Homework 1 Physics Problems
2.
The spring constant is found from the ratio of applied force to displacement.
k=
Fext mg (66 kg)(9.80 m/s2 )
=
=
= 1.294 ´ 105 N/m
-3
x
x
5.0 ´10 m
The frequency of oscillation is found from the total mass and the spring constant.
f =
5.
(a)
1
2p
The spring constant is found from the ratio of applied force to displacement.
k=
(b)
Fext mg (2.4 kg)(9.80 m/s2 )
=
=
= 653 N/m » 650 N/m
x
x
0.036 m
The amplitude is the distance pulled down from equilibrium, so A = 2.1 cm .
The frequency of oscillation is found from the oscillating mass and the spring constant.
f =
15.
k
1 1.294 ´105 N/m
=
= 1.362 Hz » 1.4 Hz
m 2p
1766 kg
1
2p
k
1
=
m 2p
653 N/m
= 2.625 Hz » 2.6 Hz
2.4 kg
(a)
At equilibrium, the velocity is its maximum, as given in Eq. 11–7.
(b)
k
A = w A = 2p fA = 2p (2.2 Hz)(0.15 m) = 2.073 m/s » 2.1 m/s
m
From Eq. 11–5b, we find the speed at any position.
umax =
u = umax 1(c)
(d)
23.
x2
A2
= (2.073 m/s) 1-
(0.10 m)2
(0.15 m)2
= 1.545 m/s » 1.5 m/s
2
Etotal = 12 mumax
= 12 (0.25 kg)(2.073 m/s)2 = 0.5372 J » 0.54 J
Since the object has a maximum displacement at t = 0, the position will be described by the cosine function.
The period of the jumper’s motion is T =
43.0 s
= 6.143 s. The spring constant can then be found from the period and the
7 cycles
jumper’s mass.
T = 2p
m
4p 2 m 4p 2 (65.0 kg)
® k=
=
= 68.004 N/m » 68.0 N/m
k
T2
(6.143 s)2
The stretch of the bungee cord needs to provide a force equal to the weight of the jumper when he is at the equilibrium point.
kDx = mg ® Dx =
mg (65.0 kg)(9.80 m/s2 )
=
= 9.37 m
k
68.004 N/m
Thus the unstretched bungee cord must be 25.0 m - 9.37 m = 15.6 m .
x = (0.15 m) cos (2p(2.2 Hz)t) ®
29.
x = (0.15 m) cos (4.4p t)
(a)
The period is given by T = 50 s/28 cycles = 1.8 s .
(b)
The frequency is given by f = 28 cycles/50 s = 0.56 Hz .