{"id":1273,"date":"2015-08-26T01:23:53","date_gmt":"2015-08-26T01:23:53","guid":{"rendered":"http:\/\/fisicaevestibular.com.br\/novo\/?page_id=1273"},"modified":"2024-08-23T13:25:50","modified_gmt":"2024-08-23T13:25:50","slug":"resolucao-comentada-dos-exercicios-de-vestibulares-sobre-pendulo-simples","status":"publish","type":"page","link":"https:\/\/fisicaevestibular.com.br\/novo\/mecanica\/dinamica\/mhs\/pendulo-simples\/resolucao-comentada-dos-exercicios-de-vestibulares-sobre-pendulo-simples\/","title":{"rendered":"P\u00eandulo Simples – Resolu\u00e7\u00e3o"},"content":{"rendered":"

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

P\u00eandulo Simples<\/b><\/span><\/span><\/span><\/p>\n

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

(01) Falsa\u00a0 — 20 vezes\u00a0 —\u00a0 10s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20T = 10\u00a0 —\u00a0 T = 10\/20\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T = 1\/2 s<\/b><\/span><\/span><\/p>\n

1 vez\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 T<\/b><\/span><\/span><\/p>\n

(02) Falsa\u00a0 —\u00a0 f = 1\/T\u00a0 —\u00a0 f =1\/1\/2<\/b><\/span><\/span>\u00a0\u00a0 —\u00a0 f = 2 Hz<\/b><\/span><\/span><\/p>\n

(04)<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/span>Verdadeira\u00a0 —\u00a0 T = 2p\u00d6L\/g\u00a0 —\u00a0 Observe que T \u00e9 diretamente proporcional \u00e0 raiz quadrada de L.<\/b><\/span><\/span><\/p>\n

(08)\u00a0 Falsa\u00a0 —\u00a0 Observe na equa\u00e7\u00e3o T = 2p\u00d6L\/g\u00a0 que o per\u00edodo T independe da massa m<\/b><\/span><\/span><\/p>\n

(16) Falsa\u00a0 —\u00a0\u00a0 Observe que o per\u00edodo T \u00e9 inversamente proporcional \u00e0 raiz quadrada de g e consequentemente diretamente proporcional \u00e0 freq\u00fc\u00eancia, pois f = 1\/T.<\/b><\/span><\/span><\/p>\n

(32) Verdadeira\u00a0 —\u00a0\u00a0 Observe na equa\u00e7\u00e3o T = 2p\u00d6L\/g\u00a0 que o per\u00edodo T independe da amplitude A\u00a0<\/b><\/span><\/span><\/p>\n

Soma \u2013 (04 +32) = 36<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

a)\u00a0 20 vezes — 30s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20T=30\u00a0 —\u00a0\u00a0 T = 1,5s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 f=1\/T\u00a0 —\u00a0 f=1\/1,5\u00a0 —\u00a0 f = 0,67Hz<\/b><\/span><\/span><\/p>\n

1\u00a0 vez\u00a0\u00a0 —\u00a0\u00a0 T<\/b><\/span><\/span><\/p>\n

b) Observando-se que em t = 0, x = + A, temos o gr\u00e1fico senoidal a seguir.<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

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

(01) Verdadeira\u00a0 —\u00a0 T = 2p\u00d6L\/g\u00a0 —\u00a0 T = 2p\u00d62,5\/10\u00a0 —\u00a0 T = 2p.0,5\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T =\u00a0p\u00a0s<\/b><\/span><\/span><\/p>\n

(02) Falsa\u00a0 —\u00a0 ponto mais alto \u2013 h=1m\u00a0 — E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=mgh\u00a0 —\u00a0 E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=30.10.1\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>=300J<\/b><\/span><\/span><\/p>\n

04) Falsa\u00a0 —\u00a0\u00a0 A energia mec\u00e2nica \u00e9 constante e vale 300J (vide 2). No ponto mais baixo\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>\u00a0= E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —<\/b><\/span><\/span><\/p>\n

300= mV<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 +\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 mgh\u00a0 —\u00a0\u00a0 300 = 30V<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 30.10.0,5\u00a0 —\u00a0 V =\u00a0\u00d6150\/15\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>V =\u00a0\u00d610 m\/s<\/b><\/span><\/span><\/p>\n

(08) Falsa. Num p\u00eandulo simples o per\u00edodo T de oscila\u00e7\u00e3o independe da massa m.<\/b><\/span><\/span><\/p>\n

(16) Falsa . O per\u00edodo independe da altura (amplitude A).<\/b><\/span><\/span><\/p>\n

Soma (\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ) 01 + 04 = 05<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

a) Como o comprimento do p\u00eandulo aumenta, pois ele se dilata devido \u00e0 eleva\u00e7\u00e3o da temperatura e como raiz quadrada de L \u00e9 diretamente proporcional \u00e0 T, o per\u00edodo aumentar\u00e1 e o rel\u00f3gio atrasar\u00e1.<\/b><\/span><\/span><\/p>\n

b) Sendo T inversamente proporcional a\u00a0\u00d6g e como g diminui (g<\/b><\/span><\/span>Lua<\/b><\/span><\/span><\/sub>\u00a0<\u00a0g<\/b><\/span><\/span>Terra<\/b><\/span><\/span><\/sub>), o per\u00edodo aumentar\u00e1 e o rel\u00f3gio atrasar\u00e1<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

O V, pois sendo T diretamente proporcional a\u00a0\u00d6L (L=comprimento do fio), para T cair pela metade (de 2s para 1s), L dever\u00e1 ser 4 vezes menor (100\/4=25)<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

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

\"\"<\/p>\n

T = 2p\u00d667\/10\u00a0 —\u00a0 T =2p\u00d667\/\u00d610\u00a0 —\u00a0 T = 2p8\/p\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>T =16s<\/b><\/span><\/span><\/p>\n

b) Permaneceria o mesmo, pois o per\u00edodo do p\u00eandulo simples n\u00e3o depende da massa pendular.<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

R: Comprimento L\u00a0 —\u00a0 T=2s<\/b><\/span><\/span><\/p>\n

Comprimento L\/4\u00a0 —\u00a0 T\u2019=2p\u00d6L\/4 \/\u00a0\u00d6g\u00a0 —\u00a0 T\u2019=2p\u00d6L\/g.1\/2\u00a0 — T\u2019=T\/2\u00a0 —\u00a0 T\u2019=2\/2\u00a0 —\u00a0 T\u2019=1s<\/b><\/span><\/span><\/p>\n

Observe na figura abaixo que o per\u00edodo (tempo que demora para ir e voltar) entre A e B \u00e9 T\/2=2\/2=1s e que o per\u00edodo entre B e C \u00e9 T\u2019\/2=1\/2=0,5s<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Assim, L demora 0,5s para ir de A at\u00e9 B; L\/4 demora 0,25s para ir de B a C; L\/4 demora 0,25s para ir de C a B e L demora 0,5s para ir de B a A.<\/b><\/span><\/span><\/p>\n

Portanto o per\u00edodo pedido \u00e9 1 + 0,5 = 1,5s<\/b><\/span><\/span><\/p>\n

 <\/p>\n

8-: C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

Sendo f = 1\/T\u00a0 —\u00a0 T<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0>\u00a0T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0>\u00a0T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 como o local \u00e9 o mesmo, g \u00e9 a mesma e como T \u00e9 diretamente proporcional a\u00a0\u00d6L\u00a0 —<\/b><\/span><\/span><\/p>\n

L<\/b><\/span><\/span>3<\/b><\/span><\/span><\/sub>\u00a0>\u00a0L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0>\u00a0L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0 \u00a0\u00a0\u00a0R: D<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

 <\/p>\n

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

a) T=2p\u00d6L\/g\u00a0 —\u00a0 0,8= 2p\u00d6L\/g\u00a0\u00a0\u00a0 —\u00a0 (0,8)<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>= 4.p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.L\/g\u00a0 —\u00a0 0,64=4g.L\/g\u00a0 —\u00a0 L=0,16m=16cm\u00a0 —\u00a0 h=16 + 5 =\u00a0<\/b><\/span><\/span>21 cm<\/b><\/span><\/span>
\nb) ) T=2p\u00d6L\/g\u00a0 —\u00a0 5= 2p\u00d6L\/g\u00a0\u00a0\u00a0 —\u00a0 (5)<\/b><\/span><\/span>2\u00a0<\/b><\/span><\/span><\/sup>= 4.p<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.L\/g\u00a0 —\u00a0 25=4g.L\/g\u00a0 —\u00a0 L=6,25m\u00a0\u00a0 —\u00a0 h=6,25 + 0,25 = 6,5m<\/b><\/span><\/span><\/p>\n

Sim, o rel\u00f3gio teria de ter mais que 6,5 m<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

6T<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 4T<\/b><\/span><\/span>2\u00a0 —\u00a0\u00a0<\/b><\/span><\/span><\/sub>6. 2p\u00d6L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\/g = 4. 2p\u00d6L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/g\u00a0 —\u00a0\u00a0\u00d6L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 6\/4\u00a0 —\u00a0 (\u00d6L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0=\u00a0 36 \/ 16\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>L<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sub>\u00a0\/ L<\/b><\/span><\/span>1<\/b><\/span><\/span><\/sub>\u00a0= 9 \/ 4<\/b><\/span><\/span><\/p>\n

 <\/p>\n

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

T=2p\u00d610\/10\u00a0 —\u00a0 T=2ps\u00a0 —\u00a0\u00a0p@3\u00a0 — T@6s<\/b><\/span><\/span><\/p>\n

86 batidas —- 60s<\/b><\/span><\/span><\/p>\n

X batidas —– 6s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

X\u00a0@\u00a09 batidas<\/b><\/span><\/span><\/p>\n

14-\u00a0<\/b><\/span><\/span>C\u00e1lculo do per\u00edodo T de cada oscila\u00e7\u00e3o\u00a0 —\u00a0 w=2\u03c0\/T\u00a0 —\u00a0 T=2\u03c0\/w=2\u03c0\/5\u03c0\/2\u00a0 —\u00a0 T=0,8s\u00a0 —\u00a0 tempo (\u2206t) gasto em n=3.600 oscila\u00e7\u00f5es\u00a0 —\u00a0 \u2206t=nT=3.600×0,8\u00a0 —\u00a0 \u2206t=2.880\/60 min\u00a0 —\u00a0 \u2206t=48 min\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

15-\u00a0<\/b><\/span><\/span>Per\u00edodo de oscila\u00e7\u00e3o (T) de um p\u00eandulo simples de comprimento L em um local onde o campo gravitacional tem intensidade g, para oscila\u00e7\u00f5es de pequenas amplitudes\u00a0 —\u00a0 T=2\u03c0\u221aL\/g\u00a0 —\u00a0 observando essa express\u00e3o voc\u00ea conclui que, quando a intensidade do campo gravitacional diminui o per\u00edodo aumenta, ou seja, o p\u00eandulo passa a oscilar mais vagarosamente\u00a0 —\u00a0 na aus\u00eancia total de gravidade, o p\u00eandulo teria per\u00edodo infinito, ou seja, deixaria de oscilar\u00a0 —\u00a0 para um sistema massa-mola (m e k) o per\u00edodo de oscila\u00e7\u00e3o (T) \u00e9 dado por\u00a0 —\u00a0 T=2\u03c0\u221am\/k\u00a0 —\u00a0 observe nessa express\u00e3o que o per\u00edodo desse sistema independe da gravidade local\u00a0 —\u00a0 assim, nesse ambiente de microgravidade, o per\u00edodo do sistema bloco-mola n\u00e3o sofrer\u00e1 altera\u00e7\u00e3o, j\u00e1 o per\u00edodo do p\u00eandulo simples deixar\u00e1 de ser o mesmo\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- B<\/b><\/span><\/span>\u00a0<\/b><\/span><\/span><\/p>\n

16-\u00a0<\/b><\/span><\/span>Novo comprimento \u2113 do p\u00eandulo quando a temperatura subir de 10<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>C\u00a0 —\u00a0 \u00a0\u2113 = \u2113<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0(1 + \u03b1(t \u2013 t<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 \u2113 = 10(1 + 10.10<\/b><\/span><\/span>-6<\/b><\/span><\/span><\/sup>.10\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u2113=1,0001m\u00a0 —\u00a0 c\u00e1lculo do per\u00edodo T que pelo enunciado vale T= \u221a(4\u2113)\u00a0 —\u00a0 T= \u221a(4.1,0001)\u00a0 —\u00a0 T= 2.\u221a1,0001 s\u00a0 —\u00a0 para meio ciclo de oscila\u00e7\u00e3o o per\u00edodo ser\u00e1 T=\u221a1,0001 s\u00a0 —\u00a0 se, para cada meia oscila\u00e7\u00e3o o per\u00edodo aumenta de T=\u221a1,0001 s, o rel\u00f3gio ir\u00e1 atrasar\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>R- C<\/b><\/span><\/span><\/p>\n

17-\u00a0<\/b><\/span><\/span>A\u00a0 —\u00a0 amplitude do p\u00eandulo\u00a0 —\u00a0 T\u00a0 —\u00a0 per\u00edodo do p\u00eandulo\u00a0 —\u00a0 T=2\u03c0\u221a\u2113\/g\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

01) Falsa\u00a0 —\u00a0 o per\u00edodo de um p\u00eandulo simples \u00e9 inversamente proporcional \u00e0 raiz quadrada da gravidade local (g) conforme mostra a express\u00e3o\u00a0 —\u00a0 T=2\u03c0\u221a\u2113\/g\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

02) Falsa\u00a0 —\u00a0 quadruplicando o comprimento (L) de um p\u00eandulo simples, seu per\u00edodo duplica, pois o per\u00edodo \u00e9 diretamente proporcional \u00e0 raiz quadrada do comprimento\u00a0 —\u00a0 veja a equa\u00e7\u00e3o\u00a0 —\u00a0 T=2\u03c0\u221a\u2113\/g\u00a0 \u00a0<\/b><\/span><\/span><\/p>\n

04) Falsa\u00a0 — observe na figura baixo a amplitude A, que \u00e9 definida como sendo a dist\u00e2ncia m\u00e1xima (A) (ou o \u00e2ngulo m\u00e1ximo \u03b8)<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

\u00a0que a part\u00edcula se afasta de sua posi\u00e7\u00e3o de equil\u00edbrio\u00a0 —\u00a0 na posi\u00e7\u00e3o central, a energia mec\u00e2nica \u00e9 igual \u00e0 energia cin\u00e9tica onde<\/b><\/span><\/span><\/p>\n

a velocidade \u00e9 m\u00e1xima e a energia potencial \u00e9 nula\u00a0 —\u00a0 a energia mec\u00e2nica \u00e9 constante em todos os pontos (sistema conservativo) e, na posi\u00e7\u00e3o central vale E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=E<\/b><\/span><\/span>c<\/b><\/span><\/span><\/sub>\u00a0+ E<\/b><\/span><\/span>p<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=mv<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2 + 0\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=mv<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 a velocidade no ponto central \u00e9 m\u00e1xima e vale\u00a0 —\u00a0 v=w.A\u00a0 —\u00a0 w=2\u03c0\/T\u00a0 —\u00a0 v= (2\u03c0\/T).A\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=m.v<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/2\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=m\/2.(4\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>).A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 E<\/b><\/span><\/span>m<\/b><\/span><\/span><\/sub>=2m\u03c0<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>.A<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\/T<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 em todos os pontos a energia mec\u00e2nica \u00e9 a mesma e observe na express\u00e3o que ela \u00e9 diretamente proporcional ao quadrado da amplitude A.<\/b><\/span><\/span><\/p>\n

09) Verdadeira\u00a0 —\u00a0 veja teoria<\/b><\/span><\/span><\/p>\n

16) Verdadeira\u00a0 —\u00a0 veja teoria.<\/b><\/span><\/span><\/p>\n

R- (08 + 16) = 24<\/b><\/span><\/span><\/p>\n

18-\u00a0<\/b><\/span><\/span>O per\u00edodo T do p\u00eandulo \u00e9 dado por T=2\u03c0\u221a(L\/g)\u00a0 —\u00a0 como o rel\u00f3gio est\u00e1 atrasando seu per\u00edodo \u00e9 maior do que deveria ser e assim, seu comprimento deve ser reduzido com o ajuste da porca sendo para a direita\u00a0 — se, para um dia (1.440min) o rel\u00f3gio atrasa 1 min, ent\u00e3o o p\u00eandulo deve registrar um tempo de 1.439min\u00a0 —\u00a0 \u2206t=n<\/b><\/span><\/span>atraso<\/b><\/span><\/span><\/sub>.T<\/b><\/span><\/span>atraso<\/b><\/span><\/span><\/sub>=n<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>.T<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>\u00a0 — \u00a01439T<\/b><\/span><\/span>atraso<\/b><\/span><\/span><\/sub>=1.440T<\/b><\/span><\/span>correto\u00a0<\/b><\/span><\/span><\/sub>\u00a0—\u00a0 1439. 2\u03c0\u221a(L<\/b><\/span><\/span>atraso<\/b><\/span><\/span><\/sub>\/g) =1.440. 2\u03c0\u221a(L<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>\/g)\u00a0 —\u00a0 L<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>atraso<\/b><\/span><\/span><\/sub>=(1.439\/1.440)<\/b><\/span><\/span>2<\/b><\/span><\/span><\/sup>\u00a0 —\u00a0 L<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>\/35 = 0,9986\u00a0 —\u00a0 L<\/b><\/span><\/span>correto<\/b><\/span><\/span><\/sub>=34,95cm\u00a0 —\u00a0<\/b><\/span><\/span><\/p>\n

\u2206L=35 \u2013 34,95=0,05cm=0,5mm\u00a0 —\u00a0 como cada rota\u00e7\u00e3o provoca varia\u00e7\u00e3o de 1 mm, para que o rel\u00f3gio funcione corretamente deve ser dada meia rota\u00e7\u00e3o \u00e0 direita\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0R- C<\/b><\/span><\/span><\/p>\n

 <\/p>\n

19-<\/b><\/span><\/span>P\u00eandulo Simples –\u00a0 consta de uma massa m, presa na extremidade inferior de um fio ideal, fixada verticalmente na sua extremidade superior (figura)\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Se o p\u00eandulo simples oscilar, com oscila\u00e7\u00f5es de pequena abertura (no m\u00e1ximo 15<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sup>), ele descreve um movimento circular de raio R=L, sendo L o comprimento do fio.<\/b><\/span><\/span><\/p>\n

Seu per\u00edodo (T), que \u00e9 o tempo que ele demora para efetuar um \u201cvai e vem\u201d completo \u00e9 fornecido pela express\u00e3o:<\/b><\/span><\/span><\/p>\n

\"\"<\/p>\n

Observe nessa express\u00e3o que o per\u00edodo T do p\u00eandulo depende da acelera\u00e7\u00e3o da gravidade g e do comprimento do p\u00eandulo L\u00a0 —\u00a0o per\u00edodo de um p\u00eandulo simples \u00e9 diretamente proporcional \u00e0 raiz quadrada de seu comprimento L\u00a0 — assim, para dobrar o per\u00edodo T de um p\u00eandulo, seu comprimento L deve ser quadruplicado e o rel\u00f3gio atrasa\u00a0 — \u00a0o per\u00edodo de um p\u00eandulo simples \u00e9 inversamente proporcional \u00e0 raiz quadrada da acelera\u00e7\u00e3o da gravidade g\u00a0 —\u00a0 assim,\u00a0 quanto maior for a acelera\u00e7\u00e3o da gravidade do local onde est\u00e1 o p\u00eandulo, menor ser\u00e1 o seu per\u00edodo e o rel\u00f3gio adianta\u00a0 —\u00a0 uma das aplica\u00e7\u00f5es do p\u00eandulo simples \u00e9 a determina\u00e7\u00e3o da acelera\u00e7\u00e3o da gravidade\u00a0 —\u00a0 per\u00edodo ao n\u00edvel do mar\u00a0 —\u00a0 T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>=2\u03c0\u221a(L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 per\u00edodo numa grande altura\u00a0 —\u00a0 T<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>==2\u03c0\u221a(L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 para que ele funcione bem nos dois locais\u00a0 —\u00a0 T<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= T<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 =2\u03c0\u221a(L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>) =2\u03c0\u221a(L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0<\/b><\/span><\/span>\u00a0L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>.<\/b><\/span><\/span><\/p>\n

\u2022 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0= L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0=g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 comprimento ao n\u00edvel do mar (L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\u00a0 —\u00a0 comprimento na altitude h (L<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>=L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(1 + \u03b1.\u2206t)\u00a0 —\u00a0 L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/[(L<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>(1 + \u03b1.\u2206t)] = g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 1\/(1 + \u03b1.\u2206t) = g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\/g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0 g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>\u00a0+ g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u03b1.\u2206t = g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0 —\u00a0\u00a0<\/b><\/span><\/span>\u03b1 = (g<\/b><\/span><\/span>h<\/b><\/span><\/span><\/sub>\u00a0\u2013 g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>)\/g<\/b><\/span><\/span>o<\/b><\/span><\/span><\/sub>.\u00a0<\/b><\/span><\/span>\u2206<\/b><\/span><\/span>t.<\/b><\/span><\/span><\/p>\n

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

Resolu\u00e7\u00e3o comentada dos exerc\u00edcios de vestibulares sobre P\u00eandulo Simples 01- (01) Falsa\u00a0 — 20 vezes\u00a0 —\u00a0 10s\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 20T = 10\u00a0 —\u00a0 T = 10\/20\u00a0 —\u00a0\u00a0T = 1\/2 s 1 vez\u00a0\u00a0\u00a0\u00a0 —\u00a0\u00a0\u00a0 T (02) Falsa\u00a0 —\u00a0 f = 1\/T\u00a0 —\u00a0 f =1\/1\/2\u00a0\u00a0 —\u00a0 f = 2 Hz (04)\u00a0Verdadeira\u00a0 —\u00a0 T = 2p\u00d6L\/g\u00a0 —\u00a0 Observe que T \u00e9 diretamente proporcional \u00e0 raiz<\/p>\n","protected":false},"author":1,"featured_media":0,"parent":1269,"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-1273","page","type-page","status-publish","hentry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1273","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=1273"}],"version-history":[{"count":3,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1273\/revisions"}],"predecessor-version":[{"id":10843,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1273\/revisions\/10843"}],"up":[{"embeddable":true,"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/pages\/1269"}],"wp:attachment":[{"href":"https:\/\/fisicaevestibular.com.br\/novo\/wp-json\/wp\/v2\/media?parent=1273"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}