| 1) A 5.0 kg mass is attached to a vertical spring and stretches the
spring 0.10 m. The spring is then pulled down |
| 0.050 m and released. Determine the: |
| (a) amplitude |
| (b) period |
| (c) frequency |
| |
| m = 5.0
kg A = 0.050 m Δx
= 0.10 m |
| |
| (a) The
amplitude, A, is 0.050 m because the mass-spring system was given a maximum
displacement of 0.050
m. |
| |
| (b) Fr
= kx |
| |
|
k = Fr/Δx = Fw/Δx = mg/Δx = 5.0
kg x 9.80 m/s2/0.10 m = 490 N/m |
| |
| T
= 2π(m/k)1/2 = 2π(5.0 kg/490 N/m)1/2 = 0.63
s |
| |
| (c) f =
1/T = T-1 = (0.63 s)-1 = 1.6 /s = 1.6 Hz |
| |
| 2) A 100. g mass hangs from a vertical spring. When it is pulled 10.
cm below its equilibrium |
| position and released, it oscillates with
a period of 2.0 s. |
| |
| (a) What is the maximum velocity of the mass? |
| (b) What is the acceleration when it is 5.0
cm below the equilibrium position? |
| (c) How much will the spring shorten if the
mass is removed? |
| |
m = 100. g A
= 10. cm x 1 m/102 cm = 0.10 m |
| |
| (a) T =
2π(m/k)1/2 |
| |
k = 4π2 x 100. g x 1 kg/103 g/(2.0
s)2 = 0.99 N/m |
| |
| (KE
+ PEe)A = (KE + PEe)eq pos |
| |
| 0
+ 1/2kx2 = 1/2mv2 + 0 |
| |
1/2
x 0.99 N/m x (0.10 m)2 = 1/2 x 100. g
x 1 kg/103 g x v2 |
| |
| v
= 0.31 m/s |
| |
| (b)
F = ma - -kx |
| |
a
= -0.99 N/m x 5.0 cm x 1 m/102 cm/(100.
g 1 kg/103 g) = -.50 m/s2 |
| |
| (c)
F = kx |
| |
x
= F/k = Fw/k = mg/k = 100. g 1kg/103 g
x 9.80 m/s2/0.99 N/m = 0.99 m |
| |
| 3) When a 0.150 kg mass is attached to a vertical spring, it stretches
by 0.0400 m. The spring is then pulled down |
| 0.300 m below the equilibrium position and released.
Determine: |
| (a) force constant of the spring |
| (b) period of the motion |
| (c) mass's displacement as a function of time |
| (d) mass's velocity as a function of time |
| (e) mass's acceleration as a function of time |
| |
| x = 0.0400 m A
= 0.300 m m = 0.150 kg |
| |
| (a) F = kx |
| |
| k
= F/x = Fw/x = mg/x = 0.150 kg x 9.80 m/s2/0.0400
m = 36.8 N/m |
| |
| (b) T = 2π(m/k)1/2
= T = 2π(0.150 kg/36.8 N/m)1/2 = 0.401 s |
| |
| (c) y = Acos(ωt) |
| |
| ω
= 2πf = 2π/T = 2 x 3.14/0.401 s = 15.7 s-1 |
| |
| y
= 0.300cos(15.7t) m |
| |
| (d) vy
= -ωAsin(ωt) |
| |
| vy
= -15.7 x 0.300sin(15.7t) = -4.71sin(15.7t) |
| |
| (e) ay
= --ω2Acos(-ωt) |
| |
| ay
= -(15.7)2 x 0.300cos(15.7t) = -73.9cos(15.7t) |
| |
| 4) A mass attached to a vertical spring is vibrating with a frequency
of 3.0 s-1 and has an amplitude of 16 cm. Determine |
| the: |
| (a) maximum velocity and the maximum acceleration |
| (b) velocity and acceleration when the mass
has a displacement of 7.0 cm |
| (c) time needed to move from the equilibrium
position to a displacement of 13 cm |
| |
| f = 3.0 s-1 A
= 16 cm |
| |
| (a) v = Aωcos(ωt
+ Φ) |
| |
v
= 16 cm x 1 m/102 cm x 2π
x 3.0 s-1 = 3.0 m/s |
| |
| a
= -Aω2sin(ωt + Φ) |
| |
a
= -16 cm x 1 m/102 cm x (2π
x 3.0 s-1)2 = -57 m/s2 |
| |
| Keep
in mind that both the cos(ωt + Φ) and the sin(ωt + Φ)
vary from a minimum of 0 to a |
| maximum
of 1. |
| |
| (b) x = Asin(ωt
+ Φ) |
| |
| 7.0
cm = 16 cm x sinθ |
| |
sin-1θ
= 7.0 cm/16 cm |
| |
| θ
= 26° |
| |
v
= Aωcosθ = 16 cm x 1 m/102 cm
x 2π x 3.0 s-1 x cos 26°
= 2.7 m/s |
| |
| a
= -Aω2sin(ωt + Φ) |
| |
a
= -16 cm x 1 m/102 cm x (2π
x 3.0 s-1)2 x sin 26°
= -25 m/s2 |
| |
| (c) x = Asin(ωt
+ Φ) |
| |
| θ
= ωt |
| |
sin-1θ
= x/A = 13cm/16 cm |
| |
| θ
= 0.948 |
| |
| θ
= ωt
|
| |
| t
= 0.948/2/π/3.0 s-1 = 0.050 s |