Problem 2:
0
The mass is at rest so focus of gravity (mg) equals the force of the spring,
4
1
4
1
4
24 = 𝑘 ( )
12
𝑘 = 72
400 = 𝑘 (2)
𝑘 = 200
4
1
16
2
3
𝑘
𝑥=0
𝑚
𝑥 ′′ + 64 𝑥 = 0
𝑚 2 + 64 = 0
𝑚= ±8i
𝑥 ′′ +
𝑥(𝑡) = 𝑐1 cos 8𝑡 + 𝑐2 sin 8𝑡
𝑥 ′ (𝑡) = −8 𝑐1 sin 8𝑡 + 8𝑐2 cos 8𝑡
𝑥(0) =
2
3
2
= 𝑐1 cos 8(0) + 𝑐2 sin 8(0)
3
2
𝑐1 =
3
𝑥′(0) = −
4
3
4
= −8 𝑐1 sin 8(0) + 8𝑐2 cos 8(0)
3
1
𝑐2 = −
6
−
2
1
3
6
𝑥(𝑡) = cos 8𝑡 − sin 8𝑡
√17
6
(-1.326)
4
π
8
2λ
ω2
4
𝑚=
𝐹
𝑔
𝑚=
8
32
(8) = 𝑘(2)
𝑚=
1
4
𝑘=4
𝐹 = 𝑘𝑠
𝛽 ′ 𝑘
𝑥 + 𝑥=0
𝑚
𝑚
2
4
𝑥 ′′ + 𝑥 ′ + 𝑥 = 0
1
1
4
4
𝑥 ′′ + 8𝑥 ′ + 16𝑥 = 0
𝑥 ′′ +
AE:
𝑚2 + 8𝑚 + 16 = 0
(𝑚 + 4)2 = 0
𝑚 = −4, −4
-3
0
16
4
General solution:
𝑥(𝑡) = 𝐶1 𝑒 −4𝑡 + 𝐶2 𝑡𝑒 −4𝑡
𝑥 ′ (𝑡 )
= −4𝐶1 𝑒 −4𝑡 + (𝐶2 𝑒 −4𝑡
− 4𝐶2 𝑡𝑒 −4𝑡 )
𝑥 (0) = 0 ,
0 = 𝐶1 𝑒 −4(0) + 𝐶2 (0)𝑒 −4(0)
𝑐1 = 0
𝑥 ′ (0) = −3
−3 = −4(0)𝑒 −4(0) + (𝐶2 𝑒 −4(0)
− 4𝐶2 (0)𝑒 −4(0) )
𝑐2 = −3
General solution:
𝑥(𝑡) = −3 𝑡𝑒 −4𝑡
0