{"id":2350,"date":"2016-08-01T23:35:28","date_gmt":"2016-08-01T23:35:28","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2350"},"modified":"2024-08-23T14:59:29","modified_gmt":"2024-08-23T14:59:29","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tubos-sonoros","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/ondulatoria\/acustica\/tubos-sonoros\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-tubos-sonoros\/","title":{"rendered":"Tubos Sonoros – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Tubos Sonoros<\/b><\/span><\/span><\/span><\/p>\n

01-<\/b><\/span><\/span><\/span> <\/b>5l\u03bb\/4=5\u00a0 —\u00a0\u03bbl=4m\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 340=4.f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f=85Hz<\/b><\/span><\/span><\/span><\/p>\n

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

03-<\/b><\/span><\/span><\/span> <\/b>V=\u03bbf\u00a0 —\u00a0 340=\u03bb.440\u00a0 —\u00a0\u00a0\u03bb=0,77m\u00a0 —\u00a0 Para atingir o teto\u00a0 o som percorre 5m\u00a0 —\u00a0 n=5\/0,77\u00a0 —\u00a0 n=6,49\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>a) Quando f=1000Hz\u00a0 —\u00a0 I=3,0.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=P\/4pr<\/b><\/span><\/span><\/span>2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sup><\/span>3.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=4.3.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>P=36.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>W<\/b><\/span><\/span><\/span><\/p>\n

b) V=<\/b><\/span><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span>f\u00a0 —\u00a0 340=\u03bb.1000\u00a0 —\u00a0\u00a0\u03bb=0,34m\u00a0 —\u00a0 0,34=2L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>L=0,17m<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

\"\"<\/p>\n

Fechado\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\/4=L\u00a0 —\u00a0\u00a0\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=4L\u00a0 —\u00a0 V=\u03bb<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span><\/p>\n

V=4Lf<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=V\/4L<\/b><\/span><\/span><\/span>\u00a0—\u00a0 R- E<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>Tubos fechados s\u00f3 emitem harm\u00f4nicos \u00edmpares\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=nV\/4L\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1X340\/4X0,5\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=11770Hz\u00a0\u00a0\u00a0\u00a0 f<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=3X340\/4X0,5\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sub><\/span>=510Hz\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span><\/span> <\/b>a) 4\u03bb\/4=7\u00a0 —\u00a0\u03bb=7m\u00a0 —\u00a0 340=7f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>f=48,6Hz<\/b><\/span><\/span><\/span><\/span><\/p>\n

b) V=d\/t\u00a0 —\u00a0 340=9045\/t\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span><\/span>t=26,6s<\/b><\/span><\/span><\/span><\/span><\/p>\n

08-<\/b><\/span><\/span><\/span> <\/b>a) M\u00faltiplas reflex\u00f5es de sons do pr\u00f3prio ambiente.<\/b><\/span><\/span><\/span><\/p>\n

b)A freq\u00fc\u00eancia predominante corresponde ao som fundamental<\/b><\/span><\/span><\/span><\/p>\n

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

2\u03bb\/4=0,3\u00a0 —\u00a0\u00a0\u03bb=0,6m\u00a0 —\u00a0 330=0,6f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f=550Hz<\/b><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>V=\u03bbf\u00a0 —\u00a0\u00a0\u03bb\/4=L\u00a0 —\u00a0\u00a0\u03bb=4L\u00a0 —\u00a0 340=4L3400\u00a0 —\u00a0 f=0,025m\u00a0 R- B<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> <\/b>a) \u00e1rea do t\u00edmpano\u00a0 —\u00a0 S=\u03c0R<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=3,14.(4.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=3,14.16.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>=50,24.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P=F\/S\u00a0 —\u00a0 1,01.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=F\/50,24.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>F\u22480,507N<\/b><\/span><\/span><\/span><\/p>\n

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

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

\u00a0<\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>\/4=2,8.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>=11,2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>m\u00a0 —\u00a0 V=<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>f\u00a0 —\u00a0 340=11,2.10<\/b><\/span><\/span><\/span>-2<\/b><\/span><\/span><\/sup><\/span>f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f=3.035,7Hz<\/b><\/span><\/span><\/span><\/p>\n

11-<\/b><\/span><\/span><\/span> <\/b>Como fio tubo est\u00e3o em resson\u00e2ncia a freq\u00fc\u00eancia \u00e9 a mesma para os dois (modo fundamental)<\/b><\/span><\/span><\/span><\/p>\n

Fio\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=n\/2L\"\"\u00a0 —\u00a0\u00a0\u00b5=m\/L\u00a0 —\u00a0 n=1 (fundamental)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f=1\/2L\"\"<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Tubo\u00a0 —\u00a0\u00a0\u03bb\/4=l\u00a0 —\u00a0\u00a0\u03bb=4l\u00a0 —\u00a0 V=c=\u03bbf\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u00a0f=c\/4\u03bb<\/b><\/span><\/span><\/span><\/p>\n

Igualando as freq\u00fc\u00eancias\u00a0 —\u00a0 f=<\/b><\/span><\/span><\/span> f=1\/2L\"\"<\/b><\/span><\/span><\/span> = c\/4\u03bb\u00a0\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>T=(c\/2\u03bb)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/\u00b5L<\/b><\/span><\/span><\/span><\/p>\n

12-<\/b><\/span><\/span><\/span> <\/b>Tubo fechado s\u00f3 tem harm\u00f4nicos \u00edmpares e sua freq\u00fc\u00eancia fundamental \u00e9 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=200Hz\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=nf<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0(n=1,3,5,…)\u00a0 —\u00a0 Quando o som for de 16.000Hz, o harm\u00f4nico n \u00e9 de\u00a0 —\u00a0 16.000=n.200\u00a0 —\u00a0 n=80. Mas, 80 \u00e9 par e o tubo n\u00e3o possui esse harm\u00f4nico mas sim, o primeiro \u00edmpar anterior que \u00e9 o 79.<\/b><\/span><\/span><\/span><\/p>\n

O primeiro harm\u00f4nico \u00e9 n=1. Assim, temos que determinar o n\u00famero de harm\u00f4nicos \u00edmpares compreendidos entre 1 e 79\u00a0 que \u00e9 38. Portanto o n\u00famero de freq\u00fc\u00eancias aud\u00edveis \u00e9 de 40.\u00a0\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

13- <\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=nV\/2L\u00a0 —\u00a0 n=1 (fundamental) \u00a0\u00a0—\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=1.340\/2.0,02\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=8.500Hz\u00a0 —\u00a0 \u00a0n=1 (fundamental)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sub><\/span>=1.340\/2.10\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>10<\/b><\/span><\/span><\/sub><\/span>=17Hz\u00a0\u00a0\u00a0\u00a0\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

14-<\/b><\/span><\/span><\/span><\/span> <\/b>I – 2\u03bb\/4=L\u00a0 —\u00a0\u00a0\u03bb=2L\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 V=2Lf\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span><\/span>I<\/b><\/span><\/span><\/span><\/sub><\/span>=V\/2L<\/b><\/span><\/span><\/span><\/span><\/p>\n

II –\u00a0\u03bb\/4=L\u00a0 —\u00a0\u00a0\u03bb=4L\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 V=4Lf\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span><\/span>II<\/b><\/span><\/span><\/span><\/sub><\/span>=V\/4L<\/b><\/span><\/span><\/span><\/span><\/p>\n

III –\u00a0\u03bb\/2=L\u00a0 —\u00a0\u00a0\u03bb=2L\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 V=2Lf\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span><\/span>III<\/b><\/span><\/span><\/span><\/sub><\/span>=V\/2L<\/b><\/span><\/span><\/span><\/span><\/p>\n

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

15-<\/b><\/span><\/span><\/span> <\/b>f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=nV\/2L\u00a0 —\u00a0 1700=n.340\/2.0,25\u00a0 —\u00a0 n=2,5\u00a0 —\u00a0 R- D\u00a0 —\u00a0 na extremidade aberta, em cima temos que ter um ventre.<\/b><\/span><\/span><\/span><\/p>\n

16-<\/b><\/span><\/span><\/span> <\/b>3\u03bb\/4=0,6\u00a0 —\u00a0\u00a0\u03bb=0,8m\u00a0 —\u00a0 V=\u03bbf\u00a0 — 340=0,8f\u00a0 —<\/b><\/span><\/span><\/span>\u00a0 f=425Hz<\/b><\/span><\/span><\/span><\/p>\n

17-<\/b><\/span><\/span><\/span> <\/b>A posi\u00e7\u00e3o dos orif\u00edcios deve coicidir com n\u00f3s (aus\u00eancia de vibra\u00e7\u00e3o) de uma onda sonora estacion\u00e1ria para que n\u00e3o haja entrada ou sa\u00edda de ar nos mesmos e considere\u00a0 a freq\u00fc\u00eancia mais baixa poss\u00edvel. (Dado: velocidade do som no ar=340m\/s).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Como na posi\u00e7\u00e3o dos orif\u00edcios temos n\u00f3s e a frequ\u00eancia sendo a mais baixa poss\u00edvel teremos o seguinte formato das ondas estacion\u00e1rias entre os dois n\u00f3s.<\/b><\/span><\/span><\/span><\/p>\n

\u03bb\/2=1\/3\u00a0 —\u00a0\u00a0\u03bb=2\/3m \u00a0—V=\u03bbf\u00a0 —\u00a0 340=(2\/3)f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>f=510Hz<\/b><\/span><\/span><\/span><\/p>\n

18-<\/b><\/span><\/span><\/span> <\/b>Tubo aberto com tr\u00eas n\u00f3s\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

6<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>\/4=L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>=2L\/3\u00a0 —\u00a0 V=<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>f\u00a0 —\u00a0 340=2L\/3.30\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>L=17m<\/b><\/span><\/span><\/span><\/p>\n

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Os tubos fechados s\u00f3 ressoam para harm\u00f4nicos \u00edmpares\u00a0 —\u00a0 se a frequ\u00eancia fundamental \u00e9 500Hz, ele ressoar\u00e1 para: 1500Hz, 2500Hz, 3500Hz, 4500Hz, etc.\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

20 –<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Dados\u00a0 —\u00a0\u00a0 L = 7 cm = 0,07 m\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>m\u00e1x<\/b><\/span><\/span><\/sub><\/span>\u00a0= 20.000 Hz\u00a0 —\u00a0\u00a0 f<\/b><\/span><\/span><\/span>m\u00edn<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>= 20 Hz;\u00a0<\/b><\/span><\/span><\/span>v<\/b><\/span><\/span><\/span>\u00a0= 340 m\/s\u00a0 —\u00a0 a figura mostra a configura\u00e7\u00e3o para o primeiro harm\u00f4nico (n = 1) de um tubo aberto\u00a0 — o comprimento do tubo \u00e9 igual a meio comprimento de onda\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2=L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2L\u00a0 —\u00a0\u00a0 frequ\u00eancia do primeiro harm\u00f4nico\u00a0 —\u00a0 V=\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 340=2L.f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=340\/2.(0,07)\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=17.000\/7 Hz\u00a0 —\u00a0 c\u00e1lculo da frequ\u00eancia do n-\u00e9simo harm\u00f4nico\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=nf<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 os harm\u00f4nicos aud\u00edveis tem freq\u00fc\u00eancias menores que 20.000Hz\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span><20.000\u00a0 —\u00a0 nf<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span><20.000\u00a0 —\u00a0 n.(17.000)\/7<20.000\u00a0 —\u00a0 n<140\/17\u00a0 —\u00a0 n<8,24\u00a0 —\u00a0 sendo n um n\u00famero inteiro\u00a0 —\u00a0 n<8\u00a0 —\u00a0 portanto os sons aud\u00edveis est\u00e3o compreendidos entre o primeiro e o oitavo harm\u00f4nico e o som aud\u00edvel mais alto (maior frequ\u00eancia) \u00e9 o do oitavo —\u00a0 f<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>=8f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>=8.(17.000\/7)\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0f<\/b><\/span><\/span><\/span>8<\/b><\/span><\/span><\/sub><\/span>=19.428 Hz<\/b><\/span><\/span><\/span>\"\"<\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considere que os tr\u00eas tubos estejam emitindo harm\u00f4nicos de mesma ordem\u00a0 —\u00a0 a velocidade de propaga\u00e7\u00e3o do som \u00e9 mesma, pois se trata do mesmo meio, o ar\u00a0 —\u00a0 equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0 —\u00a0 V=<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>f\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>=V\/f\u00a0 —\u00a0 para o<\/b><\/span><\/span><\/span>\"\"<\/p>\n

en\u00e9simo harm\u00f4nico o comprimento L do tubo \u00e9 dado por\u00a0 —\u00a0 L=n.(\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>\/2) (II)\u00a0 —\u00a0 (I) em (II)\u00a0 —\u00a0 L=n.(V\/f)\/2\u00a0 —\u00a0 L=nV\/2f\u00a0 —\u00a0 observe nessa express\u00e3o que o comprimento L do tubo \u00e9 inversamente proporcional \u00e0 frequ\u00eancia f do som emitido\u00a0 —\u00a0 observando a tabela fornecida\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>vermelho<\/b><\/span><\/span><\/sub><\/span>\u00a0< f<\/b><\/span><\/span><\/span>azul<\/b><\/span><\/span><\/sub><\/span>\u00a0< f<\/b><\/span><\/span><\/span>roxo<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 L<\/b><\/span><\/span><\/span>vermelho<\/b><\/span><\/span><\/sub><\/span>\u00a0> L<\/b><\/span><\/span><\/span>azul<\/b><\/span><\/span><\/sub><\/span>\u00a0> L<\/b><\/span><\/span><\/span>roxo<\/b><\/span><\/span><\/sub><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>\u00a0—\u00a0\u00a0<\/b><\/span><\/span><\/span>R- D\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>f<\/b><\/span><\/span><\/span>n<\/b><\/span><\/span><\/sub><\/span>=nV\/4L\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>=5V\/4L\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>=5.340\/4.2\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sub><\/span>=212,5 Hz\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

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

23-<\/b><\/span><\/span><\/span> <\/b>Trata-se de um tubo aberto nas duas extremidades cujas caracter\u00edsticas est\u00e3o fornecidas abaixo:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

\"\"<\/p>\n

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

\"\"<\/p>\n

L=50cm=0,5m\u00a0 —\u00a0 V=340m\/s\u00a0 —\u00a0 frequ\u00eancia fundamental\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V\/2L=340\/2.0,5\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=340Hz\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R-34<\/b><\/span><\/span><\/span>.\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

24- <\/b><\/span><\/span><\/span>Inicialmente o fio se comporta como uma corda sonora vibrante, com as caracter\u00edsticas a seguir:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

C\u00e1lculo da velocidade de propaga\u00e7\u00e3o V da onda no fio el\u00e9trico\u00a0 —\u00a0 \u03bc=m\/L=0,001\/2\u00a0 —\u00a0 \u03bc=0,0005kg\/m=5.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>kg\/m\u00a0 —\u00a0 V=\u221a(T\/\u03bc)\u00a0 —\u00a0 V=\u221a(80\/5.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 V=\u221a(16.10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 V=4.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>=400m\/s\u00a0 —\u00a0 1<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sup><\/span>\u00a0harm\u00f4nico ou freq\u00fc\u00eancia (som) fundamental\u00a0 —\u00a0 (dois n\u00f3s e\u00a0um<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

fuso)\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/2=L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=2L\u00a0 —\u00a0 V=\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V\/\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0\u00a0f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=V\/2L\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=400\/2.2\u00a0 —\u00a0 f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=100Hz (frequ\u00eancia do som fundamental emitido pelo fio el\u00e9trico)\u00a0 —\u00a0 essa frequ\u00eancia (100Hz) \u00e9 a mesma com que a coluna de ar no buraco vibra\u00a0 —\u00a0 agora voc\u00ea tem um tubo sonoro fechado numa das extremidades vibrando de modo fundamental (enunciado) com velocidade do ar V=330m\/s\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\"\"<\/p>\n

o som fundamental num tubo fechado possui as seguintes caracter\u00edsticas\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/4=L\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4L\u00a0 —\u00a0 V=\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.f<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 330=\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>.100\u00a0 —<\/b><\/span><\/span><\/span><\/p>\n

\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=3,3m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03b3<\/b><\/span><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=4L\u00a0 —\u00a0 3,3=4L\u00a0 —\u00a0 L=0,825m\u00a0 —\u00a0<\/b><\/span><\/span><\/span>\u00a0R- C<\/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 dos exerc\u00edcios de vestibulares sobre Tubos Sonoros 01- 5l\u03bb\/4=5\u00a0 —\u00a0\u03bbl=4m\u00a0 —\u00a0 V=\u03bbf\u00a0 —\u00a0 340=4.f\u00a0 —\u00a0\u00a0f=85Hz 02- A\u00a0\u00a0\u00a0 03- V=\u03bbf\u00a0 —\u00a0 340=\u03bb.440\u00a0 —\u00a0\u00a0\u03bb=0,77m\u00a0 —\u00a0 Para atingir o teto\u00a0 o som percorre 5m\u00a0 —\u00a0 n=5\/0,77\u00a0 —\u00a0 n=6,49\u00a0 R- E 04- a) Quando f=1000Hz\u00a0 —\u00a0 I=3,0.10-6W\/m2\u00a0 —\u00a0 I=P\/4pr2\u00a0 —\u00a0\u00a03.10-6=4.3.102\u00a0 —\u00a0\u00a0P=36.10-4W b) V=\u03bbf\u00a0 —\u00a0 340=\u03bb.1000\u00a0 —\u00a0\u00a0\u03bb=0,34m\u00a0 —\u00a0 0,34=2L\u00a0 —\u00a0\u00a0L=0,17m 05- Fechado\u00a0 —\u00a0\u00a0\u03bbA\/4=L\u00a0<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2345,"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-2350","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2350","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=2350"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2350\/revisions"}],"predecessor-version":[{"id":10910,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2350\/revisions\/10910"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2345"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2350"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}