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CJ Simple Harmonic Motion and Waves Assignment 1
10.1 The Ideal Spring & Simple Harmonic Motion 10.2 Simple Harmonic Motion & the Reference Circle
Conceptual Questions 1, 2 – page 313
Problems 3, 4, 9, 12 page 314-316
A LITTLE MORE ON SPRINGS:
Recall Hooke’s Law:
For an Ideal Spring (Spring Constant K) stretched a distance Δx the
Reaction Force will be F according the Equation F = k(Δx) .
What happens if the spring is compressed rather than being
stretched? How does this change things?
The force the mass experiences ( increases / decreases ) as the deformation (which means stretch or compression)
(increases / decreases ).
Draw the Free Body Diagram for Each Case of a Mass Attached to the End of a Spring.
Conceptual Questions 1,2 – page 313
1. Two people pull on a horizontal spring that is attached to an immovable wall. Then, they detach it from the
wall and pull on opposite ends of the horizontal spring. They pull just as hard in each case. In which situation,
if either, does the spring stretch more? Account for your answer.
A horizontal spring is attached to an immovable wall. Two people pull on the spring. They then detach it from the wall
and pull on opposite ends of the horizontal spring. They pull just as hard in each case.
Suppose that each person pulls with a force P. When the spring is attached to the wall, both people pull on the
same end, so that they pull with a total force 2P. The wall exerts a force –2P at the other end of the spring. The
tension in the spring has magnitude 2P. When the two people pull on opposite ends of the horizontal spring, one
person exerts a force P at one end of the spring, and the other person exerts a force –P at the other end of the
spring. The tension in the spring has magnitude P.
The tension in the spring is greater when it is attached to the immovable wall; therefore, the spring will stretch
more in this case.
2. The drawing shows identical springs that are attached to a box in two different ways. Initially, the springs
are unstrained. The box is then pulled to the right and released. In each case the initial displacement of the
box is the same. At the moment
of release, which box, if either,
experiences the greater net force
due to the springs? Provide a
reason for your answer.
Identical springs are attached to a box in two different ways, as shown in the drawing in the text. Initially, the springs are
unstrained. The box is then pulled to the right and released. The displacement of the box is the same in both cases.
For the box on the left, each spring is stretched and its displacement is  x , where the plus sign indicates a
displacement to the right. According to Equation 10.2, the restoring force exerted by each spring on the box
is Fx  k  x  . Thus, the net force due to both springs is Fx  2k  x  .
For the box on the right, one spring is stretched and the other is compressed. However, the displacement of
each spring is still  x . The net force exerted on the box is Fx  k  x   k  x   2k  x  .
Thus, both boxes experience the same net force.
Problems 3, 4, 9, 12 page 314-316
3. A 0.70-kg block is hung from and stretches a spring that is attached to the ceiling. A second block is attached
to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass
of the second block?
Mr. Sears Solution: Why not simply use Hooke’s Law? (Text Solution is valid but rather long)
Since Gravity is creating the force, the force is proportional to the mass of the object.
Create a Ratio of the Extensions of the Springs and use Hooke’s Law (k is the same for both situations)
𝐹
𝐹 = 𝑘 (∆𝑥) so ∆𝑥 = ∆𝑥
𝛥𝑥2
𝛥𝑥1
=
3𝑥
𝑥
=
𝐹2
𝑘
𝐹1
𝑘
=
𝑚2 𝑔
𝑘
𝑚1 𝑔
𝑘
=
𝑚2
𝑚1
so m2 = 3x1
Be careful because m2 is the total mass on the spring in situation 2. To get the added mass subtract m 1 to get
2m1 = (.7 kg) x 2 = 1.4 kg
Either method is valid.
Text Solution: The text uses Energy
The block is at equilibrium as it hangs on the spring. Therefore, the downward-directed weight of the block is
balanced by the upward-directed force applied to the block by the spring. The weight of the block is mg,
where m is the mass and g is the acceleration due to gravity. The force exerted by the spring is given by
Equation 10.2 (Fx = kx), where k is the spring constant and x is the displacement of the spring from its
unstrained length. We will apply this reasoning twice, once to the single block hanging, and again to the
two blocks. Although the spring constant is unknown, we will be able to eliminate it algebraically from the
resulting two equations and determine the mass of the second block.
SOLUTION Let upward be the positive direction. Setting the weight mg equal to the force kx exerted by
the spring in each case gives
m1 g   kx1
and
m1 g  m2 g  kx2
One hanging block
Two hanging blocks
Dividing the equation on the right by the equation on the left, we can eliminate the unknown spring
constant k and find that
m1 g  m2 g kx2

m1 g
kx1
or
1
m2 x2

m1 x1
It is given that x2/x1 = 3.0. Therefore, solving for the mass of the second block reveals that
x

m2  m1  2  1   0.70 kg  3.0  1  1.4 kg
 x1

4. A person who weighs 670 N steps onto a spring-scale in the bathroom, and the spring compresses by 0.79
cm. (a) What is the spring constant? (b) What is the weight of another person who compresses the spring by
0.34 cm?
The weight of the person causes the spring in the scale to compress. The amount x of compression, according to
Equation 10.1, depends on the magnitude FxApplied of the applied force and the spring constant k.
SOLUTION
a.
Since the applied force is equal to the person’s weight, the spring constant is
k
FxApplied
x

670 N
0.79  10
2
 8.5  104 N / m
m
(10.1)
b.
When another person steps on the scale, it compresses by 0.34 cm. The weight (or applied force) that
this person exerts on the scale is



FxApplied  k x  8.5  104 N / m 0.34  102 m  290 N
(10.1)
9. To measure the static friction coefficient between a 1.6-kg
block and a vertical wall, the setup shown in the drawing is used.
A spring is attached to the block. Someone pushes on the end of
the spring (Spring Constant = 510 N/m) in the direction
perpendicular to the wall until the block does not slip downward.
If the spring in such a setup is compressed by 0.039 m, what is the
coefficient of static friction?
The free-body diagram shows the magnitudes and
directions of the forces acting on the block. The
weight mg acts downward. The maximum force of
static friction fsmax acts upward just before the
block begins to slip. The force from the spring
FxApplied = kx (Equation 10.1) is directed to the
right. The normal force FN from the wall points to
the left. The magnitude of the maximum force of
static friction is related to the magnitude of the
normal force according to Equation 4.7
 fsmax  s FN  , where μs is the coefficient of
static friction. Since the block is at equilibrium
just before it begins to slip, the forces in the x
direction must balance and the forces in the y direction must balance. The balance of forces in the two
directions will provide us with two equations, from which we will determine the coefficient of static friction.
SOLUTION Since the forces in the x direction and in the y direction must balance, we have
FN  kx
and
mg  s FN
Substituting the first equation into the second equation gives
mg  s FN  s  kx 
or
s 
2
mg 1.6 kg   9.80 m/s 

 0.79
kx  510 N/m  0.039 m 
12. A 30.0-kg block is resting on a flat horizontal
table. On top of this block is resting a 15.0-kg block,
to which a horizontal spring is attached, as the
drawing illustrates. The spring constant of the spring
is 325 N/m. The coefficient of kinetic friction
between the lower block and the table is 0.600, and
the coefficient of static friction between the two
blocks is 0.900. A horizontal force F is applied to the
lower block as shown. The force is increasing in such
a way as to keep the lower block moving at constant
speed. At the point where the upper block begins to
slip on the lower block, determine (a) the amount by which the spring is compressed and (b) the magnitude of
the force F.
Mr. Sears Solution: Why not simply use Hooke’s Law?
Variables:
m1 = 30 kg
F=?
m2 = 15 kg
k = 325 N/m
W1 = weight of lower Block = 300 N
µk = .6 (Lower block and top block)
µs = .9 (Lower block and table)
W2 = weight of upper Block = 150 N
FN1 = Normal Force between Lower Block and Table = W1 + W2 = 450 N (∑ F = 0)
FREE BODY DIAGRAM OF TOP BLOCK
Using the First Condition of Equilibrium (Vertically and Horizontally)
Vertical: ∑ Fy= 0 , W = mg = FN = 150 N
Horizontal: ∑ Fx= 0 , Fspring = kx = Fs = µsFN = µsmg = .9 x 150 N = 135 N
135 N is the greatest Force the Friction can produce. Using Hooke’s Law the
value of Δx = F/k = { 135 N / 350 N/m } = .39 meters
FREE BODY DIAGRAM OF BOTTOM BLOCK
Using the First Condition of Equilibrium (Vertically and Horizontally)
Vertical: ∑ Fy= 0 , FN1 = W1 + W2 = 450 N
Kinetic Friction Force = Fk = µkFN1 = .6 (450 N) = 270 N
Horizontal:
∑ Fx= 0 , F = Fs + Fk = 135 N + 270 N = 405 N
TEXT BOOK SOLUTION ( Please note that the text book uses 9.8 m/s2 which was not used in the previous solution)
From the drawing given with the problem statement, we see that the kinetic frictional force on the bottom block (#1) is
given by
Ff1 = µk(m1 + m2)g
and the maximum static frictional force on the top block (#2) is
Ff2 = µsm2g
Newton’s second law applied to the bottom block gives
F – Ff1 – kx = 0
and to the top block gives
Ff2 – kx = 0
a.
To find the compression, x, we have
2
x = Ff2/k = µsm2g/k = (0.900)(15.0 kg)(9.80 m/s )/(325 N/m) = 0.407 m
b.
F = kx + µk(m1 + m2)g
2
F = (325 N/m)(0.407 m) + (0.600)(45.0 kg)(9.80 m/s ) = 397 N