Exercicios resolvidos

November 16, 2018 | Author: Simpson Bartholomew | Category: Frequency, Hertz, Velocity, Displacement (Vector), Physical Phenomena
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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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