{"id":2371,"date":"2016-08-02T00:06:06","date_gmt":"2016-08-02T00:06:06","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=2371"},"modified":"2024-08-23T14:57:46","modified_gmt":"2024-08-23T14:57:46","slug":"2371-2","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/ondulatoria\/acustica\/qualidades-fisiologicas-do-som\/2371-2\/","title":{"rendered":"Qualidades fisiol\u00f3gicas do som – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

Qualidades fisiol\u00f3gicas do som<\/b><\/span><\/span><\/span><\/p>\n

01- <\/b><\/span><\/span><\/span>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\u00a0P=36.10<\/b><\/span><\/span><\/span>-4<\/b><\/span><\/span><\/sup><\/span>W<\/b><\/span><\/span><\/span><\/p>\n

02-<\/b><\/span><\/span><\/span> <\/b>I=Kf<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 feminina \u2013 I<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>=K(200)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 masculina \u2013 I<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=F(100)<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0I<\/b><\/span><\/span><\/span>F<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>M<\/b><\/span><\/span><\/sub><\/span>=4\u00a0 —\u00a0 R- A<\/b><\/span><\/span><\/span><\/p>\n

03-<\/b><\/span><\/span><\/span> <\/b>a) Pelo gr\u00e1fico 85dB \u2013\u00a010m<\/b><\/span><\/span><\/span><\/p>\n

b) V=\u03bbf\u00a0 —\u00a0 340=\u03bb100\u00a0 —\u00a0\u00a0\u03bb=3,4m<\/b><\/span><\/span><\/span><\/p>\n

c) 50m \u2013 NS=70dB\u00a0 —\u00a0 NS=10log(I\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —7=logI \u2013 log10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 7=logI \u2013(-12)log10\u00a0 —\u00a0 7=logI + 12\u00a0 —\u00a0 logI=-5\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>=I — I=10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

04-<\/b><\/span><\/span><\/span> <\/b>90=10log(I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 9=logI \u2013 log10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 9=logI-(-12)log10\u00a0 —\u00a0 9=logI+12\u00a0 —\u00a0 -3=logI\u00a0 —I=10<\/b><\/span><\/span><\/span>-3 <\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0Turbina e Amplificador de som<\/b><\/span><\/span><\/span><\/p>\n

05-<\/b><\/span><\/span><\/span> <\/b>A diferen\u00e7a \u00e9 de 20dB=2B, que corresponde a 100 vezes maior.\u00a0\u00a0\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

06-<\/b><\/span><\/span><\/span> <\/b>a) Pelo gr\u00e1fico observamos que o indiv\u00edduo A consegue ouvir melhor que o indiv\u00edduo B sons com freq\u00fc\u00eancias compreendidas entre 20Hz e 200Hz porque a curva de seu n\u00edvel sonoro (curva tracejada) est\u00e1 mais baixa (mais pr\u00f3xima do limiar de audi\u00e7\u00e3o que \u00e9 0dB)).<\/b><\/span><\/span><\/span><\/p>\n

b) Este valor \u00e9 NS=120dB\u00a0 —\u00a0\u00a0 120=10log(I\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 12=log(I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>12<\/b><\/span><\/span><\/sup><\/span>=I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=10<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0I=1W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

c) Pelo gr\u00e1fico, quando f=100Hz \u2013 NS<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10dB e NS<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=40dB. Assim, o n\u00edvel sonoro deve passar de 10dB para 40dB<\/b><\/span><\/span><\/span><\/p>\n

NS<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10dB\u00a0 —\u00a0 10=10log(I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 1=log(I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sup><\/span>=I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>= 10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

NS<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=40dB\u00a0 —\u00a0 40=10log(I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0\u00a0 4=log(I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0\u00a0 10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>=I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>-8<\/b><\/span><\/span><\/sup><\/span>\/10<\/b><\/span><\/span><\/span>-11<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sub><\/span>=10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>I<\/b><\/span><\/span><\/span>1<\/b><\/span><\/span><\/sub><\/span>\u00a0 —\u00a0 deve ser ampliada de 1.000 vezes<\/b><\/span><\/span><\/span><\/p>\n

07-<\/b><\/span><\/span><\/span> <\/b>O n\u00edvel sonoro total de n fontes id\u00eanticas \u00e9 dado por L(t) = 10.log(n) + L, onde L \u00e9 o n\u00edvel sonoro de apenas uma fonte.<\/b><\/span><\/span><\/span><\/p>\n

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

NS(t) = 10.log(10000) + 40 —\u00a0 NS(t) = 10log10<\/b><\/span><\/span><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0+ 40\u00a0 —\u00a0 NS(t)= 10.4log10 + 40\u00a0 —\u00a0 NS(t)=10.4.1 + 40\u00a0 —\u00a0 NS(t)=80dB<\/b><\/span><\/span><\/span><\/p>\n

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

08-<\/b><\/span><\/span><\/span> <\/b>Na express\u00e3o acima, quando k=1, S-S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\u00a0est\u00e1 em bel(B) e em decib\u00e9is, ser\u00e1 S-So\/10\u00a0 —\u00a0 S-S<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/10=log(10<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

S-S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>\/10=log10<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span><\/span>DS\/10\u00a0<\/b><\/span><\/span><\/span><\/sup><\/span>= 10<\/b><\/span><\/span><\/span><\/span>3<\/b><\/span><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0DS\/10=3\u00a0 —\u00a0\u00a0DS=S-S<\/b><\/span><\/span><\/span><\/span>o<\/b><\/span><\/span><\/span><\/sub><\/span>=30dB\u00a0\u00a0 R- C<\/b><\/span><\/span><\/span><\/span><\/p>\n

09-<\/b><\/span><\/span><\/span> <\/b>70dB\u00a0 —\u00a0 NS=10log(I\/I<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>)\u00a0 —\u00a0 70\/10=logI – log10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 7=logI + 12log10\u00a0 —\u00a0 7=logI + 12\u00a0 —\u00a0 -5=logI\u00a0 —\u00a0 I=10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

120dB\u00a0 —\u00a0 120\/10=logI + 12log10\u00a0 —\u00a0 12=logI + 12\u00a0 —\u00a0 logI=0\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>0<\/b><\/span><\/span><\/sup><\/span>=I\u00a0 —\u00a0 I=1W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span><\/p>\n

Regra de tr\u00eas\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0corresponde a 100%\u00a0 — (1-10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>) corresponder\u00e1 a X\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>X=10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>(1-10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 X=(10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>-10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>)\/10<\/b><\/span><\/span><\/span>-5<\/b><\/span><\/span><\/sup><\/span><\/p>\n

X=10<\/b><\/span><\/span><\/span>7<\/b><\/span><\/span><\/sup><\/span>-10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0X=9.999.900%\u00a0 R- D<\/b><\/span><\/span><\/span><\/p>\n

10-<\/b><\/span><\/span><\/span> <\/b>a) Poderia ser de 70dB a 90dB, que s\u00e3o valores um pouco menores que o do m\u00e1ximo som musical. Entre 1.000Hz e 4.000Hz que corresponde a faixa de valores de maior sensibilidade do ouvido humano.<\/b><\/span><\/span><\/span><\/p>\n

b) 50 = 10. log(I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 5 = log(I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>)\u00a0 —\u00a0 10<\/b><\/span><\/span><\/span>5<\/b><\/span><\/span><\/sup><\/span>\u00a0= I\/10<\/b><\/span><\/span><\/span>-12<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=10<\/b><\/span><\/span><\/span>-7<\/b><\/span><\/span><\/sup><\/span>W\/m<\/b><\/span><\/span><\/span>2.<\/b><\/span><\/span><\/sup><\/span>.\u00a0\u00a0Sim, confirma.<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

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

15-<\/b><\/span><\/span><\/span> <\/b>(01) Correta\u00a0\u00a0\u00a0\u00a0\u00a0 (02) Correta\u00a0\u00a0\u00a0\u00a0\u00a0 (04) Correta\u00a0\u00a0\u00a0\u00a0\u00a0 (08) Correta\u00a0\u00a0\u00a0\u00a0\u00a0 soma=01 + 02 + 04 + 08= 15\u00a0\u00a0<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

19-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Considere I<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>\u00a0e I<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>, as intensidades dispersadas por cada uma das radia\u00e7\u00f5es azul e vermelha, respectivamente\u00a0 —\u00a0\u00a0 equa\u00e7\u00e3o fundamental da ondulat\u00f3ria\u00a0 —\u00a0 V=\u00a0<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>.f\u00a0 —\u00a0 f=V\/<\/b><\/span><\/span><\/span>\u03bb<\/b><\/span><\/span><\/span><\/span>\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>A<\/b><\/span><\/span><\/sub><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0(I)\u00a0 —\u00a0 I<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>=f<\/b><\/span><\/span><\/span>V<\/b><\/span><\/span><\/sub><\/span>4<\/b><\/span><\/span><\/sup><\/span>\u00a0(II)\u00a0 —\u00a0\u00a0 dividindo a equa\u00e7\u00e3o (I) pela (II) —\u00a0<\/b><\/span><\/span><\/span><\/p>\n

Conforme o enunciado:<\/b><\/span><\/span><\/span><\/p>\n

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

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

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

20-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Observe a figura ilustrando a situa\u00e7\u00e3o:<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

Dados\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>max\u00a0<\/b><\/span><\/span><\/sub><\/span>= 84 Pa\u00a0 —\u00a0\u00a0\u00a0r\u00a0= 10 m\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>\u00a0= 1,2 kg\/m<\/b><\/span><\/span><\/span>3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0\u00a0v\u00a0= 350 m\/s\u00a0 —\u00a0 substituindo esses dados na express\u00e3o fornecida\u00a0 —\u00a0 I=P<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>max<\/b><\/span><\/span><\/sub><\/span>\/2\u00a0<\/b><\/span><\/span><\/span>\u03c1<\/b><\/span><\/span><\/span><\/span>V=84<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\/2.(1,2).(350)\u00a0 —\u00a0 I=8,4Pa\u00a0 —\u00a0 mas a intensidade de uma onda num ponto de uma superf\u00edcie \u00e9 a raz\u00e3o entre a pot\u00eancia da fonte (P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>) e a \u00e1rea (S) de abrang\u00eancia (no caso, uma superf\u00edcie esf\u00e9rica de raio 10m\u00a0 —\u00a0 I=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/S (I)\u00a0 —\u00a0 \u00e1rea da superf\u00edcie esf\u00e9rica de raio r=10m\u00a0 —\u00a0 S=4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0(II)\u00a0 —\u00a0 (II) em (I)\u00a0 —\u00a0 I=P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>\/4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>r<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.I=4.<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>.l0<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>.8,4\u00a0 —\u00a0 P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>=3.360<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>\u00a0W ou<\/b><\/span><\/span><\/span><\/p>\n

P<\/b><\/span><\/span><\/span>o<\/b><\/span><\/span><\/sub><\/span>= 3,36<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>kW\u00a0 —\u00a0\u00a0R- B<\/b><\/span><\/span><\/span><\/p>\n

21-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Se o alto-falante possui pot\u00eancia sonora m\u00e9dia constante, ent\u00e3o o produto da intensidade sonora a uma dada dist\u00e2ncia da fonte pela \u00e1rea da superf\u00edcie esf\u00e9rica com raio igual a esta dist\u00e2ncia n\u00e3o varia\u00a0 —\u00a0 considerando desprez\u00edvel o efeito das reflex\u00f5es das ondas sonoras, voc\u00ea tem que 10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>.(4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>.6<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>) = I.4<\/b><\/span><\/span><\/span>\u03c0<\/b><\/span><\/span><\/span><\/span>.10<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=(36\/100).10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=0,36.10<\/b><\/span><\/span><\/span>-6<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 I=0,36mW\/m<\/b><\/span><\/span><\/span>2<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0\u00a0R- C<\/b><\/span><\/span><\/span><\/p>\n

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

22-<\/b><\/span><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Primeira\u00a0 —\u00a0 correta\u00a0 —\u00a0 O timbre \u00e9 uma qualidade sonora que permite distinguir dois sons de mesma altura (mesma freq\u00fc\u00eancia) e mesma intensidade, emitidos por instrumentos diferentes tocando a mesma nota musical ou acorde.<\/b><\/span><\/span><\/span><\/p>\n

\"\"\"\"<\/p>\n

Assim, conseguimos distinguir se a mesma nota musical \u00e9 emitida por um viol\u00e3o ou por um piano pelo timbre, pois ele 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><\/span><\/p>\n

Segunda\u00a0 — correta\u00a0 —\u00a0 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 acima).<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Terceira\u00a0 —\u00a0 falsa\u00a0 —\u00a0 \u00e9 pelo timbre (veja a primeira).<\/b><\/span><\/span><\/span><\/p>\n

Quarta\u00a0 —\u00a0 correta\u00a0 —\u00a0 a velocidade de propaga\u00e7\u00e3o das ondas depende do meio em que elas est\u00e3o se propagando.<\/b><\/span><\/span><\/span><\/p>\n

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

23-<\/b><\/span><\/span><\/span> <\/b>01. Falsa\u00a0 —\u00a0 a \u00fanica grandeza invari\u00e1vel de uma onda e que independe do meio de propaga\u00e7\u00e3o \u00e9 sua frequ\u00eancia.<\/b><\/span><\/span><\/span><\/p>\n

02. Falsa\u00a0 —\u00a0 quando ela se move em dire\u00e7\u00e3o ao viol\u00e3o a frequ\u00eancia do som percebido aumenta (efeito Dopler).<\/b><\/span><\/span><\/span><\/p>\n

04. Falsa\u00a0 — \u00a0O timbre \u00e9 uma qualidade sonora que permite distinguir dois sons de mesma altura (mesma freq\u00fc\u00eancia) e mesma intensidade, emitidos por instrumentos diferentes tocando a mesma nota musical ou acorde<\/b><\/span><\/span><\/span><\/p>\n

08. Correta\u00a0 —\u00a0 Altura do som est\u00e1 relacionada com sua freq\u00fc\u00eancia, ou seja, a altura (tom) \u00e9 a qualidade do som que permite ao ouvido distinguir um som grave, de baixa frequ\u00eancia, de um som agudo, de alta freq\u00fc\u00eancia\u00a0 —\u00a0 se voc\u00ea aumentar a tens\u00e3o na corda, estar\u00e1 esticando mais o fio aumentando assim sua frequ\u00eancia, tornando-o mais agudo.<\/b><\/span><\/span><\/span><\/p>\n

16. Correta\u00a0 —\u00a0 densidade linear da corda\u00a0 —\u00a0 \u03bc=m\/\u2113=25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\/1=25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>kg\/m\u00a0 —\u00a0 V=\u221aT\/\u03bc=\u221a(10\/25.10<\/b><\/span><\/span><\/span>-3<\/b><\/span><\/span><\/sup><\/span>\u00a0 —\u00a0 V=20m\/s<\/b><\/span><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0—\u00a0 segundo harm\u00f4nico —\u00a0 (tr\u00eas n\u00f3s e\u00a0dois fusos)\u00a0 —\u00a0 2\u03bb\/2= \u2113\u00a0 —\u00a0\u00a0 \u03bb=\u2113\u00a0 —\u00a0\u00a0\u03bb=\u00a0=1m —\u00a0 V=\u03bb.f\u00a0 —\u00a0 20=1.f\u00a0 —\u00a0 f=20Hz.<\/b><\/span><\/span><\/span><\/p>\n

32. Falsa \u00a0—\u00a0 \u00a0volume \u00a0do som – pode ser apresentado tocando a mesma nota mais baixo e, a seguir, de novo, mais alto, ou seja, usando mais for\u00e7a ou energia ao tocar, sem mudar, contudo, a nota ou a corda\u00a0 — est\u00e1 relacionado com a energia transportada pela onda que, quanto mais perto da fonte maior volume ter\u00e1 o som e mais afastado da fonte, menor volume\u00a0 —\u00a0 dessa maneira, o som emitido, por exemplo, por uma britadeira,\u00a0 ser\u00e1 mais forte para uma pessoa perto da mesma e mais fraco para uma pessoa afastada da mesma.<\/b><\/span><\/span><\/span>\"\"<\/p>\n

Corretas: 08 e 16\u00a0 —\u00a0\u00a0Soma=24.<\/b><\/span><\/span><\/span><\/p>\n

24-<\/b><\/span><\/span><\/span>Cada nota musical tem sua frequ\u00eancia caracter\u00edstica que \u00e9 sempre a mesma independente do meio em que est\u00e1 se propagando\u00a0 — \u00a0assim, por exemplo, se um violinista, com o violino afinado,\u00a0 est\u00e1 emitindo um som r\u00e9, de frequ\u00eancia 293,7Hz, este som ser\u00e1 emitido em todas as dire\u00e7\u00f5es e sentidos, e ser\u00e1 ouvido por cada espectador com esta frequ\u00eancia (293,7Hz). R- B<\/b><\/span><\/span><\/p>\n

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

25-<\/b><\/span><\/span><\/span> <\/b>A\u00a0altura\u00a0\u00e9 uma caracter\u00edstica do som que nos permite classific\u00e1-lo em grave ou agudo\u00a0 —\u00a0 essa propriedade do som \u00e9 caracterizada pela frequ\u00eancia da onda sonora\u00a0 —\u00a0 um som com baixa frequ\u00eancia \u00e9 dito som grave e o som com altas frequ\u00eancias \u00e9 dito som agudo\u00a0 —\u00a0 na linguagem t\u00e9cnica ou musical dizemos que o som grave \u00e9 baixo e o agudo \u00e9 alto\u00a0 —\u00a0 a frequ\u00eancia est\u00e1 relacionada com a maior ou menor for\u00e7a de tra\u00e7\u00e3o (que estica) a corda que, por sua vez est\u00e1 relacionada com as diferentes dilata\u00e7\u00f5es das cordas ou da madeira\u00a0 —\u00a0\u00a0R- D.<\/b><\/span><\/span><\/p>\n

26-<\/b><\/span><\/span><\/span> <\/b>I. Correta\u00a0 —\u00a0 reverbera\u00e7\u00e3o \u00e9 um fen\u00f4meno de prolongamento de um som ap\u00f3s o fim de emiss\u00e3o por parte de uma mesma fonte sonora\u00a0 —\u00a0 quando o obst\u00e1culo refletor do som est\u00e1 um pouco mais afastado, de modo que a som emitido e o som refletido t\u00eam um intervalo de tempo menor que 0,1s ocorre o fen\u00f4meno da\u00a0reverbera\u00e7\u00e3o\u00a0 —\u00a0 nesse caso ao receber dois est\u00edmulos do mesmo tipo em menos de 0,1s o ouvinte tem a sensa\u00e7\u00e3o que o som ainda n\u00e3o foi extinto.<\/b><\/span><\/span><\/span><\/p>\n

II. Falsa\u00a0 —\u00a0 veja justificativa da I.<\/b><\/span><\/span><\/span><\/p>\n

III. Correta\u00a0 — \u00a0eco\u00a0\u00e9 uma\u00a0reflex\u00e3o\u00a0de som que chega ao ouvinte pouco tempo depois do som direto\u00a0 —\u00a0 um\u00a0eco verdadeiro\u00a0\u00e9 uma \u00fanica reflex\u00e3o do som emitido pela fonte, que voc\u00ea ouve instantes ap\u00f3s ouvir o som id\u00eantico inicial.<\/b><\/span><\/span><\/span><\/p>\n

R- D<\/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 Qualidades fisiol\u00f3gicas do som 01- 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 02- I=Kf2\u00a0 —\u00a0 feminina \u2013 IF=K(200)2\u00a0 —\u00a0 masculina \u2013 IM=F(100)2\u00a0 —\u00a0IF\/IM=4\u00a0 —\u00a0 R- A 03- a) Pelo gr\u00e1fico 85dB \u2013\u00a010m b) V=\u03bbf\u00a0 —\u00a0 340=\u03bb100\u00a0 —\u00a0\u00a0\u03bb=3,4m c) 50m \u2013 NS=70dB\u00a0 —\u00a0 NS=10log(I\/Io)\u00a0 —7=logI \u2013 log10-12\u00a0 —\u00a0 7=logI \u2013(-12)log10\u00a0 —\u00a0 7=logI +<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":2366,"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-2371","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2371","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=2371"}],"version-history":[{"count":4,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2371\/revisions"}],"predecessor-version":[{"id":10908,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2371\/revisions\/10908"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/2366"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=2371"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}