a0 a freq\u00fc\u00eancia da fonte \u00e9 a mesma<\/b><\/span><\/span><\/span><\/p>\n

(2) Falsa<\/b><\/span><\/span><\/span><\/p>\n

(3) Verdadeira\u00a0 —\u00a0 a dist\u00e2ncia fonte-observador \u00e9 a mesma<\/b><\/span><\/span><\/span><\/p>\n

(4) Falsa\u00a0 —\u00a0 vale tamb\u00e9m para a luz que n\u00e3o necessita de um meio material para se propagar.<\/b><\/span><\/span><\/span><\/p>\n

F F V F<\/b><\/span><\/span><\/span><\/p>\n

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

04-<\/b><\/span><\/span><\/span> <\/b>A\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span> <\/b>C\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>D\u00a0<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> <\/b>Apenas a componente horizontal de U, U<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>\u00a0influi no movimento, pois as frentes de onda s\u00e3o planas (vide figura)<\/b><\/span><\/span><\/span>\"\"<\/p>\n

U<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=Ucos60<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=6,6.0,5\u00a0 —\u00a0 U<\/b><\/span><\/span><\/span>x<\/b><\/span><\/span><\/sub><\/span>=3,3m\/s\u00a0 —\u00a0 c\u00e1lculo da freq\u00fc\u00eancia da fonte\u00a0 —\u00a0 V=lf\u00a0 —\u00a0 330=0,165f\u00a0 —\u00a0 f=2.000Hz<\/b><\/span><\/span><\/span><\/p>\n

f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=2000(330-3,3)\/330\u00a0 —\u00a0 f=653.400\/330\u00a0 —\u00a0\u00a0f=1.980Hz\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

a) Maior. \u00c0 medida que a pessoa se aproxima da fonte, ele observa um aumento do n\u00famero de frentes de onda passando por ele por unidade de tempo em rela\u00e7\u00e3o a situa\u00e7\u00e3o em que a pessoa se encontra parada, implicando num aumento da freq\u00fc\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

b) Se aproximando\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=735(343+V)\/343\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=735 + 2,1V<\/b><\/span><\/span><\/span><\/p>\n

\u00a0Se afastando\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=735(343-V)\/343\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=735-2,1V<\/b><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>\u0394f=f<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/span><\/sub><\/span>-f<\/b><\/span><\/span><\/span><\/span>2<\/b><\/span><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 78=735+2,1V \u2013 (735-2,1V)\u00a0 —\u00a0 78=735+2,1V -735 +2,1V\u00a0 —\u00a0V=18,6m\/s<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

– f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 a fonte se afasta do observador que est\u00e1 em repouso<\/b><\/span><\/span><\/span><\/p>\n

. f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=f.V\/(V+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=400.340\/(340+1,7)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=136.000\/341,7\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=398Hz<\/b><\/span><\/span><\/span><\/p>\n

– f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 som se aproxima do observador que est\u00e1 parado<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=V\/(V-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=400.340\/(340-1,7)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=136.000\/338,3\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=402Hz<\/b><\/span><\/span><\/span><\/p>\n

– a frequ\u00eancia de batimento, por defini\u00e7\u00e3o, \u00e9 fornecida por\u00a0f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>maior<\/b><\/span><\/span><\/sub><\/span>\u00a0– f<\/b><\/span><\/span><\/span>menor<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=402-398\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=4Hz\u00a0\u00a0 R- C
\n<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> \u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=?\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=436Hz\u00a0 —\u00a0 f=440Hz<\/b><\/span><\/span><\/span><\/p>\n

\u00a0C\u00e1lculo de V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0(velocidade com que o diapas\u00e3o chega ao solo)\u00a0 —\u00a0 Dopper\u00a0 — \u00a0\u00a0f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=f.V\/V+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 436=440.330\/(330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 436\/440=330\/330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 145.200=143.880 + 436V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u22483m\/s<\/b><\/span><\/span><\/span><\/p>\n

Queda livre\u00a0 —\u00a0 aplicando Torricelli com V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=0 e a=g=10m\/s<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.a.DS\u00a0 —\u00a0 3<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 2.10.h\u00a0 —\u00a0\u00a0h\u22480,45m<\/b><\/span><\/span><\/span><\/p>\n

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

11-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=80\/3,6 m\/s\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>=0\u00a0 —\u00a0\u00a0 f=700Hz\u00a0 —\u00a0 V=340m\/s\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=f(V+V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/V\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=700(340+80\/3,6)\/340\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=700(1 + 80\/3,6.340) — f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=700 + 56.000\/1.224\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>\u2248744Hz\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=125\/3,6=34,7m\/s\u00a0 —\u00a0 as freq\u00fc\u00eancias aparentes s\u00e3o iguais, ou seja, f<\/b><\/span><\/span><\/span>aA<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>aB<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

Ambul\u00e2ncia A \u2013 est\u00e1 se afastando do observador parado\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>aA<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\/(V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>+V<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>) (I)<\/b><\/span><\/span><\/span><\/p>\n

Ambul\u00e2ncia B \u2013 est\u00e1 se aproximando do observador parado\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>aB<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>..V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\/(V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>-V<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>) (II)<\/b><\/span><\/span><\/span><\/p>\n

Dividindo membro a membro I por II:<\/b><\/span><\/span><\/span><\/p>\n

F<\/b><\/span><\/span><\/span>aA<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>aB<\/b><\/span><\/span><\/sub><\/span>\u00a0= f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>.340\/340+34,7 X 340-34,7\/f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>.340\u00a0 —\u00a0 1=305,3.f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0\/ 374,7.f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=374,7\/305,3\u00a0 —\u00a0\u00a0f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/f<\/b><\/span><\/span><\/span>B<\/b><\/span><\/span><\/sub><\/span>=1,23<\/b><\/span><\/span><\/span><\/p>\n

13-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Autom\u00f3vel se aproximando do observador parado\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=f.V\/(V-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 284=f.330\/(330-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 284\/f=330\/330-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0\u00a0I<\/b><\/span><\/span><\/span><\/p>\n

Autom\u00f3vel se afastando do observador parado\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>a<\/b><\/span><\/span><\/sub><\/span>=f.V\/(V+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 266\/f=330\/330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00a0 II\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Dividindo I por II:<\/b><\/span><\/span><\/span><\/p>\n

284\/f X f\/266=330\/330-Vf X 330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/330\u00a0 —\u00a0 284\/266=330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\/330-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 266(330+V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>) = 284(330-V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)<\/b><\/span><\/span><\/span><\/p>\n

87.780 + 266V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0= 93.720-284V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 550V<\/b><\/span><\/span><\/span>f\u00a0<\/b><\/span><\/span><\/sub><\/span>= 5.940\u00a0 —\u00a0 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0= 10,8m\/s\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

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

<\/a> 15-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe\u00a0o esquema da figura abaixo\u00a0 —\u00a0 no instante t = 0 a jovem (ouvinte ou observador) est\u00e1 na posi\u00e7\u00e3o\u00a0mostrada na figura\u00a0 —\u00a0\u00a0 nesse ponto, a frequ\u00eancia ouvida (f) \u00e9 igual a frequ\u00eancia da sirene (f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) tal que f\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=1\u00a0 — A partir desse instante, at\u00e9T\/2, h\u00e1 afastamento relativo entre a jovem e a sirene, de modo que a frequ\u00eancia ouvida por ela \u00e9 menor que a frequ\u00eancia emitida pela sirene f\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span><1\u00a0 —\u00a0 quando t =T\/2, ela volta a ouvir a frequ\u00eancia emitida pela fonte\u00a0 —\u00a0 De T\/2 at\u00e9 T, ocorre aproxima\u00e7\u00e3o relativa entre a jovem e a sirene e ela passa a perceber uma frequ\u00eancia maior que a emitida pela sirene f\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>>1\u00a0 —\u00a0 as velocidades m\u00e1ximas de afastamento e de aproxima\u00e7\u00e3o ocorrem em t = T\/4 e em t =3T\/4, instantes em f\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>atinge valores m\u00ednimo e m\u00e1ximo, respectivamente\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span>\"\"<\/p>\n

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

(01) Errada\u00a0 —\u00a0 a velocidade de propaga\u00e7\u00e3o de uma onda s\u00f3 depende da caracter\u00edstica da pr\u00f3pria onda e das condi\u00e7\u00f5es do meio.<\/b><\/span><\/span><\/span><\/p>\n

(02) Errada —\u00a0 \u00a0velocidade de propaga\u00e7\u00e3o da onda independe da velocidade da fonte.<\/b><\/span><\/span><\/span><\/p>\n

(04) Correta\u00a0 —\u00a0 quando a onda sonora se reflete na parede, esta funciona como fonte\u00a0 —\u00a0 como o morcego est\u00e1 se aproximando da fonte, ocorre o efeito Doppler e ele detecta um som mais agudo que o emitido, ou seja, de maior frequ\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

(08) Correta\u00a0 — \u00a0o meio \u00e9 o mesmo.<\/b><\/span><\/span><\/span><\/p>\n

(16) Errada\u00a0 —\u00a0 como j\u00e1 especificado, chama-se efeito Doppler.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\u00a0R- (04 + 08)=12<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>a) Dados\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\u00a0=\u00a0v<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>= 340 m\/s\u00a0 —\u00a0\u00a0\u00a0\u03bb\u00a0= 50 cm = 0,5 m\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0 —\u00a0 v= \u03bbf\u00a0 —\u00a0 340=0,5f\u00a0 —\u00a0\u00a0f=680Hz<\/b><\/span><\/span><\/span><\/p>\n

b e c) Dados\u00a0 —\u00a0\u00a0 v<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>\u00a0= 0; v<\/b><\/span><\/span><\/span>ouvinte<\/b><\/span><\/span><\/sub><\/span>\u00a0= 72 km\/h = 20 m\/s\u00a0 —\u00a0 a frequ\u00eancia aparente (f<\/b><\/span><\/span><\/span>ap<\/b><\/span><\/span><\/sub><\/span>) percebida pelo motorista da ambul\u00e2ncia (ouvinte) \u00e9 dada pela express\u00e3o do efeito Doppler\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>ap<\/b><\/span><\/span><\/sub><\/span>={(v<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\u00a0+ v<\/b><\/span><\/span><\/span>ouvinte<\/b><\/span><\/span><\/sub><\/span>)\/(V<\/b><\/span><\/span><\/span>som<\/b><\/span><\/span><\/sub><\/span>\u00a0+ V<\/b><\/span><\/span><\/span>fonte<\/b><\/span><\/span><\/sub><\/span>) }.f\u00a0 —\u00a0 colocando os valores\u00a0 —\u00a0f<\/b><\/span><\/span><\/span>ap<\/b><\/span><\/span><\/sub><\/span>=(340 + 20)\/(340 + 0).680\u00a0 —\u00a0\u00a0f<\/b><\/span><\/span><\/span>ap<\/b><\/span><\/span><\/sub><\/span>=720Hz\u00a0 —\u00a0 esse valor significa que o motorista recebe 720 frentes de onda por segundo\u00a0 — em tr\u00eas segundos, a quantidade de frentes de ondas (N) recebidas \u00e9\u00a0 —\u00a0 N = 3<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/sub><\/span>(720)\u00a0 —\u00a0\u00a0N = 2.160.<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>A quest\u00e3o refere-se ao efeito Doppler, que d\u00e1 a frequ\u00eancia aparente percebida por um ouvinte quando entre ele e a fonte ocorre movimento relativo. Quando h\u00e1 aproxima\u00e7\u00e3o relativa, a frequ\u00eancia aparente \u00e9 maior que a emitida pela fonte e, menor, quando h\u00e1 afastamento relativo.<\/b><\/span><\/span><\/span><\/p>\n

a) Errada. A velocidade do som num fluido depende da velocidade desse fluido. A frequ\u00eancia aparente para o golfinho \u00e9 maior que 9,74 kHz, como se a baleia se movesse com a mesma velocidade da correnteza em \u00e1guas paradas.<\/b><\/span><\/span><\/span><\/p>\n

b) Errada. Nada se pode afirmar, pois n\u00e3o foi fornecida a velocidade do som na \u00e1gua.<\/b><\/span><\/span><\/span><\/p>\n

c) Correta. Haveria aproxima\u00e7\u00e3o relativa entre o golfinho e a baleia. Assim, a frequ\u00eancia percebida pelo golfinho seria maior que a frequ\u00eancia emitida pela baleia.<\/b><\/span><\/span><\/span><\/p>\n

d) Errada. O golfinho emitir\u00e1 som na mesma frequ\u00eancia recebida.\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>=72km\/h=20m\/s\u00a0 —\u00a0 conforme o enunciado, quando o carro se aproxima do radar, a frequ\u00eancia da onda refletida<\/b><\/span><\/span><\/span><\/p>\n

(f) \u00e9 maior que a da onda emitida (f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 o que implica que a diferen\u00e7a \u2206f = f \u2013 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0deve ser maior que zero\u00a0 —\u00a0 para isso, voc\u00ea deve escolher o valor positivo para a velocidade m\u00e9dia V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\u00a0 na equa\u00e7\u00e3o \u2206f = f \u2013 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0\u00b1(V<\/b><\/span><\/span><\/span>m<\/b><\/span><\/span><\/sub><\/span>\/c).f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 substituindo\u00a0 —\u00a0\u00a0 \u2206f=+ (20\/3.10<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sup><\/span>).2,4.10<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0 \u2206f= + 1.600 Hz\u00a0 —\u00a0\u00a0R- A<\/b><\/span><\/span><\/span><\/p>\n

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Equa\u00e7\u00e3o do efeito Doppler\u00a0 —\u00a0 f\/f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=(V \u2013 V<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\/(V \u2013 V<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 225\/235=(340 \u2013 0)\/(340 \u2013 V)\u00a0 —\u00a0 V=-15,1m\/s\u00a0 —\u00a0 supondo a velocidade da pessoa nula quando a mola estiver esticada em 20m\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0= E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 80.10.20=k.4<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 k=2.000N\/m\u00a0 —\u00a0 condi\u00e7\u00e3o em que a corda n\u00e3o est\u00e1 esticada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 10h=(-15,1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 h=11,4m\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Condi\u00e7\u00e3o em que a corda se encontra tracionada\u00a0 —\u00a0 E<\/b><\/span><\/span><\/span>mi<\/b><\/span><\/span><\/sub><\/span>=E<\/b><\/span><\/span><\/span>mf<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 mgh\u2019=mV<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + kx<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 80.10.(16 + x)=80.(-15,1)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2 + 2.000.x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2\u00a0 —\u00a0 x<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2,4m\u00a0 —\u00a0 x<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=-1,6m (n\u00e3o conv\u00e9m)\u00a0 —\u00a0 dist\u00e2ncia pedida\u00a0 —\u00a0 h\u2019=16 + 2,4\u00a0 —\u00a0 h\u2019=18,4m<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Como o carro de pol\u00edcia que emite som de frequ\u00eancia f se aproxima do observador que est\u00e1 parado, este perceber\u00e1<\/b><\/span><\/span><\/span><\/p>\n

um som de frequ\u00eancia maior que a real emitida pelo carro\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

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

22-<\/b><\/span><\/span><\/span>\u00a0Voc\u00ea deve sempre seguir a orienta\u00e7\u00e3o de regras de sinais indicadas abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

0. Correta\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0negativa (observador se afasta da fonte, suposta parada)\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0negativa\u00a0 —\u00a0 a fonte se aproxima do observador, suposto parado)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(u \u00b1v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = f<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/(u\u00b1v<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(u – v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = f<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/(u – v<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(340 \u2013 30) = 500\/(340 \u2013 60)\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/310 = 500\/280\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=155000\/280\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=553,57Hz.<\/b><\/span><\/span><\/span><\/p>\n

1. Falsa\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0negativa (observador se afasta da fonte, suposta parada)\u00a0 —\u00a0 v<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/sub><\/span>\u00a0negativa\u00a0 —\u00a0 a fonte se aproxima do observador, suposto parado)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(u \u00b1v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = f<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/(u\u00b1v<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(u – v<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) = f<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/(u – v<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/(340 \u2013 30) = 500\/(340 \u2013 60)\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/310 = 500\/280\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=155000\/280\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=553,57Hz. (observe que \u00e9 a mesma situa\u00e7\u00e3o que a anterior, apenas inverteram-se os sentidos e, \u00e9 claro que a frequ\u00eancia percebida pelo observador deve ser a mesma).)<\/b><\/span><\/span><\/span><\/p>\n

23-<\/b><\/span><\/span><\/span> <\/b>Esse fen\u00f4meno que provoca altera\u00e7\u00f5es na frequ\u00eancia das ondas chama-se efeito Doppler\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0 —\u00a0 v=\u03bb.F\u00a0 —\u00a0 1540 = \u03bb.2,2.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 \u03bb = 1540\/2,2.10<\/b><\/span><\/span><\/span>6<\/b><\/span><\/span><\/sup><\/span>=700.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 \u03bb=7.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>.10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>mm\u00a0 —\u00a0 \u03bb=0,7mm<\/b><\/span><\/span><\/span><\/p>\n

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

24-<\/b><\/span><\/span><\/span> <\/b>O efeito Doppler refere-se \u00e0 varia\u00e7\u00e3o da freq\u00fc\u00eancia notada por um observador quando a dist\u00e2ncia entre ele e uma fonte de ondas est\u00e1 aumentando ou diminuindo.<\/b><\/span><\/span><\/span><\/p>\n

Na aproxima\u00e7\u00e3o entre fonte e observador, o mesmo perceber\u00e1 o som emitido pela fonte mais agudo (maior freq\u00fc\u00eancia, recebe maior n\u00famero de frentes de onda na unidade de tempo) do que perceberia se fonte e observador estivessem parados. Nesse caso, o comprimento de onda aparente percebido pelo observador ser\u00e1 menor que o comprimento da onda emitido pela fonte (observador O1\u00a0da figura abaixo).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

No afastamento entre fonte e observador, o mesmo perceber\u00e1 o som emitido pela fonte mais grave (menor freq\u00fc\u00eancia, recebe menor n\u00famero de frentes de onda na unidade de tempo) do que perceberia se fonte e observador estivessem parados. Nesse caso, o comprimento de onda aparente percebido pelo observador ser\u00e1 maior que o comprimento da onda emitido pela fonte (observador O2\u00a0da figura ao lado).<\/b><\/span><\/span><\/span><\/p>\n

Observe que o motorista da ambul\u00e2ncia n\u00e3o percebe nenhuma altera\u00e7\u00e3o no som emitido pela sirene, pois eles se movem juntos.<\/b><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Resolu\u00e7\u00e3o comentada das quest\u00f5es de vestibulares sobre Efeito Doppler \u00a0 01- I \u2013 Errada\u00a0 —\u00a0 mais agudo\u00a0 —\u00a0 ambul\u00e2ncia se aproximando\u00a0 —\u00a0 recebe mais frentes de onda. II \u2013 Correta\u00a0 —\u00a0 a dist\u00e2ncia entre o observador e a fonte \u00e9 sempre a mesma (raio da circunfer\u00eancia). III \u2013 Errada\u00a0 —\u00a0 mais grave\u00a0 —\u00a0 ambul\u00e2ncia se afastando\u00a0 —\u00a0 recebe menos frentes<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2356,"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-2360","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2360","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=2360"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2360\/revisions"}],"predecessor-version":[{"id":10902,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2360\/revisions\/10902"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2356"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2360"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}