NT9A-RT1: MASS ON A VERTICAL SPRING—TIME
The figures below show six identical masses attached to springs and hung vertically. The masses are
pulled down various distances and then released. The spring constant (k), which measures the stiffness of
the spring, and the distance (d) that the mass is pulled down are given for each case in the figures.
A
B
C
D
E
F
d = 60 cm
k = 240 N/m
d = 90 cm
k = 180 N/m
d = 50 cm
k = 150 N/m
d = 90 cm
k = 150 N/m
d = 50 cm
k = 240 N/m
d = 50 cm
k = 180 N/m
Rank these situations on the basis of the time for the mass to get from the maximum distance below
the equilibrium point to the maximum distance above the equilibrium point.
Greatest
1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least
OR, The times are the same for all these cases.
___
OR, The times are zero for all these cases.
___
OR, We cannot determine the ranking for the times of these cases.
___
Please explain your reasoning.
Answer: C = D > B = F > A = E
The time needed is half of the period of oscillation. Factors that affect the period for this system are the
mass of the oscillating object and the stiffness of the spring, characterized by the spring constant. For a
mass on a spring the period is proportional to the inverse square root of the spring constant. The
amplitude and period are independent of each other for simple harmonic motion (SHM) so the
distance d does not matter.
NT9A-RT2: MASS ON A VERTICAL SPRING—AMPLITUDE
The figures below show six metal pieces attached to identical springs and hung vertically. The metal
pieces are pulled down various distances and released. In each figure the kinetic energy of the metal piece
as it passes through the equilibrium point and the mass of the piece are given.
A
KE = 12 J
m= 130 g
B
KE = 15 J
m = 140 g
C
KE = 10 J
m= 140 g
D
KE = 15 J
m = 130 g
E
KE = 12 J
m= 150 g
F
KE = 10 J
m = 150 g
Rank these situations on the basis of the amplitude of the oscillation, i.e., the maximum distance the
metal pieces move down from the equilibrium point.
Greatest
1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least
OR, The amplitudes are the same for all these situations.
___
OR, The amplitudes are zero for all these situations.
___
OR, We cannot determine the ranking for these situations.
___
Please explain your reasoning.
Answer: B = D > A = E > C = F
The given kinetic energies are the maximum kinetic energies since the metal pieces will have their
maximum speed when they pass through the equilibrium point. If we take the gravitational potential
energy to be zero when the masses are at their equilibrium points, then the maximum kinetic energy
will be equal to the maximum spring potential energy. Since the maximum spring potential energy is
1/2kA2, the ranking here will be on the basis of the kinetic energies.
NT9A-RT4: SPHERE ON A STRING—STRING LENGTH
The six figures below show metal spheres hung on the ends of strings. The spheres have been pulled to
the side and released so that they are swinging back and forth. For each sphere-string system the diagrams
give the mass of the sphere, the frequency of the swing, and how far, in terms of the angle from the
vertical, that the spheres were initially pulled to the side.
A
B
 = 2°
m = 500 g
f = 0.75 Hz
 = 2°
m = 500 g
f = 0.6 Hz
C
 = 1°
m = 800 g
f = 0.75 Hz
D
 = 3°
m = 300 g
f = 0.6 Hz
E
 = 3°
m = 800 g
f = 0.45 Hz
F
 = 1°
m = 300 g
f = 0.45 Hz
Rank these systems on the basis of the length of the string.
Greatest
1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least
OR, The string lengths are the same for all these systems.
___
OR, We cannot determine the ranking for the lengths of the strings.
___
Please explain your reasoning.
Answer: E = F > D = B > A = C
The frequency is the only variable that relates to the length of the string: neither angle (for small
angles such as are given here) nor mass affect the period of oscillation. The length is inversely
proportional to the square of the frequency, so the greater the frequency, the smaller the length of the
string.
NT9A-CT5: MASS HANGING ON A SPRING—PERIOD OF OSCILLATION
A mass M hangs from a spring as shown. If the mass is pulled down a small distance y
from its equilibrium position and released, the system will oscillate up and down.
a) If the mass is instead pulled down a distance 2y, will the period of oscillation for
the spring-mass system be greater than, less than, or equal to the period of
oscillation of the same spring-mass system when it was pulled down a distance y?
Explain.
The factors that affect the period of oscillation of this spring-mass system are the
stiffness of the spring (characterized by the spring constant) and the mass of the
oscillating object. The amplitude has no effect on the period or frequency, so the
period is unaffected by this change.
M
b) If this spring-mass system were given a larger total energy, would the period of oscillation (the
time for one complete cycle of motion) increase, decrease, or remain the same?
Explain.
To increase the total energy of the system the amplitude of oscillation must be increased. As in part (1),
this has no effect on the period or frequency.
c) If an additional mass m is added to the original mass M, will the period of
oscillation for the spring-mass system be greater than, less than, or equal to
the period of the spring-mass system with only mass M?
Explain.
The frequency will be smaller for the larger mass so the period will be larger,
since the frequency is proportional to the square root of the ratio of the spring
constant to the mass.
M
m
NT9A-RT8: SIMPLE PENDULA—OSCILLATION FREQUENCY
A simple pendulum consists of a mass M attached to a massless string of length L as shown in figure A
below. If the mass is pulled to one side a small distance and released, it will swing back and forth with a
frequency fo and swing amplitude A. Cases B, C, and D shown below are variations of this system.
A
B
L
C
L
Swing
amplitude
A
Swing
amplitude
A
M
D
L
Swing
amplitude
2A
2M
L
2
Swing
amplitude M
A
M
Rank these cases on the basis of the oscillation frequency of the masses.
Greatest 1 ______ 2 ______ 3 ______ 4 ______ Least
OR, The frequencies are the same for all these cases.
___
OR, We cannot determine the ranking for the frequencies.
___
Please explain your reasoning.
Answer: D > A = B = C
For pendula, the period is dependent on the length of the string and independent of both the amplitude
and the mass, so A = B = C. When the length of a pendulum is reduced the frequency increases, so the
frequency of D will be greater.
NT9A-CRT12: VELOCITY VS. TIME GRAPH—FREQUENCY AND PERIOD
A cart attached to a spring is displaced from equilibrium and then
released. A graph of velocity as a function of time for the cart is
shown below. There is no friction.
4
0
–
+
Velocity, cm/s
2
0
–2
2
4
6
8
10
12
14
16
18
20
22 Time, s
–4
(a) What is the period of the motion for this cart?
Explain.
Answer: The period is the time for the complete cycle of the oscillation so it is 8 seconds from the
graph.
(b) What is the frequency of the motion for this cart?
Explain.
Answer: The period is 8 seconds from the graph. The frequency is the reciprocal of the period or
1/8 Hz.
(c) In which direction was the cart displaced from equilibrium before it was released?
Explain.
Answer: Since the velocity just after the cart was released is positive, the cart must be moving toward
the right. It was displaced to the left.
NT9B-LMCT16: OSCILLATION GRAPH OF DISPLACEMENT VS. TIME—KINEMATIC QUANTITIES
A cart attached to a spring is displaced from equilibrium and then
0
released. A graph of displacement as a function of time for the cart is
–
+
shown below. There is no friction. Eight points are labeled A – H in the
graph.
4
Displacement, cm
A
2
0
–2
–4
2
4
6
8
B
C
10
G
D
12 F 14
E
H
16 Time, s
For each question below, choose from the labeled points above or state “none.”
1. At which point or points is the acceleration positive?
Explain.
Points D, E, and F. The acceleration is positive when the velocity is positive and increasing, or when
the velocity is negative and decreasing. The velocity at point D is negative (the slope of the position
graph is negative) and decreasing (the magnitude of the slope is smaller just after point D than it is just
before point D); at point F the velocity is positive and increasing; and at point E the slope of the graph
is instantaneously zero, but the slope is negative just before and positive just after, so the acceleration
is positive.
2. At which point or points does the cart have zero velocity but nonzero net force?
Explain.
Points A and E. Since the velocity is zero at these points, but the velocity is changing, there is an
acceleration, and therefore a net force.
3. At which point or points is the net force on the cart equal to zero?
Explain.
Points C and G. At these points the slope of the position graph is constant, so the velocity is constant.
The acceleration is therefore zero, and the net force must be zero. From a dynamics perspective, at
these points the cart is passing through the equilibrium position, and the spring exerts no force.
4. At which point or points are the acceleration, velocity, and displacement all positive?
Explain.
None. The acceleration is always negative when the position is positive. The force on the cart is always
acting toward the equilibrium position; negative when the displacement is positive, and positive when
the displacement is negative.
5. At which point or points is the acceleration nonzero and opposite in sign to the position?
Explain.
A, B, D, E, F, H. The acceleration is always opposite in sign to the position.
6. At which point or points is the velocity nonzero and opposite in sign to the acceleration?
Explain.
Points D and H. Whenever the cart is slowing down, the velocity and acceleration are in opposite
directions.
NT9A-QRT9: POSITION VS. TIME GRAPH OF A CART ATTACHED TO A SPRING—MASS & PERIOD
Rest
A frictionless cart of mass m is attached to a spring with spring
position
constant k. When the cart is displaced from its rest position and
released, it oscillates with a period  that is given by
 = 2 m .
k
k
Cart
Track
The graph of the position of the cart as a function of time is shown below for Experiment A. Graphs for
two other experiments are shown below this. The same spring is used in all three experiments.

(1) Compared to Experiment A, in Experiment B the cart has
a) twice as much mass.
Position, cm
b) four times as much mass.
c) one-half the mass.
Experiment A
d) one-fourth the mass.
Time, s
Position, cm
e) the same mass.
Explain.
Answer: e) The cart has the same mass. The periods
of the systems in the two experiments are the
same, so the masses have to be the same also.
Experiment B
Time, s
Position, cm
(2) Compared to Experiment A, in Experiment C the
cart has
Experiment C
Time, s
a) twice as much mass.
c) one-half the mass.
b) four times as much mass.
d) one-fourth the mass.
e) the same mass.
Explain.
Answer: d) The cart has one-fourth the mass. The period of experiment C is half that of experiment A,
so the mass is less for experiment C. Since the period is proportional to the square root of the mass,
the mass in C has to be one-fourth that of A.
(3) Suppose that for a fourth experiment (Experiment D), the mass used in Experiment A was
doubled and the spring was replaced with a spring with double the spring constant. The period
in Experiment D would be
a) the same as the period in Experiment A.
c) four times the period in Experiment A.
b) double the period in Experiment A.
d) one-half the period in Experiment A.
e) one-fourth the period in Experiment A.
Explain.
Answer: a) The period would be the same in experiment D as it was in experiment A. Since the period
is determined by the square root of the mass dievided by the spring constant, doubling both will not
change the period.
NT9D-BCT25: OSCILLATING MASS-SPRING GRAPH OF DISPLACEMENT VS. TIME—ENERGY
A cart attached to a spring
0
–
+
is given an initial push,
4
displacing it from its
2
equilibrium position. A
graph of displacement as a
0
function of time for the
–2
cart is shown at right. The
system has a total initial energy of 12 J and there is no –4
friction. Five points are labeled A – E in the graph.
Displacement, cm
B
A
C
2
4
D 6
E
8
10 Time, s
For each labeled point, complete the bar chart below for the kinetic energy and the potential energy
for the cart-spring system.
Point A
KE
PEsp
Point B
KE
PEsp
Point C
KE
PEsp
Point D
KE
PEsp
Point E
KE
PEsp
Bar chart key
KE
Kinetic energy
Spring
PEspring potential energy
Explain.
At point A the cart is instantaneously at rest, so all of the energy is spring potential energy. At point B,
which is half way been the equilibrium point and the point of maximum displacement the spring
potential energy is one-quarter of its maximum, so the kinetic energy is three-fourths of the maximum.
At point C the spring is neither compressed nor stretched so there is no spring potential energy and the
kinetic energy is at a maximum. At point D the energy values are the same as at B and at point E all of
the energy is again spring potential.
Point A
KE
PEsp
Point B
KE
PEsp
Point C
KE
PEsp
Point D
KE
PEsp
Point E
KE
PEsp
Bar chart key
KE
Kinetic energy
Spring
PEspring potential energy