This learning object focuses on the topics of simple harmonic motion and energy conservation in horizontal mass-spring systems. It is in the form of a word problem that has two parts that each focus on one of the two topics above.

1. Motion and Energy in Horizontal Mass-Spring Systems
Part 1: Simple Harmonic Motion in a Horizontal Mass-Spring
System
Considering the above horizontal mass-spring system, calculate:
1) The value of the spring constant (k) if the acceleration of the block at X=A is
-8.00 m/s2
2) The displacement of the spring when the acceleration is 3.95 m/s2
3.00 kg
X=0 X=A=0.250 m
k=? N/m

2. Part 2: Conservation of Energy in a Horizontal Mass-Spring
System
Considering the above horizontal mass-spring system with the value of the spring
constant (k) found in Part 1, calculate (ignoring friction):
1) The velocity of the spring at equilibrium (X=0)
2) The kinetic energy of the spring at X=0.250 m
3) The kinetic energy of the spring at X=A/2
4) The potential energy of the spring when the velocity of the spring is Vmax/2
3.00 kg
X=0 X=A=0.250 m
k=? N/m

3. Solutions for Part 1
Background Information: Because

Fnet  kx  max for a mass on a spring, the
equation

ax  
k
m





x describes the acceleration of the mass.
1) Using the equation

ax  
k
m





x you can rearrange it to get

k  
ax  m
x





which
will give you the spring constant. To solve, plug in the numbers given:

k  
ax  m
x







k  
(8.00m/s2
)  (3.00kg)
(0.250m)






k  96.0N /m
2) Using the relationship described previously,

ax  
k
m





x , you can rearrange it to
solve for x, the displacement:

x  
ax  m
k





. To solve, plug in the numbers
given:

x  
ax  m
k







x  
(3.95m/s2
)  (3.00kg)
(96.0N /m)






x  0.123m
Therefore, the displacement is 0.123 m to the left of the equilibrium position.

4. Solutions for Part 2
Background Information: The conservation of energy equation for this system
(K=kinetic energy, U=potential energy) is

E K U or

E 
1
2
mv2

1
2
kx2
. The
total energy of the system is also equal to

1
2
kA2
when the velocity is zero at the
point of maximum displacement at X=A, so we can combine this with the
previous equation to get

1
2
kA2

1
2
mv2

1
2
kx2
1) In the situation in question 1, the mass-spring system is at equilibrium (X=0),
which means that the potential energy

1
2
kx2
is zero. This simplifies the
conservation of energy equation for a mass-spring system to

1
2
kA2

1
2
mv2
.
Rearranging this to find velocity, you get

v 
kA2
m
. To solve, plug in the
numbers given:

v 
kA2
m
v 
(96.0N /m)  (0.250m)2
(3.00kg)
v 1.41m/s
2) In the situation in question 2, the mass-spring system is at X=A, which means
that the velocity, and therefore the kinetic energy, are zero because the
oscillation is at its maximum displacement. To see this using the conservation
of energy equation plug in the numbers into the equation and solve:

1
2
kA2

1
2
mv2

1
2
kx2
1
2
kA2

1
2
kx2

1
2
mv2
 K
1
2
k(A2
 x2
)  K
1
2
(96.0N /m)((0.250m)2
 (0.250m)2
)  K
1
2
(96.0N /m)(0)  K
0  K

5. 3) Using the equation

1
2
kA2

1
2
mv2

1
2
kx2
, rearrange to find kinetic energy,
and then put in the numbers given to solve:

1
2
kA2

1
2
mv2

1
2
kx2
1
2
kA2

1
2
kx2

1
2
mv2
 K
1
2
k(A2
 x2
)  K
1
2
(96.0N /m) (0.250m)2

0.250
2
m






2




 K
2.25J  K
4) For this question, one must realize that the velocity calculated in question 1 is
the maximum velocity because the velocity experienced at X=0 is maximum
velocity in simple harmonic motion. Using this knowledge and again using the
conservation of energy equation for this system, the problem can be solved:

1
2
kA2

1
2
mv2

1
2
kx2
1
2
kA2

1
2
mv2

1
2
kx2
U
1
2
(96.0N /m)(0.250m)2

1
2
(3.00kg)
1.41m/s
2






2
U
2.25J U