{"id":3372,"date":"2017-11-05T22:58:06","date_gmt":"2017-11-05T22:58:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=3372"},"modified":"2024-09-04T15:01:41","modified_gmt":"2024-09-04T15:01:41","slug":"resolucao-comentada-ondulatoria-2017-2016-2015","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/vestibulares-recentes\/ondulatoria\/vestibulares-recentes-ondulatoria-2017-2016-2015\/resolucao-comentada-ondulatoria-2017-2016-2015\/","title":{"rendered":"Resolu\u00e7\u00e3o comentada Ondulat\u00f3ria \u2013 2017- 2016- 2015"},"content":{"rendered":"

Resolu\u00e7\u00e3o comentada das quest\u00f5es de F\u00cdSICA de Vestibulares recentes sobre <\/span><\/span><\/span><\/p>\n

Ondulat\u00f3ria \u2013 2017- 2016- 2015<\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

\"\"Os\u00a0<\/b><\/span><\/span>sonares emitem e recebem ondas sonoras<\/b><\/span><\/span>que possuem as<\/b><\/span><\/span>seguintes carcter\u00edsticas:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"S\u00e3o\u00a0<\/b><\/span><\/span>ondas mec\u00e2nicas\u00a0<\/b><\/span><\/span>que<\/b><\/span><\/span>necessitam<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>de um\u00a0<\/b><\/span><\/span>meio material\u00a0para se propagar, transportando energia.\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Exemplos: Ondas no mar\u00a0<\/b><\/span><\/span>(o meio \u00e9 a \u00e1gua),\u00a0<\/b><\/span><\/span>ondas sonoras\u00a0<\/b><\/span><\/span>(o meio pode ser ar, \u00e1gua, etc.),<\/b><\/span><\/span>ondas em uma corda\u00a0<\/b><\/span><\/span>(o meio \u00e9 a corda),\u00a0<\/b><\/span><\/span>etc.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Todas\u00a0<\/b><\/span><\/span>essas perturba\u00e7\u00f5es\u00a0<\/b><\/span><\/span>s\u00e3o\u00a0<\/b><\/span><\/span>causadas em\u00a0meios materiais.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"S\u00e3o ondas longitudinais<\/b><\/span><\/span>nas quais\u00a0a\u00a0<\/b><\/span><\/span>dire\u00e7\u00e3o de propaga\u00e7\u00e3o coincide com a dire\u00e7\u00e3o de oscila\u00e7\u00e3o.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Exemplos:\u00a0<\/b><\/span><\/span>ondas<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>sonoras,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>ondas<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>numa mola,<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>etc.<\/b><\/span><\/span>\u00a0(<\/b><\/span><\/span>a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>membrana do alto falante vibra no eixo x<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>e as\u00a0<\/b><\/span><\/span>ondas seguem essa mesma dire\u00e7\u00e3o).<\/b><\/span><\/span>
\n<\/b><\/span><\/span>\u00a0\"\"\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"A\u00a0<\/b><\/span><\/span>frequ\u00eancia\u00a0<\/b><\/span><\/span>\u00e9\u00a0<\/b><\/span><\/span>caracter\u00edstica da fonte que est\u00e1 emitindo a onda\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>\u00a0independe do meio onde a onda sonora ou eletromagn\u00e9tica estiver se propagando.\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim,\u00a0<\/b><\/span><\/span>por exemplo,\u00a0<\/b><\/span><\/span>se uma\u00a0<\/b><\/span><\/span>fonte\u00a0<\/b><\/span><\/span>emitir uma\u00a0<\/b><\/span><\/span>onda (sonora ou luminosa) <\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>frequ\u00eancia\u00a0f <\/b><\/span><\/span>num\u00a0<\/b><\/span><\/span>dado meio,<\/b><\/span><\/span>\u00a0em\u00a0<\/b><\/span><\/span>qualquer outro meio\u00a0<\/b><\/span><\/span>que esta onda estiver\u00a0<\/b><\/span><\/span>se propagando ela\u00a0<\/b><\/span><\/span>ter\u00e1\u00a0<\/b><\/span><\/span>sempre essa mesma\u00a0frequ\u00eancia\u00a0f.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

VP\u00a0=\u00a0\"\"\u00a0\"\"<\/b><\/span><\/span>\u00a08 =\u00a0\"\"\u00a0\"\"\u00a0t\u2019 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>t\u2019 = 150 s<\/b><\/span><\/span><\/p>\n

VS\u00a0=\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a05 =\u00a0\"\"\u00a0\"\"\u00a0t\u2019\u2019 =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>t\u2019\u2019 = 240 s<\/b><\/span><\/span><\/p>\n

\u2206<\/span>t =\u00a0<\/b><\/span><\/span>240 \u2013 150\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t = 90 s<\/b><\/span><\/span><\/p>\n

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

B)\u00a0<\/b><\/span><\/span>O\u00a0<\/b><\/span><\/span>texto afirma\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>cada lado do quadrado grande equivale a 0,20 s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No\u00a0<\/b><\/span><\/span>ECG do paciente\u00a0<\/b><\/span><\/span>voc\u00ea observa que,\u00a0<\/b><\/span><\/span>entre dois picos sucessivos<\/b><\/span><\/span>existem\u00a0<\/b><\/span><\/span>5 quadrados grandes,\u00a0<\/b><\/span><\/span>o que corresponde a\u00a0<\/b><\/span><\/span>um per\u00edodo T =<\/b><\/span><\/span>5\u00d70,20s\u00a0<\/b><\/span><\/span>= 1,0 s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A frequ\u00eancia f \u00e9 o inverso do per\u00edodo T\u00a0\"\"\u00a0f = 1\/T\u00a0<\/b><\/span><\/span>=<\/b><\/span><\/span> 1\/1\"\"<\/b><\/span><\/span>\u00a0f = 1,0 Hz\u00a0<\/b><\/span><\/span>= 1\u00d760<\/b><\/span><\/span>= 60 bpm.<\/b><\/span><\/span><\/span><\/span><\/p>\n

04- <\/span><\/span><\/span>As ondas eletromagn\u00e9ticas\u00a0<\/b><\/span><\/span>possuem a caracter\u00edstica de n\u00e3o dependerem de um meio para se propagarem e se propagam de maneira transversal ao seu modo de vibra\u00e7\u00e3o<\/b><\/span><\/span>. Sendo assim a\u00a0<\/b><\/span><\/span>alternativa B est\u00e1 errada e a A correta<\/b><\/span><\/span>, mas vamos analisar os erros das outras tamb\u00e9m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A velocidade de propaga\u00e7\u00e3o de uma onda \u00e9 da ordem de\u00a0<\/b><\/span><\/span>299.792.458 m\/s<\/b><\/span><\/span>, enquanto que a do som \u00e9 de\u00a0<\/b><\/span><\/span>340 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Marte\u00a0<\/b><\/span><\/span>n\u00e3o emite radia\u00e7\u00e3o danosa<\/b><\/span><\/span>, ele apenas\u00a0<\/b><\/span><\/span>absorve e emite radia\u00e7\u00e3o assim como a Terra<\/b><\/span><\/span>, essa troca \u00e9\u00a0<\/b><\/span><\/span>respons\u00e1vel pela manuten\u00e7\u00e3o da temperatura\u00a0<\/b><\/span><\/span>em ambos os planetas, por exemplo.<\/b><\/span><\/span><\/span><\/span>\"\"<\/p>\n

O espectro eletromagn\u00e9tico \u00e9\u00a0<\/b><\/span><\/span>muito maior que apenas as faixas de ultravioleta e infravermelho<\/b><\/span><\/span>, tanto \u00e9 que ele engloba essas duas.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

05- <\/span><\/span><\/span>Observe<\/b><\/span><\/span>\u00a0na\u00a0<\/b><\/span><\/span>figura 1<\/b><\/span><\/span>\u00a0que, quando as dist\u00e2ncias\u00a0<\/b><\/span><\/span>do auto falante ao microfone est\u00e3o aumentando de 3 mm at\u00e9 60 mm<\/b><\/span><\/span>\u00a0a\u00a0<\/b><\/span><\/span>intensidade da onda sonora em decib\u00e9is (dB) est\u00e1 diminuindo<\/b><\/span><\/span>\u00a0e,\u00a0<\/b><\/span><\/span>durante<\/b><\/span><\/span>\u00a0essa\u00a0<\/b><\/span><\/span>diminui\u00e7\u00e3o a frequ\u00eancia<\/b><\/span><\/span>\u00a0(eixo horizontal) est\u00e1\u00a0<\/b><\/span><\/span>aumentando de aproximadamente 50 Hz at\u00e9 250 Hz.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Na\u00a0<\/b><\/span><\/span>figura 2<\/b><\/span><\/span>\u00a0voc\u00ea observa que essa\u00a0<\/b><\/span><\/span>diminui\u00e7\u00e3o dos decib\u00e9is ocorre na faixa das freq\u00fc\u00eancias graves<\/b><\/span><\/span>.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

Sendo\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0a dist\u00e2ncia do celular \u00e0 primeira torre T<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span>o celular pode estar em\u00a0<\/b><\/span><\/span>qualquer ponto da circunfer\u00eancia de centro na primeira torre T<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0e raio\u00a0R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0(figura I).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Sendo\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0a dist\u00e2ncia do celular \u00e0 segunda torre T<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>, o\u00a0<\/b><\/span><\/span>celular pode estar num dos dois pontos\u00a0<\/b><\/span><\/span>em que\u00a0<\/b><\/span><\/span>as circunfer\u00eancias se interceptam (figura II).<\/b><\/span><\/span><\/p>\n

Sendo\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>3<\/b><\/span><\/sub><\/span>\u00a0a dist\u00e2ncia do celular \u00e0 terceira torre\u00a0T<\/b><\/span><\/span>3<\/b><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span>a <\/b><\/span><\/span>intersec\u00e7\u00e3o das tr\u00eas circunfer\u00eancias de raios R<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>,\u00a0<\/b><\/span><\/span>R<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>\u00a0e R<\/b><\/span><\/span>3<\/b><\/span><\/sub><\/span>\u00a0 <\/b><\/span><\/span>ocorre\u00a0<\/b><\/span><\/span>num \u00fanico ponto onde deve se localizar o celular (figura III),<\/b><\/span><\/span>\u00a0conforme pedido no exerc\u00edcio.<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

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

\"\"<\/p>\n

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

\"\"\u00a0S\u00e3o dados\u00a0\"\"\u00a0<\/b><\/span><\/span>velocidade de propaga\u00e7\u00e3o das ondas de ultrassom nos\u00a0<\/b><\/span><\/span>tecidos humanos\u00a0<\/b><\/span><\/span>V = 1540 m\/s\u00a0\"\"<\/b><\/span><\/span>\u00a0essa\u00a0<\/b><\/span><\/span>dimens\u00e3o\u00a0<\/b><\/span><\/span>da estrutura a ser percorrida\u00a0<\/b><\/span><\/span>d =<\/b><\/span><\/span>\u00a01,5 mm\u00a0<\/b><\/span><\/span>= 1,5.10-3\u00a0m<\/b><\/span><\/span>,\u00a0<\/b><\/span><\/span>corresponde<\/b><\/span><\/span>\u00a0ao\u00a0<\/b><\/span><\/span>comprimento de onda do ultrassom emitido pelo transdutor\u00a0<\/b><\/span><\/span>(pesquisei).<\/b><\/span><\/span><\/p>\n

Utilizando a\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o fundamental da ondulat\u00f3ria<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>V =\u00a0.f\u00a0\"\"<\/b><\/span><\/span>\u00a01540 =<\/b><\/span><\/span>1,5.10-3.f\u00a0\"\"\u00a0<\/b><\/span><\/span>f =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>f =<\/b><\/span><\/span>\u00a01026.103<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 1,026.106\u00a0Hz = <\/b><\/span><\/span>1,026 MHz.<\/b><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

10- <\/b><\/span><\/span><\/span>Observe na\u00a0<\/b><\/span><\/span>figura que o comprimento de onda (<\/b><\/span><\/span>),\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>corresponde \u00e0 dist\u00e2ncia entre dois pontos<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

consecutivos em\u00a0<\/b><\/span><\/span>concord\u00e2ncia de fase, como por exemplo, duas cristas, vale\u00a0\u00a0= 8 cm = 0,08 m.<\/b><\/span><\/span><\/p>\n

Aplicando a\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o fundamental da ondulat\u00f3ria V =\u00a0<\/b><\/span><\/span>.f\u00a0\"\"<\/b><\/span><\/span>\u00a00,48 = 0,08.f\u00a0\"\"f =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>f = 6 Hz<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

11- <\/b><\/span><\/span><\/span>Observe na figura abaixo que a<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>amplitude vale A = 1,2 cm<\/b><\/span><\/span>\u00a0e o <\/b><\/span><\/span>comprimento de onda, \u03bb = 2 cm.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

12- <\/b><\/span><\/span><\/span>C\u00e1lculo dos\u00a0<\/b><\/span><\/span>comprimentos de onda das frequ\u00eancias extremas da faixa UV-B.
\nV = c = \u03bb.f<\/b><\/span><\/span>\u00a0\"\"\u00a0 \u03bb\u00a0=\u00a0\"\"\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\u03bbmin\u00a0=<\/b><\/span><\/span>\u00a03,0.108\/1,03.1015\u00a0\u00a0\"\"\u00a0\u00a0\u03bbmin\u00a0<\/b><\/span><\/span>\u2245<\/b><\/span><\/span><\/span>\u00a02,9.10-7\u00a0m\u00a0<\/b><\/span><\/span>= 290 nm.
\n\u03bbm\u00e1x\u00a0=<\/b><\/span><\/span>\u00a03,0.108\/9,34.1014\u00a0\u00a0\"\"\u00a0\u00a0\u03bbm\u00e1x\u00a0<\/b><\/span><\/span>\u2245<\/b><\/span><\/span><\/span>\u00a03,2.10-7\u00a0m\u00a0<\/b><\/span><\/span>= 320 nm.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Entre\u00a0<\/b><\/span><\/span>290 nm e 320 nm,<\/b><\/span><\/span>\u00a0o filtro que apresenta\u00a0<\/b><\/span><\/span>absor\u00e7\u00e3o m\u00e1xima \u00e9 o filtro solar IV.
\n<\/b><\/span><\/span>R- B<\/b><\/span><\/span><\/span><\/p>\n

 <\/p>\n

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

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

 <\/p>\n

14- <\/b><\/span><\/span><\/span>Vamos usar a\u00a0<\/b><\/span><\/span>equa\u00e7\u00e3o fundamental da ondulat\u00f3ria para esse exerc\u00edcio<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Onde:<\/b><\/span><\/span><\/span><\/span><\/p>\n

V \u00e9 a velocidade de propaga\u00e7\u00e3o<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u03bb \u00e9 o comprimento de onda<\/b><\/span><\/span><\/span><\/span><\/p>\n

f \u00e9 a frequ\u00eancia<\/b><\/span><\/span><\/span><\/span><\/p>\n

Isolando o comprimento de onda:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Substituindo,\u00a0<\/b><\/span><\/span>n\u00e3o podemos esquecer que M \u00e9 de mega que \u00e9 o equivalente a 106<\/b><\/span><\/span>:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

R \u2013 D<\/b><\/span><\/span><\/span><\/span><\/span><\/p>\n

15- <\/b><\/span><\/span><\/span>Observe na\u00a0<\/b><\/span><\/span>figura\u00a0<\/b><\/span><\/span>que o\u00a0<\/b><\/span><\/span>comprimento de onda\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span>(dist\u00e2ncia entre dois pontos consecutivos em<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

concord\u00e2ncia de fase, por exemplo, duas cristas)\u00a0<\/b><\/span><\/span>vale\u00a0\u00a0= 5,0 cm = 5.10-2m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>frequ\u00eancia da onda<\/b><\/span><\/span>, fornecida, vale\u00a0<\/b><\/span><\/span>f = 28 kHz = 28.103\u00a0Hz.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>V =\u00a0.f\u00a0<\/b><\/span><\/span>= 5.10-2.28.103\u00a0\"\"\u00a0V = 140.101\u00a0\"\"\u00a0<\/b><\/span><\/span>V = 1400 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>intervalo de tempo\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>\u2206t = 1,2 s\u00a0<\/b><\/span><\/span>corresponde \u00e0\u00a0<\/b><\/span><\/span>ida e volta da onda,\u00a0<\/b><\/span><\/span>logo, para ir e voltar ela\u00a0<\/b><\/span><\/span>percorreu a dist\u00e2ncia de 2d,\u00a0<\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>velocidade constante V = 1400 m\/s.<\/b><\/span><\/span><\/span><\/span><\/p>\n

V =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a01400 =\u00a0\"\"\u00a0\"\"\u00a0d =\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>d = 840 m.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

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

\"\"<\/p>\n

 <\/p>\n

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

\"\"<\/p>\n

Para esse exerc\u00edcio utilizaremos a equa\u00e7\u00e3o fundamental da ondulat\u00f3ria:<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>ar<\/b><\/span><\/sub><\/span> = <\/b><\/span><\/span>\u0001<\/b><\/span><\/span>ar<\/b><\/span><\/sub><\/span>.f<\/b><\/span><\/span>ar<\/b><\/span><\/sub><\/span><\/span><\/span><\/p>\n

Onde: V \u00e9 a velocidade de propaga\u00e7\u00e3o da onda<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0<\/span>\u00e9 o comprimento de onda<\/b><\/span><\/span><\/span><\/span><\/p>\n

f \u00e9 a sua frequ\u00eancia<\/b><\/span><\/span><\/span><\/span><\/p>\n

Primeiramente precisamos calcular a\u00a0<\/b><\/span><\/span>frequ\u00eancia da onda\u00a0<\/b><\/span><\/span>com os valores dispon\u00edveis, substituindo na equa\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

340 = 34.f<\/b><\/span><\/span>ar<\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/sub><\/span>isolando o f \"\"<\/b><\/span><\/span> <\/b>f = 10 Hz<\/b><\/span><\/span><\/span><\/p>\n

Agora que j\u00e1 temos a\u00a0<\/b><\/span><\/span>frequ\u00eancia, que ser\u00e1 igual pois a onda \u00e9 a mesma<\/b><\/span><\/span>, podemos usar equa\u00e7\u00e3o da onda na \u00e1gua:<\/b><\/span><\/span><\/span><\/span><\/p>\n

V<\/b><\/span><\/span>\u00e1gua <\/b><\/span><\/sub><\/span> = <\/b><\/span><\/span>\u0001<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/sub><\/span>.f<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/sub><\/span> \"\"<\/b><\/span><\/sub><\/span> 1360 = <\/b><\/span><\/span>\u0001<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/sub><\/span>.10 \"\"<\/b><\/span><\/span> <\/b>\u0001<\/b><\/span><\/span>\u00e1gua<\/b><\/span><\/sub><\/span> = 136 m<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

18- <\/b><\/span><\/span><\/span>Refra\u00e7\u00e3o<\/b><\/span><\/span>\u00a0\u00e9 um fen\u00f4meno que ocorre quando\u00a0<\/b><\/span><\/span>uma onda deixa um meio com certa caracter\u00edstica\u00a0<\/b><\/span><\/span>e entra em outro, com\u00a0<\/b><\/span><\/span>caracter\u00edstica diferente, variando sua velocidade.\u00a0<\/b><\/span><\/span>No caso da luz, a\u00a0<\/b><\/span><\/span>caracter\u00edstica do meio que importa \u00e9 o \u00edndice de refra\u00e7\u00e3o.\u00a0<\/b><\/span><\/span>E no caso de\u00a0<\/b><\/span><\/span>uma onda no mar<\/b><\/span><\/span>\u00a0pr\u00f3ximo \u00e0s praias a\u00a0<\/b><\/span><\/span>caracter\u00edstica do meio\u00a0<\/b><\/span><\/span>que importa \u00e9 a profundidade da \u00e1gua, fazendo\u00a0<\/b><\/span><\/span>variar a velocidade de propaga\u00e7\u00e3o.<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Ondas circulares em meio bi e tridimensionais<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Pontos\u00a0<\/b><\/span><\/span>onde se\u00a0<\/b><\/span><\/span>encontram uma crista e um vale (interfer\u00eancia destrutiva)<\/b><\/span><\/span>s\u00e3o chamados de\u00a0<\/b><\/span><\/span>n\u00f3s (linhas nodais).<\/b><\/span><\/span><\/p>\n

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

 <\/p>\n

20- <\/b><\/span><\/span><\/span>Observe atentamente<\/b><\/span><\/span>\u00a0nas figuras abaixo que a\u00a0<\/b><\/span><\/span>primeira vez que o perfil da corda ser\u00e1 como na figura 2\u00a0<\/b><\/span><\/span>ser\u00e1 quando houver\u00a0<\/b><\/span><\/span>interfer\u00eancia destrutiva do vale do pulso 1 (ponto A) com a crista do pulso 2 (ponto B) se anulando fornecendo o perfil de onda pedido e mostrado pela linha preta..\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim,\u00a0<\/b><\/span><\/span>cada ponto A ou B, X ou Y\u00a0<\/b><\/span><\/span>deslocaram-se de\u00a0\"\"<\/b><\/span><\/span>\u00a0= 60 cm em sentidos contr\u00e1rios, cada um com velocidade<\/b><\/span><\/span>\u00a0fornecida de\u00a0<\/b><\/span><\/span>V = 60 cm\/s\u00a0<\/b><\/span><\/span>tal que\u00a0<\/b><\/span><\/span>V =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a060 =\u00a0\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>\u2206t<\/b><\/span><\/span>\u00a0<\/b><\/span>=\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span>= 1 s<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

21- <\/b><\/span><\/span><\/span>A\u00a0<\/b><\/span><\/span>luz que atinge essa fenda F\u00a0<\/b><\/span><\/span>se\u00a0<\/b><\/span><\/span>espalha sofrendo difra\u00e7\u00e3o (fen\u00f4meno\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>permite<\/b><\/span><\/span>\u00a0com que uma\u00a0<\/b><\/span><\/span>onda atravesse fendas ou contorne obst\u00e1culos, atingindo regi\u00f5es\u00a0<\/b><\/span><\/span>onde, segundo a\u00a0<\/b><\/span><\/span>propaga\u00e7\u00e3o retil\u00ednea da luz, n\u00e3o conseguiria chegar).\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Atr\u00e1s<\/b><\/span><\/span>\u00a0da\u00a0<\/b><\/span><\/span>primeira tela,\u00a0<\/b><\/span><\/span>ele colocou\u00a0<\/b><\/span><\/span>outra tela opaca\u00a0<\/b><\/span><\/span>com\u00a0<\/b><\/span><\/span>duas\u00a0fendas muito estreitas e<\/b><\/span><\/span><\/span><\/span><\/p>\n

convenientemente pr\u00f3ximas (S1\u00a0e S2),\u00a0<\/b><\/span><\/span>sendo que\u00a0<\/b><\/span><\/span>cada uma delas funciona como uma fonte prim\u00e1ria de ondas exatamente iguais\u00a0<\/b><\/span><\/span>(mesma frequ\u00eancia, mesmo comprimento de onda, mesma velocidade e em fase), ou seja,\u00a0<\/b><\/span><\/span>ondas coerentes,\u00a0<\/b><\/span><\/span>que \u00e9\u00a0<\/b><\/span><\/span>condi\u00e7\u00e3o necess\u00e1ria para que ocorra a <\/b><\/span><\/span>interfer\u00eancia.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Ent\u00e3o ele\u00a0<\/b><\/span><\/span>observou no anteparo\u00a0<\/b><\/span><\/span>uma\u00a0<\/b><\/span><\/span>figura de interfer\u00eancia<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>formada por\u00a0<\/b><\/span><\/span>franjas brilhantes coloridas\u00a0<\/b><\/span><\/span>(<\/b><\/span><\/span>interfer\u00eancia construtiva)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>alternadas por franjas escuras\u00a0<\/b><\/span><\/span>(interfer\u00eancia destrutiva).<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Veja um resumo da teoria:<\/b><\/span><\/span><\/span><\/span><\/p>\n

O\u00a0<\/b><\/span><\/span>ponto P fica em refor\u00e7o\u00a0<\/b><\/span><\/span>(interfer\u00eancia construtiva), o que\u00a0<\/b><\/span><\/span>ocorre quando, no mesmo instante, duas cristas ou dois vales est\u00e3o se superpondo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo do\u00a0<\/b><\/span><\/span>comprimento de onda\u00a0\"\"<\/b><\/span><\/span>\u00a0dessas ondas que possuem<\/b><\/span><\/span>frequ\u00eancia de f = 4,0 Hz\u00a0<\/b><\/span><\/span>e se\u00a0<\/b><\/span><\/span>propagam<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>velocidade de V = 2,0 m\/s\u00a0\"\"V =\u00a0\"\"<\/b><\/span><\/span>.f\u00a0\"\"\u00a0<\/b><\/span><\/span>2 =\u00a0\"\"<\/b><\/span><\/span><\/span><\/span><\/p>\n

D2\u00a0= D\u00a0\"\"\u00a0d1\u00a0= 2, 8 m\u00a0\"\"\u00a0\"\"\"\"\"\"<\/b><\/span><\/span>para que D seja m\u00ednimo n = 1\u00a0\"\"d2\u00a0\u2013 d1\u00a0= (2n).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"\u00a0\"\"\u00a0<\/span>D \u2013 2,8 = (2.1).\"\"<\/b><\/span><\/span>\u00a0\"\"\u00a0<\/b><\/span><\/span>D = 2,8 + 2.0,25\u00a0\"\"<\/b><\/span><\/span>\u00a0D = 3,3 m<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Breve teoria:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No caso do exerc\u00edcio a\u00a0<\/b><\/span><\/span>manteiga derreter\u00e1 nas regi\u00f5es de interfer\u00eancia construtiva\u00a0<\/b><\/span><\/span>que corresponde \u00e0\u00a0<\/b><\/span><\/span>m\u00e1xima vibra\u00e7\u00e3o ou varia\u00e7\u00e3o de amplitude<\/b><\/span><\/span>(ventres ou fusos).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, a\u00a0<\/b><\/span><\/span>manteiga derrete\u00a0<\/b><\/span><\/span>nos pontos\u00a0<\/b><\/span><\/span>I, III e V.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, os\u00a0<\/b><\/span><\/span>pontos consecutivos ser\u00e3o ou I e III ou III e V.<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

 <\/p>\n

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

Resson\u00e2ncia<\/b><\/span><\/span><\/span><\/p>\n

Qualquer\u00a0<\/b><\/span><\/span>sistema f\u00edsico possui uma ou mais frequ\u00eancias naturais de vibra\u00e7\u00e3o<\/b><\/span><\/span>. Quando ele \u00e9\u00a0<\/b><\/span><\/span>\u201cexcitado\u201d por algum agente externo, agindo no ritmo de uma dessas frequ\u00eancias<\/b><\/span><\/span>, surge o fen\u00f4meno da\u00a0<\/b><\/span><\/span>resson\u00e2ncia<\/b><\/span><\/span>, ou seja, ele come\u00e7a a\u00a0<\/b><\/span><\/span>oscilar gradativamente at\u00e9 atingir uma dessas frequ\u00eancias, onde sua amplitude \u00e9 m\u00e1xima.<\/b><\/span><\/span><\/p>\n

Exemplos:<\/b><\/span><\/span><\/p>\n

–\u00a0<\/b><\/span><\/span>Uma crian\u00e7a,\u00a0<\/b><\/span><\/span>num balan\u00e7o<\/b><\/span><\/span>\u00a0j\u00e1 em movimento, sabe qual \u00e9 o momento certo, para que, com o<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

movimento do corpo provoque\u00a0um impulso, que faz aumentar a amplitude do movimento, obtendo assim, a resson\u00e2ncia. Trata-se de\u00a0<\/b><\/span><\/span>resson\u00e2ncia mec\u00e2nica.<\/b><\/span><\/span><\/p>\n

–\u00a0<\/b><\/span><\/span>Ao sintonizar uma\u00a0<\/b><\/span><\/span>emissora de r\u00e1dio ou TV<\/b><\/span><\/span>\u00a0estamos fazendo com que nosso\u00a0<\/b><\/span><\/span>aparelho receptor<\/b><\/span><\/span>\u00a0entre em\u00a0<\/b><\/span><\/span>resson\u00e2ncia com a mesma freq\u00fc\u00eancia que a das ondas eletromagn\u00e9ticas da esta\u00e7\u00e3o que as emitiu. Trata-se de <\/b><\/span><\/span>resson\u00e2ncia eletr\u00f4nica.<\/b><\/span><\/span><\/p>\n

–<\/b><\/span><\/span>\u00a0As\u00a0<\/b><\/span><\/span>vibra\u00e7\u00f5es da corda de um viol\u00e3o entram em resson\u00e2ncia\u00a0<\/b><\/span><\/span>com o ar contido em sua\u00a0 caixa de<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

madeira \u201ccaixa de resson\u00e2ncia\u201d ou com o a\u00a0<\/b><\/span><\/span>nota emitida por um diapas\u00e3o<\/b><\/span><\/span>, que vibram com a\u00a0<\/b><\/span><\/span>mesma frequ\u00eanci<\/b><\/span><\/span>a da nota emitida pelo diapas\u00e3o, afinando a corda do mesmo. Trata-se de\u00a0<\/b><\/span><\/span>resson\u00e2ncia sonora.<\/b><\/span><\/span><\/p>\n

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

25-<\/span><\/span><\/span><\/p>\n


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

\"\"<\/p>\n

S\u00e3o dados: f =\u00a0<\/b><\/span><\/span>640 Hz\u00a0\"\"\u00a0<\/b><\/span><\/span>Vo<\/b><\/span><\/span>\u00a0= 0\u00a0\"\"\u00a0<\/b><\/span><\/span>V =\u00a0<\/b><\/span><\/span>340 m\/s\u00a0\"\"\u00a0<\/b><\/span><\/span>Vf<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 20 m\/s\u00a0\"\"\u00a0<\/b><\/span><\/span>fa<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= ?.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>fonte<\/b><\/span><\/span>\u00a0se\u00a0<\/b><\/span><\/span>aproxima do observador\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>fa\u00a0= 640.(340 + 0)\/(340 \u2013 20)<\/b><\/span><\/span>\u00a0= 640.\u00a0340\/320\u00a0\"\" <\/b><\/span><\/span>fa<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 640.1,0625\u00a0<\/b><\/span><\/span>= <\/b><\/span><\/span>680 Hz.<\/b><\/span><\/span><\/span><\/p>\n

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

26- <\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Para o\u00a0<\/b><\/span><\/span>c\u00e1lculo das frequ\u00eancias percebidas pelo observador<\/b><\/span><\/span>\u00a0voc\u00ea deve utilizar o\u00a0<\/b><\/span><\/span>Efeito Doppler\u00a0<\/b><\/span><\/span>cuja\u00a0<\/b><\/span><\/span>express\u00e3o<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>regras de sinais<\/b><\/span><\/span>\u00a0est\u00e3o fornecidas a seguir:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\u00c9 pedida a\u00a0<\/b><\/span><\/span>raz\u00e3o f<\/b><\/span><\/span>B<\/b><\/span><\/sub><\/span>\u2019\/f<\/b><\/span><\/span>A<\/b><\/span><\/sub><\/span>\u2019 =\u00a0\"\"\u00a0= 1<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

27-<\/span><\/span><\/span><\/span><\/span><\/p>\n

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

Refere-se \u00e0\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o da frequ\u00eancia notada<\/b><\/span><\/span>\u00a0por um\u00a0<\/b><\/span><\/span>observador<\/b><\/span><\/span>\u00a0quando a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre ele e uma fonte de ondas est\u00e1 aumentando ou diminuindo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>aproxima\u00e7\u00e3o entre fonte e observador,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>mesmo perceber\u00e1 o som emitido pela fonte mais agudo\u00a0<\/b><\/span><\/span>(maior frequ\u00eancia, recebe maior n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Nesse caso, o\u00a0<\/b><\/span><\/span>comprimento de onda aparente percebido pelo observador ser\u00e1 menor\u00a0<\/b><\/span><\/span>que o <\/b><\/span><\/span>comprimento da onda emitido pela fonte (observador O1\u00a0das figuras abaixo).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No\u00a0<\/b><\/span><\/span>afastamento entre fonte e observador,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>mesmo perceber\u00e1 o som emitido pela fonte mais grave\u00a0<\/b><\/span><\/span>(menor frequ\u00eancia, recebe menor n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Nesse caso, o\u00a0<\/b><\/span><\/span>comprimento de onda aparente percebido pelo observador ser\u00e1 maior que o comprimento da onda emitido pela fonte (observador O2\u00a0das figuras acima).<\/b><\/span><\/span><\/span><\/span><\/p>\n

Observe que o\u00a0<\/b><\/span><\/span>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><\/span><\/p>\n

\"\"<\/p>\n

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

28- <\/span><\/span><\/span>Vamos analisar as afirma\u00e7\u00f5es individualmente:<\/b><\/span><\/span><\/p>\n

A primeira est\u00e1 correta,<\/b><\/span><\/span>\u00a0pois o\u00a0<\/b><\/span><\/span>timbre \u00e9 nada mais que uma quantidade de harm\u00f4nicos,<\/b><\/span><\/span>\u00a0ao se\u00a0<\/b><\/span><\/span>modificar essa quantidade o timbre muda.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>segunda alternativa est\u00e1 incorreta,<\/b><\/span><\/span>\u00a0pois\u00a0<\/b><\/span><\/span>n\u00e3o h\u00e1 rela\u00e7\u00e3o entre intensidade e comprimento de onda.\u00a0<\/b><\/span><\/span>A intensidade \u00e9 a energia transportada pela onda sonora.<\/b><\/span><\/span><\/p>\n

A\u00a0<\/b><\/span><\/span>terceira alternativa est\u00e1 correta,<\/b><\/span><\/span>\u00a0pois\u00a0<\/b><\/span><\/span>a altura depende diretamente da frequ\u00eancia, sendo que quanto mais alto, mais agudo e maior a frequ\u00eancia.<\/b><\/span><\/span><\/p>\n

Bom,\u00a0<\/b><\/span><\/span>pot\u00eancia \u00e9 quantidade de energia no tempo estabelecido<\/b><\/span><\/span>, sendo assim,\u00a0<\/b><\/span><\/span>uma pot\u00eancia maior d\u00e1 mais energia para a onda e consequentemente mais intensidade.<\/b><\/span><\/span><\/p>\n

Fica V \u2013 F \u2013 V \u2013 V. <\/b><\/span><\/span><\/p>\n

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

29- <\/span><\/span><\/span>Altura do som<\/b><\/span><\/span>\u00a0est\u00e1 relacionada com sua\u00a0<\/b><\/span><\/span>freq\u00fc\u00eancia,<\/b><\/span><\/span>\u00a0ou seja, a\u00a0<\/b><\/span><\/span>altura (tom)<\/b><\/span><\/span>\u00a0\u00e9 a\u00a0<\/b><\/span><\/span>qualidade do som que permite ao ouvido distinguir um som grave, de baixa frequ\u00eancia, de um som agudo, de alta<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

frequ\u00eancia.<\/b><\/span><\/span>\u00a0 Cada\u00a0<\/b><\/span><\/span>nota musical (tom)<\/b><\/span><\/span>\u00a0possui\u00a0<\/b><\/span><\/span>frequ\u00eancia caracter\u00edstica, \u00fanica para essa nota.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

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

30- \"\"<\/span><\/span><\/span>Intensidade\u00a0\"\"<\/b><\/span><\/span>como a\u00a0<\/b><\/span><\/span>mariposa se afasta do morcego em repouso a intensidade<\/b><\/span><\/span>\u00a0(caracter\u00edstica do som\u00a0que\u00a0<\/b><\/span><\/span>nos permite\u00a0distinguir um som forte de um som fraco<\/b><\/span><\/span>\u00a0e est\u00e1\u00a0relacionada\u00a0com a\u00a0<\/b><\/span><\/span>energia transportada pela onda<\/b><\/span><\/span>\u00a0que,\u00a0quanto\u00a0<\/b><\/span><\/span>mais perto da fonte mais forte ser\u00e1 o som e mais afastado da fonte, mais fraco),dos ecos, recebidos pelo morcego, diminui.
\nExemplo\"\"<\/b><\/span><\/span>\u00a0o\u00a0<\/b><\/span><\/span>som emitido,<\/b><\/span><\/span>\u00a0por exemplo, por\u00a0uma\u00a0<\/b><\/span><\/span>britadeira,<\/b><\/span><\/span>\u00a0\u00a0ser\u00e1\u00a0<\/b><\/span><\/span>mais forte para uma pessoa<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

perto da mesma e mais fraco para uma pessoa afastada da mesma.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Intervalo de tempo\"\"<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>intervalo de tempo\u00a0de retorno dos ecosaumenta\u00a0<\/b><\/span><\/span>em virtude das\u00a0<\/b><\/span><\/span>dist\u00e2ncias\u00a0 cada vez maiores que as ondas devem percorrer.<\/b><\/span><\/span>
\n<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"Frequ\u00eancia\u00a0\"\"Efeito Doppler\u00a0\"\"<\/b><\/span><\/span>Refere-se \u00e0\u00a0<\/b><\/span><\/span>varia\u00e7\u00e3o da frequ\u00eancia notada<\/b><\/span><\/span>\u00a0por um\u00a0<\/b><\/span><\/span>observador<\/b><\/span><\/span>\u00a0quando a\u00a0<\/b><\/span><\/span>dist\u00e2ncia entre ele e uma fonte de ondas est\u00e1 aumentando ou diminuindo.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Na\u00a0<\/b><\/span><\/span>aproxima\u00e7\u00e3o entre fonte e observador,\u00a0<\/b><\/span><\/span>o\u00a0<\/b><\/span><\/span>mesmo perceber\u00e1 o som emitido pela fonte mais agudo\u00a0<\/b><\/span><\/span>(maior freq\u00fc\u00eancia, recebe maior n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Nesse caso, o\u00a0<\/b><\/span><\/span>comprimento de onda aparente percebido pelo observador ser\u00e1 menor\u00a0<\/b><\/span><\/span>que o\u00a0<\/b><\/span><\/span>comprimento da onda emitido pela fonte (observador O1\u00a0das figuras abaixo).<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

No\u00a0<\/b><\/span><\/span>afastamento entre fonte e observador,<\/b><\/span><\/span>\u00a0o\u00a0mesmo\u00a0<\/b><\/span><\/span>perceber\u00e1 o som emitido pela fonte mais grave\u00a0<\/b><\/span><\/span>(<\/b><\/span><\/span>menor frequ\u00eancia,<\/b><\/span><\/span>\u00a0recebe menor n\u00famero de frentes de onda na unidade de tempo) do que\u00a0<\/b><\/span><\/span>perceberia se fonte e observador estivessem parados.<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio, como a\u00a0<\/b><\/span><\/span>mariposa se afasta do morcego,<\/b><\/span><\/span>\u00a0conforme o <\/b><\/span><\/span>efeito Doppler<\/b><\/span><\/span>, a\u00a0<\/b><\/span><\/span>frequ\u00eancia\u00a0dos ecos recebidos pelo morcego \u00e9\u00a0menor que a frequ\u00eancia dos ultrassons emitidos.
\n<\/b><\/span><\/span>R- A<\/b><\/span><\/span><\/span><\/span><\/p>\n

31- <\/span><\/span><\/span>O\u00a0<\/b><\/span><\/span>timbre \u00e9 uma qualidade sonora\u00a0<\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>permite distinguir dois sons de mesma altura\u00a0<\/b><\/span><\/span>e<\/b><\/span><\/span>mesma intensidade (mesma nota musical)<\/b><\/span><\/span>, emitidos\u00a0<\/b><\/span><\/span>por instrumentos diferentes tocando a mesma nota\u00a0<\/b><\/span><\/span>musical ou acorde.<\/b><\/span><\/span>\"\"<\/p>\n

Assim,\u00a0<\/b><\/span><\/span>conseguimos distinguir se a mesma nota musical \u00e9 emitida por um viol\u00e3o ou por um piano\u00a0<\/b><\/span><\/span>pelo\u00a0<\/b><\/span><\/span>timbre,<\/b><\/span><\/span>pois ele\u00a0<\/b><\/span><\/span>difere nos dois instrumentos e nos fornece sensa\u00e7\u00f5es sonoras diferentes, devido \u00e0s diferentes composi\u00e7\u00f5es de harm\u00f4nicos gerados por cada instrumento.\u00a0<\/b><\/span><\/span><\/p>\n

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

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

 <\/p>\n

32- <\/b><\/span><\/span><\/span>Velocidade de propaga\u00e7\u00e3o dos pulsos nos meios unidimensionais (corda)<\/b><\/span><\/span><\/span><\/span><\/p>\n

\u00a0<\/span>Considere um\u00a0<\/b><\/span><\/span>pulso ou v\u00e1rios pulsos sucessivos (onda) <\/b><\/span><\/span>se\u00a0<\/b><\/span><\/span>propagando<\/b><\/span><\/span>\u00a0com\u00a0<\/b><\/span><\/span>velocidade V\u00a0<\/b><\/span><\/span>numa <\/b><\/span><\/span>corda<\/b><\/span><\/span><\/span><\/span>\"\"<\/p>\n

tracionada\u00a0<\/b><\/span><\/span>(esticada) por uma\u00a0<\/b><\/span><\/span>for\u00e7a de intensidade F.<\/b><\/span><\/span><\/span><\/span><\/p>\n

Denominamos<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>densidade linear de massa (\u00b5)\u00a0<\/b><\/span><\/span>de uma\u00a0<\/b><\/span><\/span>corda homog\u00eanea,\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>se\u00e7\u00e3o transversal\u00a0 constante,\u00a0<\/b><\/span><\/span>que possui\u00a0<\/b><\/span><\/span>massa (m)\u00a0e\u00a0comprimento (L),\u00a0<\/b><\/span><\/span>\u00e0 express\u00e3o:<\/b><\/span><\/span><\/span><\/span><\/p>\n

Significado f\u00edsico de\u00a0\u00b5\u00a0\"\"<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>significa quanto\u00a0<\/b><\/span><\/span>de\u00a0<\/b><\/span><\/span>massa a corda tem por unidade de comprimento.<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/span><\/p>\n

Assim, uma\u00a0<\/b><\/span><\/span>corda de 0,5 kg\/m\u00a0<\/b><\/span><\/span>possui, no SI,\u00a0<\/b><\/span><\/span>0,5kg de massa para cada 1 metro de comprimento.<\/b><\/span><\/span><\/span><\/span>\"\"<\/p>\n

A\u00a0<\/b><\/span><\/span>velocidade (V)\u00a0de propaga\u00e7\u00e3o do pulso na corda \u00e9 dada,\u00a0<\/b><\/span><\/span>tamb\u00e9m<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>por:<\/b><\/span><\/span><\/span><\/span>\"\"<\/p>\n

Onde\u00a0<\/b><\/span><\/span>T \u00e9 a intensidade da for\u00e7a <\/b><\/span><\/span>que\u00a0<\/b><\/span><\/span>traciona\u00a0<\/b><\/span><\/span>(estica)\u00a0<\/b><\/span><\/span>a corda e \u00b5 \u00e9 a densidade linear de massa da mesma. (F\u00f3rmula de Taylor).<\/b><\/span><\/span><\/span><\/span><\/p>\n

No caso do exerc\u00edcio:<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Como a\u00a0<\/b><\/span><\/span>densidade linear\u00a0<\/b><\/span><\/span>da\u00a0<\/b><\/span><\/span>primeira corda (\u00b5<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span>\u00a0\u00e9\u00a0<\/b><\/span><\/span>maior do que a da segunda (\u00b5<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>)<\/b><\/span><\/span>, ou seja,\u00a0<\/b><\/span><\/span>\u00b5<\/b><\/span><\/span>1<\/b><\/span><\/sub><\/span>\u00a0>\u00a0\u00b5<\/b><\/span><\/span>2<\/b><\/span><\/sub><\/span>,<\/b><\/span><\/span><\/span><\/span><\/p>\n

quando o\u00a0<\/b><\/span><\/span>pulso chega ao ponto de jun\u00e7\u00e3o das cordas,\u00a0<\/b><\/span><\/span>ocorre ao\u00a0<\/b><\/span><\/span>mesmo tempo refra\u00e7\u00e3o e reflex\u00e3o.<\/b><\/span><\/span>\u00a0Esse ponto (jun\u00e7\u00e3o) funciona\u00a0<\/b><\/span><\/span>como uma\u00a0<\/b><\/span><\/span>extremidade livre\u00a0\u00a0e o\u00a0pulso refletido retorna sem invers\u00e3o de fase.<\/b><\/span><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

 <\/p>\n

33- <\/span><\/span><\/span>\u00a0<\/b><\/span><\/span>C\u00e1lculo do\u00a0<\/b><\/span><\/span>comprimento de onda<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>\u00a0de uma das\u00a0<\/b><\/span><\/span>ondas formadas no interior do tubo\u00a0\"\"\u00a0V = c =\u00a0.f\u00a0\"\"\u00a0<\/b><\/span><\/span>3.108\u00a0=\u00a0.2,5.109\u00a0\"\"\u00a0\u00a0=\u00a0\"\"\u00a0\"\"\u00a0<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span>= 0,12 m\u00a0<\/b><\/span><\/span>= 12 cm.<\/b><\/span><\/span><\/p>\n

Como o\u00a0<\/b><\/span><\/span>tubo tem comprimento de 36 cm,<\/b><\/span><\/span>\u00a0em seu interior voc\u00ea ter\u00e1\u00a0<\/b><\/span><\/span>n =\u00a0\"\"= 3 comprimentos de onda.<\/b><\/span><\/span><\/p>\n

Sendo que\u00a0<\/b><\/span><\/span>cada onda estacion\u00e1ria possui meio comprimento de onda<\/b><\/span><\/span>voc\u00ea ter\u00e1 no\u00a0<\/b><\/span><\/span>total (3\u00d72) = 6 ondas estacion\u00e1rias<\/b><\/span><\/span>\u00a0e\u00a0<\/b><\/span><\/span>excluindo os dois n\u00f3s das paredes met\u00e1licas,\u00a0<\/b><\/span><\/span>entre as paredes\u00a0<\/b><\/span><\/span>existir\u00e3o (7 \u2013 2) = 5 n\u00f3s\u00a0<\/b><\/span><\/span>(veja figura abaixo).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

34- <\/span><\/span><\/span>Voc\u00ea pode interpretar\u00a0<\/b><\/span><\/span>a folha de papel enrolada<\/b><\/span><\/span>\u00a0como um\u00a0<\/b><\/span><\/span>tubo aberto nas duas extremidades<\/b><\/span><\/span>\u00a0com as\u00a0<\/b><\/span><\/span>seguintes caracter\u00edsticas:<\/b><\/span><\/span><\/p>\n

Tubo aberto nas duas extremidades<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

Generalizando:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

35- <\/span><\/span><\/span>Observe na figura abaixo<\/b><\/span><\/span>\u00a0que entre a primeira\u00a0<\/b><\/span><\/span>tecla (nota) D\u00f31 e a segunda D\u00f32 voc\u00ea tem 13 teclas (notas).<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

O\u00a0<\/b><\/span><\/span>enunciado pede o comprimento de onda, no ar, da\u00a0<\/b><\/span><\/span>nota Sol<\/b><\/span><\/span>, fornecendo que a\u00a0<\/b><\/span><\/span>frequ\u00eancia<\/b><\/span><\/span>\u00a0da nota\u00a0<\/b><\/span><\/span>L\u00e1\u00a0<\/b><\/span><\/span>do teclado da figura\u00a0<\/b><\/span><\/span>\u00e9 440 Hz.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/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 F\u00cdSICA de Vestibulares recentes sobre Ondulat\u00f3ria \u2013 2017- 2016- 2015   01- Os\u00a0sonares emitem e recebem ondas sonorasque possuem asseguintes carcter\u00edsticas: S\u00e3o\u00a0ondas mec\u00e2nicas\u00a0quenecessitam\u00a0de um\u00a0meio material\u00a0para se propagar, transportando energia.\u00a0 Exemplos: Ondas no mar\u00a0(o meio \u00e9 a \u00e1gua),\u00a0ondas sonoras\u00a0(o meio pode ser ar, \u00e1gua, etc.),ondas em uma corda\u00a0(o meio \u00e9 a corda),\u00a0etc. Todas\u00a0essas perturba\u00e7\u00f5es\u00a0s\u00e3o\u00a0causadas em\u00a0meios materiais.<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":3370,"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-3372","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3372","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=3372"}],"version-history":[{"count":2,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3372\/revisions"}],"predecessor-version":[{"id":3374,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3372\/revisions\/3374"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/3370"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=3372"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}