Exercicios resolvidos
Short Description
Fisica II Cap 15...
Description
f =
1 1 = = 2, 0H z. T 0, 50s 50s
x = (6, (6, 0 ) cos[( cos[(33π
/ t) + π/3 π/ 3
] t = 2, 0
t = 2, 0 π 19π 19π x(2, (2, 0) = 6, 6, 0 cos[ cos[33π ·2, 0+ ] = 6, 0cos( ) = 6, 0·0, 5 ⇒ x(2, (2, 0) = 3, 3, 0m 3 3
v=
dx π π = −3π · 6, 0 sin[ sin[(3 (3πt πt)) + ] = −18π 18π sin[(3πt sin[(3πt)) + ] dt 3 3
v (2, (2, 0) = −18π 18π sin(
19π 19π ) ≈ −18 · 0, 866 ⇒ v (2, (2, 0) = −16m/s 16m/s 3
dv = −54π2 cos[(3πt) + π/3)] dt
a =
a(2, 0) = −54π2 cos[(3π · 2, 0) + π/3)] = −54 · 0, 5 a(2, 0) = −27m/s2 φ0 =
π 3
f =
ω 3π = ⇒= 1, 5 H z 2π 2π
T =
1 1 = ⇒ T = 0, 67 s f 1, 5
T = 0, 75
f =
ω =
1 1 = ⇒ f = 1, 3 Hz T 0, 75
2π 2π = ⇒ ω = 8, 4 rad/s T 0, 75
E = 12 kx2m 1 N E = · 1, 3 · (2, 4 cm)2 = 3, 744N cm ⇒ E = 0, 037J 2 cm
·
τ = −κφ τ φ
κ = | | κ =
0, 20 N m ⇒ κ = 0, 235 Nm/rad 0, 85 rad
I = 2mR2 /5 I = T = 2π
2 · 95 kg · (0, 15m)2 ⇒ I = 0, 855 kgm2 5
i κ
T = 2π
0, 855 kgm2 ⇒ T = 12 s 0, 235 Nm/rad
T = 2π
L g
gT 2 L = 4π 4
L = L − d
T = 2π
L L d T 2 d = 2π − = 2π − = 2π g g g 4π 2 g
T = 8, 77s
d = 0, 35 m
(8, 85 s)2 0, 35 m − 4π2 9, 8 m/s2
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