3/22/23, 3:21 AM
Homework 1
Homework 1
Due: 11:59pm on Wednesday, January 18, 2023
To understand how points are awarded, read the Grading Policy for this assignment.
Exercise 14.7 - Enhanced - with Solution
A 2.39 kg ball is attached to an unknown spring and allowed to oscillate. shows a graph of the ball's position x as a function of
time t .
For related problem-solving tips and strategies, you may want to view
a Video Tutor Solution of Angular frequency, frequency, and period in
shm.
Part A
For this motion, what is the period?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
T
= 0.800 s
Correct
IDENTIFY and SET UP: The period is the time for one cycle. A is the maximum value of x.
EXECUTE: From the figure with the problem, T
= 0.800 s
.
Part B
What is the frequency?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
f
= 1.25 Hz
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Homework 1
Correct
f =
1
T
= 1.25 Hz
Part C
What is the angular frequency?
Express your answer in rad/s to three significant figures.
ANSWER:
= 7.85
ω
rad/s
Correct
ω = 2πf = 7.85 rad/s
Part D
What is the amplitude?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
A
= 3.00×10−2 m
Correct
From the figure with the problem, A
= 3.0 cm
.
Part E
What is the force constant of the spring?
Express your answer to three significant figures and include the appropriate units.
ANSWER:
k
= 147
N
m
Correct
−
−
T = 2π√
m
k
, so k
= m(
2π
T
2
) = (2.39 kg)(
2π
0.800 s
2
) = 147 N/m.
EVALUATE: The amplitude shown on the graph does not change with time, so there must be little or no friction in
this system.
Exercise 14.11 - Enhanced - with Solution
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2/23
3/22/23, 3:21 AM
Homework 1
An object is undergoing SHM with period 0.980 s and amplitude 0.320 m. At t = 0, the object is at x = 0.320 m and is
instantaneously at rest.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Angular frequency, frequency,
and period in shm.
Part A
Calculate the time it takes the object to go from x = 0.320 m, to x = 0.160 m.
Express your answer with the appropriate units.
ANSWER:
t
= 0.163 s
Correct
IDENTIFY: For SHM the motion is sinusoidal.
SET UP: x(t)
.
= A cos(ωt)
EXECUTE: x(t)
x = 0.320 m
, where A
= A cos(ωt)
at t 1
0.160 m = (0.320
cos(ωt 2 ) = 0.500
It takes t 2 − t 1
= 0.320 m
. Let t 2 be the instant when x
m) cos(ωt2 ) .
= 0
. ωt 2
.
= 1.047 rad t 2 =
= 0.163 s
and ω
2π
T
= 0.160 m
1.047 rad
6.41
=
=
2π
0.980 s
= 6.41 rad/s
.
. Then we have
= 0.163 rad/s
.
.
Part B
Calculate the time it takes the object to go from x = 0.160 m, to x = 0.
Express your answer with the appropriate units.
ANSWER:
t
= 8.2×10−2 s
Correct
Let t 3 be when x
t3 =
1.570 rad
6.41 rad/s
= 0
. Then we have cos(ωt 3 )
= 0.245 s
. It takes t 3 − t 2
= 0
and ωt 3
= 1.571 rad
.
= 0.245 s − 0.163 s = 8.2 × 10
−2
s
.
EVALUATE: Note that it takes twice as long to go from x = 0.320 m to x = 0.160 m than to go from
x = 0.160 m to x = 0 , even though the two distances are the same, because the speeds are different over the
two distances.
Exercise 14.14 - Enhanced - with Feedback
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3/23
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Homework 1
A 2.00 kg , frictionless block is attached to an ideal spring with force constant 300 N/m . At t = 0 the block has velocity
-4.00 m/s and displacement +0.200 m away from equilibrium.
Part A
Find the amplitude.
Express your answer to three significant figures and include the appropriate units.
ANSWER:
A
= 0.383 m
Correct
Part B
Find the phase angle.
Express your answer in radians.
ANSWER:
ϕ
= 1.02
rad
Correct
Part C
Choose an equation for the position as a function of time.
ANSWER:
x(t) = 0.383 ⋅ cos (12.2t − 1.02)
x(t) = −0.383 ⋅ cos (12.2t − 1.02)
x(t) = 0.383 ⋅ cos (12.2t + 1.02)
x(t) = −0.383 ⋅ cos (12.2t + 1.02)
Correct
Exercise 14.15
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Homework 1
A block of mass m is undergoing SHM on a horizontal, frictionless surface while attached to a light, horizontal spring. The spring
has force constant k, and the amplitude of the SHM is A . The block has v = 0, and x = +A at t = 0. It first reaches x = 0 when
t = T /4 , where T is the period of the motion.
Part A
In terms of T , what is the time t when the block first reaches x
= A/2
?
Express your answer in terms of the variable T .
ANSWER:
t1
T
=
6
Correct
Part B
The block has its maximum speed when t
value v max /2?
= T /4
. What is the value of t when the speed of the block first reaches the
Express your answer in terms of the variable T .
ANSWER:
t2
T
=
12
Correct
Part C
Does v
= v max /2
when x
= A/2
?
ANSWER:
Yes, it does.
No, it doesn't.
Correct
Matching Initial Position and Velocity of Oscillator
Learning Goal:
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Homework 1
Understand how to determine the constants in the general equation for simple harmonic motion, in terms of given initial
conditions.
A common problem in physics is to match the particular initial conditions - generally given as an initial position x0 and velocity v )
at t = 0 - once you have obtained the general solution. You have dealt with this problem in kinematics, where the formula
1. x(t)
= x 0 + v0 t +
at
2
2
has two arbitrary constants (technically constants of integration that arise when finding the position given that the acceleration is
a constant). The constants in this case are the initial position and velocity, so "fitting" the general solution to the initial conditions
is very simple.
For simple harmonic motion, it is more difficult to fit the initial conditions, which we take to be
x0
, the position of the oscillator at t = 0 , and
, the velocity of the oscillator at t = 0 .
v0
There are two common forms for the general solution for the position of a harmonic oscillator as a function of time t :
2. x(t) = A cos (ωt + ϕ) and
3. x(t) = C cos (ωt) + S sin (ωt),
where A , ϕ, C , and S are constants, ω is the oscillation frequency, and t is time.
Although both expressions have two arbitrary constants--parameters that can be adjusted to fit the solution to the initial
conditions--Equation 3 is much easier to use to accommodate x0 and v 0 . (Equation 2 would be appropriate if the initial
conditions were specified as the total energy and the time of the first zero crossing, for example.)
Part A
Find C and S in terms of the initial position and velocity of the oscillator.
Give your answers in terms of x0 , v 0 , and ω. Separate your answers with a comma.
Hint 1. The only good way to start
Which of the following procedures would solve this problem in the most straightforward manner?
ANSWER:
Differentiate x (t) twice to find a (t). Then integrate it twice. Plug in v 0 and x0 as the constants of
integration.
Differentiate x (t) once to find v (t) . Evaluate x(t
quantities.
= 0)
and v(t
= 0)
and then solve for the desired
Dimensional analysis suffices since x0 and v 0 have different dimensions.
Use Equation 1. Plug in a
= −kx(t)/m
where k/m
= ω
2
.
Hint 2. Using kinematic relationships
Find v (t) , the velocity as a function of time from Equation 3.
Hint 1. Derivative of a trig function
From the chain rule of calculus, find the derivative of cos (mt) with respect to time.
ANSWER:
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Homework 1
d cos (mt)
dt
=
−msin (mt)
ANSWER:
=
v (t)
−C ωsin (ωt) + Sωcos (ωt)
Hint 3. Initial position
Now you have general expressions for x (t) and v (t) . Find the position at t
= 0
.
ANSWER:
x(t = 0)
=
C
Hint 4. Initial velocity
Find the velocity at time t
= 0
.
ANSWER:
v(t = 0)
=
Sω
ANSWER:
C
,S=
x0 ,
v0
ω
Correct
Relating Two General Simple Harmonic Motion Solutions
Learning Goal:
To understand how the two standard ways to write the general solution to a harmonic oscillator are related.
There are two common forms for the general solution for the position of a harmonic oscillator as a function of time t:
1. x(t)
2. x(t)
= A cos (ωt + ϕ)
and
.
= C cos (ωt) + S sin (ωt)
Either of these equations is a general solution of a second-order differential equation (F ⃗ = ma⃗); hence both must have at least
two--arbitrary constants--parameters that can be adjusted to fit the solution to the particular motion at hand. (Some texts refer to
these arbitrary constants as boundary values.)
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Homework 1
Part A
What are the arbitrary constants in Equation 1?
Hint 1. What is considered a constant?
A constant is something that is defined by the physical situation under consideration (e.g., a spring constant,
acceleration due to gravity, frequency of oscillation, or mass) and it does not change even if the motion is different
owing to different initial conditions.
Hint 2. What is considered arbitrary?
Arbitrary constants are used to "fit" the general solution to a particular set of initial conditions, such as how far you
pull the oscillator from its equilibrium position, and how fast it is moving when you let it go.
ANSWER:
ω
only
A
only
A
and ϕ
A
and ω
ω
A
and ϕ
and ω and ϕ
Correct
Part B
What are the arbitrary constants in Equation 2?
ANSWER:
S
only
ω
only
C
only
S
and C
S
and ω
C
and ω
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Homework 1
Correct
Because both Equation 1 and Equation 2 are general solutions, they can both represent any set of initial
conditions (i.e., initial position and velocity). Therefore, one equation could be expressed in terms of the other.
Part C
Find analytic expressions for the arbitrary constants C and S in Equation 2 (found in Part B) in terms of the constants A
and ϕ in Equation 1 (found in Part A), which are now considered as given parameters.
Give your answers for the coefficients of cos(ωt) and sin(ωt) , separated by a comma. Express your answers in
terms of A and ϕ.
Hint 1. A useful trig identity
What is the angle-sum trigonometric identity for cos (a + b)? Hint: If you can remember the general form but are
unsure of the sign or whether a particular term is cos or sin, try your expression for simple values like 0 and/or π/2 .
Give your answer in terms of cos(a), cos(b), sin(a) , and sin(b).
ANSWER:
cos (a + b)
=
cos (a) cos (b) − sin (a) sin (b)
Hint 2. How to use the trig indentity
The left side of the equation, cos (a + b), is in the form of Equation 1, whereas the right side is in the form of
Equation 2. If you make a proper substitution for a and b first on the left and then correspondingly on the right side
you can solve for the arbitrary constants in Equation 2.
ANSWER:
C
,S=
Acos (ϕ) , −Asin (ϕ)
Correct
Part D
Find analytic expressions for the arbitrary constants A and ϕ in Equation 1 (found in Part A) in terms of the constants C
and S in Equation 2 (found in Part B), which are now considered as given parameters.
Express the amplitude A and phase ϕ (separated by a comma) in terms of C and S.
Hint 1. Find a relationship between C , S and A
Examine the sum C 2
+S
2
where you substitute the preceding answers for C and S.
Give your answer in terms of A and other given quantities.
Hint 1. Useful trig identity
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Homework 1
Recall that sin
2
2
(ϕ) + cos (ϕ) = 1
.
ANSWER:
C
2
+S
2
=
A
2
Hint 2. Useful trig indentityfor finding ϕ
Try using
tan(ϕ) =
sin(ϕ)
cos ϕ
.
Of course you don't want functions of ϕ on the right; however, you do want C and S on the right.
ANSWER:
A
,ϕ=
−
−−
−
−−
−
2
2
√ C + S , tan−1
S
C
Correct
This problem was very mathematical. To understand its utility, realize that for a typical mechanical oscillator, the
variables have the following meaning:
is the amplitude of oscillation,
is the initial phase angle,
C is the initial position at t = 0 , and
S is the initial velocity at t = 0 divided by ω .
A
ϕ
Therefore, if you are given the initial amplitude and phase, you can find the initial position and velocity. Similarly, if
you are given the initial position and velocity you can find the initial amplitude (whose square is related to the total
energy) and the phase angle, which permits you to answer questions like "When does the particle reach its
maximum displacement?" or "When does the particle first return to x = 0 ?"
Energy of a Spring
An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum
displacement from equilibrium is A and the total mechanical energy of the system is E .
Part A
What is the system's potential energy when its kinetic energy is equal to
3
4
E?
Hint 1. How to approach the problem
Since the sum of kinetic and potential energies of the system is equal to the system's total energy, if you know the
fraction of total energy corresponding to kinetic energy you can calculate how much energy is potential energy.
Moreover, using conservation of energy you can calculate the system's total energy in terms of the given quantities k
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Homework 1
and A . At this point you simply need to combine those results to find the potential energy of the system in terms of k
and A .
Hint 2. Find the fraction of total energy that is potential energy
When the kinetic energy of the system is equal to
3
4
E,
what fraction of the total energy E is potential energy?
Express your answer numerically.
Hint 1. Conservation of mechanical energy
In a system where no forces other than gravitational and elastic forces do work, the sum of kinetic energy K
and potential energy U is conserved. That is, the total energy E of the system, given by E = K + U , is
constant.
ANSWER:
0.250
Hint 3. Find the total energy of the system
What is the total mechanical energy of the system, E ?
Express your answer in terms of some or all of the variables m , k, and A .
Hint 1. How to approach the problem
If you apply conservation of energy to the system when the object reaches its maximum displacement, you
can calculate the system's total energy E in terms of the given quantities A and k. In fact, when the object is
at its maximum displacement from equilibrium, its speed is momentarily zero and so is its kinetic energy. It
follows that the system's energy at this point is entirely potential, that is, E = U , where U is the spring's
elastic potential energy.
Hint 2. Elastic potential energy
The elastic potential energy U of a spring that has been compressed or stretched by a distance x is given by
U =
where k is the force constant of the spring.
1
2
kx
2
,
ANSWER:
E
=
kA
2
2
ANSWER:
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Homework 1
kA
2
kA
2
2
kA
2
4
kA
2
8
Correct
Part B
What is the object's velocity when its potential energy is
2
3
E?
Hint 1. How to approach the problem
You can calculate the object's velocity using energy considerations. Calculate the fraction of the system's total
energy that is kinetic energy and then find the object's velocity from the definition of kinetic energy. To simplify your
expression write the total energy in terms of k and A . Alternatively, you could directly use the formula for the object's
velocity in terms of the variables k, m , A , and displacement x derived from energy considerations. The only
unknown quantity in such a formula would be the object's displacement x, which can be calculated from the system's
potential energy.
Hint 2. Find the kinetic energy
If the system's potential energy is
2
3
E,
what is the system's kinetic energy?
Hint 1. Conservation of mechanical energy
In a system where no forces other than gravitational and elastic forces do work, the sum of kinetic energy K
and potential energy U is conserved. That is, the total energy E of the system, given by E = K + U , is
constant.
Hint 2. Total energy of the system
The total energy of a system consisting of an object attached to a horizontal spring of force constant k is
given by
E =
1
2
kA
2
,
where A is the maximum displacement of the object from its equilibrium position.
ANSWER:
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Homework 1
1
2
1
3
1
6
2
3
kA
kA
kA
kA
2
2
2
2
Hint 3. Formula for the velocity in terms of position
The velocity of an object of mass m undergoing simple harmonic motion is given by
−
−
−−−−−
−
k
2
√A − x2 ,
v = ±√
m
where k is the force constant of the system, x is the object's position, and A is maximum displacement.
Hint 4. Find the object's position
When the system's potential energy is
2
3
E,
what is the displacement x of the object from its equilibrium position?
Hint 1. Elastic potential energy
The elastic potential energy U of a spring that has been compressed or stretched by a distance x is given by
U =
where k is the force constant of the spring.
1
2
kx
2
,
Hint 2. Total energy of the system
The total energy of a system consisting of an object attached to a horizontal spring of force constant k is
given by
E =
1
2
kA
2
,
where A is the maximum displacement of the object from its equilibrium position.
ANSWER:
2
3
A
−
−
±√
2
3
A
−−
±√
2
3k
A
−
−
1
±√ A
3
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Homework 1
ANSWER:
−
−
±√
k
m
A
−
−
±√
k
m
−
−
±√
k
2
3
A
A
–
m √3
−
−
±√
−
−
√
k
A
–
m √6
Correct
Energy of Harmonic Oscillators
Learning Goal:
To learn to apply the law of conservation of energy to the analysis of harmonic oscillators.
Systems in simple harmonic motion, or harmonic oscillators, obey the law of conservation of energy just like all other systems
do. Using energy considerations, one can analyze many aspects of motion of the oscillator. Such an analysis can be simplified if
one assumes that mechanical energy is not dissipated. In other words,
E = K + U = constant,
where E is the total mechanical energy of the system, K is the kinetic energy, and U is the potential energy.
As you know, a common example of a harmonic oscillator is a mass attached to a spring. In this problem, we will consider a
horizontally moving block attached to a spring. Note that, since the gravitational potential energy is not changing in this case, it
can be excluded from the calculations.
For such a system, the potential energy is stored in the spring and is given by
U =
1
2
kx
2
,
where k is the force constant of the spring and x is the distance from the equilibrium position.
The kinetic energy of the system is, as always,
K =
1
2
2
mv
,
where m is the mass of the block and v is the speed of the block.
We will also assume that there are no resistive forces; that is, E
= constant
.
Consider a harmonic oscillator at four different moments, labeled A, B, C, and D, as shown in the figure . Assume that the force
constant k, the mass of the block, m , and the amplitude of vibrations, A , are given. Answer the following questions.
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Homework 1
Part A
Which moment corresponds to the maximum potential energy of the system?
Hint 1. Consider the position of the block
Recall that U
=
1
2
kx
2
, where x is the distance from equilibrium. Thus, the farther the block is from equilibrium,
the greater the potential energy. When is the block farthest from equilibrium?
ANSWER:
A
B
C
D
Correct
Part B
Which moment corresponds to the minimum kinetic energy of the system?
Hint 1. How does the velocity change?
Recall that K
=
1
2
2
mv
, where v is the speed of the block. When is the speed at a minimum? Keep in mind that
speed is the magnitude of the velocity, so the lowest value that it can take is zero.
ANSWER:
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Homework 1
A
B
C
D
Correct
When the block is displaced a distance A from equilibrium, the spring is stretched (or compressed) the most, and
the block is momentarily at rest. Therefore, the maximum potential energy is Umax
of course, K
= Kmin = 0
. Recall that E
= K +U
=
1
2
kA
2
. At that moment,
. Therefore,
E =
1
2
kA
2
.
In general, the mechanical energy of a harmonic oscillator equals its potential energy at the maximum or minimum
displacement.
Part C
Consider the block in the process of oscillating.
ANSWER:
at the equilibrium position.
at the amplitude displacement.
If the kinetic energy of the block is increasing, the block must be
moving to the right.
moving to the left.
moving away from equilibrium.
moving toward equilibrium.
Correct
Part D
Which moment corresponds to the maximum kinetic energy of the system?
Hint 1. Consider the velocity of the block
As the block begins to move away from the amplitude position, it gains speed. As the block approaches equilibrium,
the force applied by the spring—and, therefore, the acceleration of the block—decrease. The speed of the block is at
a maximum when the acceleration becomes zero. At what position does the object begin to slow down?
ANSWER:
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Homework 1
A
B
C
D
Correct
Part E
Which moment corresponds to the minimum potential energy of the system?
Hint 1. Consider the distance from equilibrium
The smallest potential energy corresponds to the smallest distance from equilibrium.
ANSWER:
A
B
C
D
Correct
When the block is at the equilibrium position, the spring is not stretched (or compressed) at all. At that moment, of
course, U = Umin = 0. Meanwhile, the block is at its maximum speed (v max ). The maximum kinetic energy can
then be written as Kmax
=
1
2
2
mvmax .
Recall that E
= K +U
and that U
= 0
at the equilibrium position.
Therefore,
E =
1
2
2
mvmax .
Recalling what we found out before,
E =
1
2
kA
2
,
we can now conclude that
1
2
kA
2
=
1
2
2
mvmax ,
or
−
−
k
A = ωA .
vmax = √
m
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Homework 1
Part F
At which moment is K
= U
?
Hint 1. Consider the potential energy
At this moment, U
=
1
2
Umax .
Use the formula for Umax to obtain the corresponding distance from equilibrium.
ANSWER:
A
B
C
D
Correct
Part G
Find the kinetic energy K of the block at the moment labeled B.
Express your answer in terms of k and A .
Hint 1. How to approach the problem
Find the potential energy first; then use conservation of energy.
Hint 2. Find the potential energy
Find the potential energy U of the block at the moment labeled B.
Express your answer in terms of kand A .
ANSWER:
U
1
=
8
kA
2
ANSWER:
K
=
3
8
kA
2
Correct
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Homework 1
Exercise 14.26 - Enhanced - with Feedback
A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. The amplitude of the
motion is 0.290 m and the period is 3.09 s .
Part A
What is the acceleration of the block when x = 0.160 m?
Express your answer with the appropriate units.
ANSWER:
ax
= -0.662
m
2
s
Correct
Part B
What is the speed of the block when x = 0.160 m?
Express your answer with the appropriate units.
ANSWER:
vx
= 0.492
m
s
Correct
Exercise 14.43 - Enhanced - with Solution
You pull a simple pendulum of length 0.250 m to the side through an angle of 3.50∘ and release it.
For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of A simple pendulum.
Part A
How much time does it take the pendulum bob to reach its highest speed?
Express your answer with the appropriate units.
ANSWER:
t
= 0.25 s
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19/23
3/22/23, 3:21 AM
Homework 1
Correct
IDENTIFY: T
−
−
−
= 2π√ L/g
is the time for one complete swing.
SET UP: The motion from the maximum displacement on either side of the vertical to the vertical position is onefourth of a complete swing.
EXECUTE: To the given precision, the small-angle approximation is valid. The highest speed is at the bottom of
the arc, which occurs after a quarter period,
T
4
=
π
2
−
−
√
L
g
= 0.251s
.
Part B
How much time does it take if the pendulum is released at an angle of 1.75∘ instead of 3.50∘ ?
Express your answer with the appropriate units.
ANSWER:
t
= 0.25 s
Correct
The same as calculated in (a), 0.251 s . The period is independent of amplitude.
EVALUATE: For small amplitudes of swing, the period depends on L and g.
Exercise 14.48 - Enhanced - with Feedback
In the laboratory, a student studies a pendulum by graphing the angle θ that the string makes with the vertical as a function of
time t , obtaining the graph shown in the figure .
Part A
What is the period of the pendulum's motion?
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20/23
3/22/23, 3:21 AM
Homework 1
Express your answer to two significant figures and include the appropriate units.
ANSWER:
T
= 1.6 s
Correct
Part B
What is the frequency of the pendulum's motion?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
f
= 0.63 Hz
Correct
Part C
What is the angular frequency of the pendulum's motion?
Express your answer in radians per second to two significant figures.
ANSWER:
ω
= 3.9
rad/s
Correct
Part D
What is the amplitude of the pendulum's motion?
Express your answer in degrees to one significant figure.
ANSWER:
θmax
= 6
∘
Correct
Part E
How long is the pendulum?
Express your answer to two significant figures and include the appropriate units.
ANSWER:
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21/23
3/22/23, 3:21 AM
L
Homework 1
= 0.64 m
Correct
Part F
Is it possible to determine the mass of the bob?
ANSWER:
yes
no
Correct
Problem 14.87
A slender, uniform, metal rod with mass M is pivoted without friction about an axis through its midpoint and perpendicular to the
rod. A horizontal spring with force constant k is attached to the lower end of the rod, with the other end of the spring attached to
a rigid support.
If the rod is displaced by a small angle Θ from the vertical and
released, it moves in angular SHM. (Hint: Too see this, assume that
the angle Θ is small enough for the approximations sinΘ ≈ Θ and
cosΘ ≈ 1 to be valid. The motion is simple harmonic if
2
2
2
d θ/dt = −ω θ , and the period is then T = 2π/ω.)
Part A
Calculate the period.
Express your answer in terms of the given quantities.
ANSWER:
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22/23
3/22/23, 3:21 AM
=
T
Homework 1
−−
2π√
M
3k
Correct
Exercise 14.55
An object is moving in damped SHM, and the damping constant can be varied.
Part A
If the angular frequency of the motion is ω when the damping constant is zero, what is the angular frequency, expressed in
terms of ω, when the damping constant is one-half the critical damping value?
Express your answer in terms of ω.
ANSWER:
ω
′
=
–
√3
2
ω
Correct
Exercise 14.56 - Enhanced - with Feedback
A 55.0 g hard-boiled egg moves on the end of a spring with force constant k = 25.0 N/m . Its initial displacement is 0.500 m. A
damping force Fx = −bv x acts on the egg, and the amplitude of the motion decreases to 0.100 m in a time of 5.00 s .
Part A
Calculate the magnitude of the damping constant b .
Express your answer in kilograms per second.
ANSWER:
b
= 3.54×10−2
kg/s
Correct
Score Summary:
Your score on this assignment is 100%.
You received 14 out of a possible total of 14 points.
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