# Física teórica e experimental

Páginas: 7 (1721 palavras) Publicado: 13 de novembro de 2011
Trabalho de campo (1/Av2)

Equação geral do Oscilador Harmônico Simples.

x(t) = A . cos (wt + Φ)

Quando temos:

Amplitude (A) = 28 m

Frequência angular (w) = 19π/2 . rad⁄s
Ângulo de fase (Φ) = 10π/13 .rad

Aplicando os valores na equação geral:

x(t) = 28 . cos (19π/2 .t+ 10π/13)

A amplitude é máxima em x(t) quando o cos (19π/2 .t+ 10π/13) = 1(19π/2 .t+ 10π/13) = Ө

cos Ө = 1 quando Ө = 0, 2π, 4π, (...).

Logo os valores de t para:

Ө = 0
19π/2 .t+ 10π/13 = 0
19π/2 .t = -10π/13
t= -10π/13 . 2/19π
t= (-20π)/247π
t= (-20)/247 s

Ө = 2π
19π/2 .t+ 10π/13 = 2π
19π/2 .t = 2π-10π/13
t= 16π/13 . 2/19π
t= 32π/247π
t= 32/247 s

Ө = 4π
19π/2 .t+ 10π/13 = 4π
19π/2 .t = 4π-10π/13
t= 42π/13 . 2/19π
t=84π/247π
t= 84/247 s

A amplitude é mínima em x(t) quando cos (19π/2 .t+ 10π/13) = -1
(19π/2 .t+ 10π/13) = Ө

cos Ө = -1 quando Ө = π, 3π, 5π, (...).

Logo os valores de t para:

Ө = π
19π/2 .t+ 10π/13 = π
19π/2 .t = π-10π/13
t= 3π/13 . 2/19π
t= 6π/247π
t= 6/247 s

Ө = 3π
19π/2 .t+ 10π/13 = 3π
19π/2 .t = 3π-10π/13
t= 29π/13 . 2/19π
t= 58π/247π
t= 58/247 sӨ = 5π
19π/2 .t+ 10π/13 = 5π
19π/2 .t = 5π-10π/13
t= 55π/13 . 2/19π
t= 110π/247π
t= 110/247

A amplitude é zero em x(t) quando cos (19π/2 .t+ 10π/13) = 0
(19π/2 .t+ 10π/13) = Ө

cos Ө = 0 quando Ө = π/2, 3π/2, 5π/2, 7π/2, 9π/2,(...).

Logo os valores de t para:

Ө = π/2
19π/2 .t+ 10π/13 = π/2
19π/2 .t = π/2-10π/13
t= (-7π)/26 . 2/19π
t= (-14π)/494π
t=〖-14〗^(÷2)/〖494〗^(÷2) s
t= (-7)/247 s

Ө = 3π/2
19π/2 .t+ 10π/13 = 3π/2
19π/2 .t = 3π/2-10π/13
t= 19π/26 . 2/19π
t= 38π/494π
t= 〖38〗^(÷2)/〖494〗^(÷2)
t= 1/13 s

Ө = 5π/2
19π/2 .t+ 10π/13 = 5π/2
19π/2 .t = 5π/2-10π/13
t= 45π/26 . 2/19π
t= 90π/494π
t= 〖90〗^(÷2)/〖494〗^(÷2)
t= 45/247 s

Ө = 7π/2
19π/2 .t+ 10π/13 = 7π/2
19π/2 .t = 7π/2-10π/13
t= 71π/26 . 2/19π
t=142π/494π
t= 〖142〗^(÷2)/〖494〗^(÷2)
t= 71/247 s
Ө = 9π/2
19π/2 .t+ 10π/13 = 9π/2
19π/2 .t = 9π/2-10π/13
t= 97π/26 . 2/19π
t= 194π/494π
t= 〖194〗^(÷2)/〖494〗^(÷2)
t= 97/247 s

Tabela para construção do gráfico x(t):

t . (s) Ө = (19π/2 .t+ 10π/13) . (rad) cos Ө . (rad/s) x(t) = A . cos Ө . (m)
t = (-20)/247 ≡ -0,081 0 1 28
t = (-7)/247 ≡ -0,028 π/2 0 0
t = 6/247 ≡ 0,024 π -1-28
t = 1/13 ≡ 0,077 3π/2 0 0
t = 32/247 ≡ 0,129 2π 1 28
t = 45/247 ≡ 0,182 5π/2 0 0
t = 58/247 ≡ 0,235 3π -1 -28
t = 71/247 ≡ 0,287 7π/2 0 0
t = 84/247 ≡ 0,340 4π 1 28
t = 97/247 ≡ 0,393 9π/2 0 0
t = 110/247 ≡ 0,445 5π -1 -28

dx/dt(t) = -Aw . sen (19π/2 .t+ 10π/13)

dx/dt(t) = -835,66 . sen (19π/2 .t+ 10π/13)

Avelocidade é máxima em x(t) quando o sen (19π/2 .t+ 10π/13) = -1
(19π/2 .t+ 10π/13) = Ө

sen Ө = -1 quando Ө = 3π/2, 7π/2, 11π/2, (...).
Logo os valores de t para:

Ө = 3π/2
19π/2 .t+ 10π/13 = 3π/2
19π/2 .t = 3π/2-10π/13
t= 19π/26 . 2/19π
t= 38π/494π
t= 〖38〗^(÷2)/〖494〗^(÷2)
t= 1/13 s

Ө = 7π/2
19π/2 .t+ 10π/13 = 7π/2
19π/2 .t = 7π/2-10π/13
t= 71π/26 . 2/19π
t=142π/494π
t= 〖142〗^(÷2)/〖494〗^(÷2)
t= 71/247 s

Ө = 11π/2
19π/2 .t+ 10π/13 = 11π/2
19π/2 .t = 11π/2-10π/13
t= 123π/26 . 2/19π
t= 246π/494π
t= 〖246〗^(÷2)/〖494〗^(÷2)
t= 123/247 s

A velocidade é mínima em dx/dt(t) quando sen (19π/2 .t+ 10π/13) = 1
(19π/2 .t+ 10π/13) = Ө

sen Ө = 1 quando Ө = π/2, 5π/2, 9π/2, (...).

Logo os valores de t para:

Ө = π/2
19π/2 .t+ 10π/13= π/2
19π/2 .t = π/2-10π/13
t= (-7π)/26 . 2/19π
t= (-14π)/494π
t= 〖-14〗^(÷2)/〖494〗^(÷2) s
t= (-7)/247 s

Ө = 5π/2
19π/2 .t+ 10π/13 = 5π/2
19π/2 .t = 5π/2-10π/13
t= 45π/26 . 2/19π
t= 90π/494π
t= 〖90〗^(÷2)/〖494〗^(÷2)
t= 45/247 s

Ө = 9π/2
19π/2 .t+ 10π/13 = 9π/2
19π/2 .t = 9π/2-10π/13
t= 97π/26 . 2/19π
t= 194π/494π
t= 〖194〗^(÷2)/〖494〗^(÷2)
t=...

Por favor, assinar para o acesso.