Homework # 2, PHYS 122 Spring 2009
P14.7. Model: The air-track glider attached to a spring is in simple harmonic motion. The glider is
pulled
to the right and released from rest at t = 0 s. It then oscillates with a period T = 2.0 s and a maximum
speed 4v max = 0 cm/s = 0.40 m/s. While the amplitude of the oscillation can be obtained from Equation 14.13,
2π t
the position of the glider can be obtained from Equation 14.10, x(t) = A cos ( T ).
Solve: (a)
vmax = (2π A/T ) ⇒ A =
vmaxT (0.40 m/s)(2.0 s)
=
= 0.127 m = 0.13 m
2π
2π
(b) The glider’s position at t = 0.25 s is
⎡ 2 π (0.25 s) ⎤
x0. 25 s = (0.127 m ) cos ⎢
⎥ = 0.090 m = 9.0 c m
2.0 s ⎥⎦
⎢⎣
Assess: At t = 0.25 s, which is less than one quarter of the time period, the object has not reached the
equilibrium position and is still moving toward the left.
P14.9. Prepare: Please refer to Figure P14.9. The oscillation is the result of simple harmonic motion.
As the graph shows, the time to complete one cycle (or the period) is T = 4.0 s. We will use Equation 14.1
to find frequency.
Solve: (a) The amplitude A = 20 cm.
(b) The period T = 4.0 s, thus
f =
Assess:
1
1
=
= 0.25 Hz
T 4.0 s
It is important to know how to find information form a graph.
Treating the building as an oscillator, the magnitude of the maximum displacement
is the amplitude, the magnitude of the maximum velocity is determined by vmax = 2π fA, and the magnitude
P14.12.
Prepare:
2 2
of the maximum acceleration is determined by amax = 2π f A.
The magnitude of the maximum displacement is x max = A = 0.30 m.
The magnitude of the maximum velocity is vmax = 2π fA = 2π (1.2 Hz)(0.30 m) = 2.3m/s.
Solve:
2 2
2
2
2
The magnitude of the maximum acceleration is amax = 2π f A = 2π (1.2 Hz) (0.30 m) = 8.5 m/s .
Assess: These are reasonable values for the magnitude of the maximum displacement, velocity, and
acceleration.
Comment: This problem was optional.
1 Homework # 2, PHYS 122 Spring 2009
P14.14. Prepare: In order to get a good handle on what is happening to the x-component of the
position, velocity and force, let’s look at the figures below. Figure (a) shows the merry-go-round rotating
clockwise and the time every quarter of a period, (b) shows the displacement at each quarter of a period,
(c) shows the velocity at each quarter of a period, and (d) shows the force at each quarter of a period.
Finally, Figure (e) keeps track of the x-component of these quantities for a time interval of 2T.
The magnitude of the maximum displacement is determined by x max = A = 3.0 m.
The magnitude of the maximum velocity is determined by vmax = 2π A/T = 2π (3.0 m)/10 s = 1.9 m/s.
The magnitude of the maximum force is determined by
Solve:
Fmax = m amax = m4π 2 A/T 2 = 4π 2 (25.0 kg)(3.0 m)/(10 s) 2 = 30 N
2 Homework # 2, PHYS 122 Spring 2009
Assess: This problem helps us visualize that the x-component of the motion of a particle on the
reference circle is simple harmonic motion.
P14.15. Prepare: The spring undergoes simple harmonic motion. The elastic potential energy in a
spring stretched by a distance x from its equilibrium position is given by Equation 14.20, and the total
mechanical energy of the object is the sum of kinetic and potential energies as in Equation 14.21. At
1
maximum displacement, the total energy is simply E = 2 kA2, Equation 14.22.
Solve:
1
(a). When the displacement is x = 2 A, the potential energy is
U=
1 2 1 ⎛ 1 ⎞2 1 ⎛ 1 2 ⎞ 1
3
kx = k ⎜ A⎟ = ⎜ kA ⎟ = E ⇒ K = E − U = E
⎠ 4
2
2 ⎝2 ⎠
4⎝ 2
4
Thus, one quarter of the energy is potential and three-quarters is kinetic.
1
(b) To have U = 2 E requires
U=
1 2 1
1 ⎛1
A
⎞
kx = E = ⎜ kA2 ⎟ ⇒ x =
2
2
2 ⎝2
⎠
2
P14.17. Prepare: The block attached to the spring is in simple harmonic motion. The period of an
oscillating mass on a spring is given by Equation 14.27.
Solve: The period of an object attached to a spring is
T = 2π
where m is the mass and k is the spring constant.
3 m
= T0 = 2.0 s
k
Homework # 2, PHYS 122 Spring 2009
(a) For mass = 2m,
T = 2π
2m
= ( 2 )T0 = 2.8 s
k
1
(b) For mass 2 m,
T = 2π
1
2
m
= T0/ 2 = 1.4 s
k
(c) The period is independent of amplitude. Thus T = T0 = 2.0 s.
(d) For a spring constant = 2k,
T = 2π
Assess:
m
= T0/ 2 = 1.4 s
2k
As would have been expected, increase in mass leads to slower simple harmonic motion.
Comment: This problem was optional.
P14.16. Prepare: The block attached to the spring is in simple harmonic motion. At maximum
displacement position, x = A, and the kinetic energy is zero. We can use the energy conservation Equation
14.21 to find the amplitude.
Solve:
(a) The conservation of mechanical energy equation Kf = Uf = Ki = Ui is
1
2
mv21 +
1
2
2
k (Δx ) =
1
2
mv20 + 0 J ⇒ 0 J +
⇒ A=
m
v =
k 0
1
2
1
2
mv02 + 0 J
1.0 kg
(0.40 m/s) = 0.10 m = 10.0 cm
16 N/m
(b) We have to find the velocity at a point where x
equation K2 + U2 = Ki + Ui is
4 2
kA =
A/2. The conservation of mechanical energy
Homework # 2, PHYS 122 Spring 2009
2
1 2 1 ⎛ A⎞ 1 2
⎞ 3⎛1
⎞
1
1
1 ⎛1 2⎞ 1
1⎛1
mv2 + k ⎜ ⎟ = mv0 + 0 J ⇒ mv 22 = mv02 − ⎜ kA ⎟ = mv02 − ⎜ mv20 ⎟ = ⎜ mv20 ⎟
2
⎠ 2
⎠ 4⎝2
⎠
2 ⎝2⎠ 2
2
2
4 ⎝2
4⎝2
⇒ v2 =
3
3
v =
(0.40 m/s) = 0.346 m/s = 35 cm/s
4 0
4
Assess: Speed decreases as an object moves away from the equilibrium position, so a decreased speed
of 35 m/s compared to 40 m/s at the equilibrium position is reasonable.
5