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Week 13: Chapter 14
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Week 13: Chapter 14
Due: 11:59pm on Sunday, April 26, 2015
To understand how points are awarded, read the Grading Policy for this assignment.
Exercise 14.2
Description: If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it
will oscillate. If it is displaced a distance ## m from its equilibrium position and released with zero initial speed. Then
after a time ## s...
If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released, it will oscillate. If it
is displaced a distance 0.125m from its equilibrium position and released with zero initial speed. Then after a time 0.795
s its displacement is found to be a distance 0.125m on the opposite side, and it has passed the equilibrium position
once during this interval.
Part A
Find the amplitude.
ANSWER:
A
= = 0.125 m Part B
Find the period.
ANSWER:
T
= 1.59 s = Part C
Find the frequency.
ANSWER:
f
= = 0.629 Hz Exercise 14.7
Description: A body of unknown mass is attached to an ideal spring with force constant ## N/m. It is found to
vibrate with a frequency of ## Hz. (a) Find the period. (b) Find the angular frequency. (c) Find the mass of the body.
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A body of unknown mass is attached to an ideal spring with force constant 130N/m . It is found to vibrate with a
frequency of 6.50Hz .
Part A
Find the period.
ANSWER:
T
= = 0.154 s Part B
Find the angular frequency.
ANSWER:
ω
= = 40.8 rad/s Part C
Find the mass of the body.
Express your answer using two significant figures.
ANSWER:
m
= = 7.8×10−2 kg Exercise 14.9
Description: An object is undergoing SHM with period T and amplitude 0.320 m. At t = 0, the object is at x = 0.320
m and is instantaneously at rest. (a) Calculate the time it takes the object to go from x = 0.320 m, to x = 0.160 m.
(b) Calculate the time it...
An object is undergoing SHM with period 0.860s and amplitude 0.320 m. At t = 0, the object is at x = 0.320 m and is
instantaneously at rest.
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:
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t
= = 0.143
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
= = 7.17×10−2
Exercise 14.11
Description: A frictionless block of mass ## kg is attached to an ideal spring with force constant ## N/m. At t=0 the
spring is neither stretched nor compressed and the block is moving in the negative direction at a speed of ## m/s.
(a) Find the amplitude. (b)...
A frictionless block of mass 2.45kg is attached to an ideal spring with force constant 270N/m . At t = 0 the spring is
neither stretched nor compressed and the block is moving in the negative direction at a speed of 12.7m/s .
Part A
Find the amplitude.
ANSWER:
A
= = 1.21 m Part B
Find the phase angle.
ANSWER:
ϕ
= 1.57 rad Part C
Write an equation for the position as a function of time.
ANSWER:
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x = (−
1.21m )sin(( 10.5rad/s )t)
x = (−
1.21m )cos(( 10.5rad/s )t)
x = (−
10.5 m)sin((1.21 rad/s)t)
x = (−
10.5 m)cos(( 1.21 rad/s)t)
Exercise 14.12
Description: A 2.00kg, 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. (a) Find (a) the amplitude and (b) the phase angle. (b) ...
(c) Write an equation for the ...
A 2.00kg, frictionless block is attached to an ideal spring with force constant 300 N/m . At t
4.00 m/s and displacement +0.200 m.
= 0
the block has velocity
Part A
Find (a) the amplitude and (b) the phase angle.
ANSWER:
A
= 0.383 m Part B
ANSWER:
ϕ
= 1.02 rad Part C
Write an equation for the position as a function of time.
Assume x (t) in meters and t in seconds.
ANSWER:
x (t)
= Also accepted: m , Exercise 14.19
Description: A m mass on a spring has displacement as a function of time given by the equation x( t ) = ( (7.40 (
cm)) ) cos ( (( (4.16 ( rad/s) ) )t 2.42 (rad)) ). (a) Find the time for one complete vibration. (b) Find the force constant
of the spring. (c...
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A 2.10kg mass on a spring has displacement as a function of time given by the equation x(t) = (7.40 cm) cos [(4.16 rad/s)t − 2.42 rad] .
Part A
Find the time for one complete vibration.
ANSWER:
T
= 1.51 s Part B
Find the force constant of the spring.
ANSWER:
k
= 36.4 N/m = Part C
Find the maximum speed of the mass.
ANSWER:
v max
= 0.308 m/s Part D
Find the maximum magnitude of force on the mass.
ANSWER:
F max
= = 2.69 N Part E
Find the position of the mass at t
= 1.00 s;
ANSWER:
x
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Part F
Find the speed of the mass at t
= 1.00 s;
ANSWER:
v
= 0.303 m/s Part G
Find the magnitude of acceleration of the mass at t
= 1.00 s;
ANSWER:
a
= 0.216 m/s 2 Part H
Find the magnitude of force on the mass at t
= 1.00 s;
ANSWER:
F
= = 0.453 N Also accepted: = 0.454
Exercise 14.30
Description: A toy of mass ## kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring
with force constant ## N/m. When the object is a distance ## m from its equilibrium position, it is observed to have
a speed of ## m/s. (a) What is the...
A toy of mass 0.150kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force
constant 295N/m . When the object is a distance 1.15×10−2m from its equilibrium position, it is observed to have a
speed of 0.305m/s .
Part A
What is the total energy of the object at any point of its motion?
ANSWER:
E
= = 2.65×10−2 J Part B
What is the amplitude of the motion?
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ANSWER:
A
= 1.34×10−2 m = Part C
What is the maximum speed attained by the object during its motion?
ANSWER:
v max
= = 0.594 m/s Exercise 14.38
Description: A thrillseeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass
and oscillates vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its
natural unstretched length...
A thrillseeking cat with mass 4.00 kg is attached by a harness to an ideal spring of negligible mass and oscillates
vertically in SHM. The amplitude is 0.050 m, and at the highest point of the motion the spring has its natural unstretched
length. Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring), the kinetic
energy of the cat, the gravitational potential energy of the system relative to the lowest point of the motion, and the sum
of these three energies when the cat is
Part A
Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at
its highest point.
ANSWER:
Uspring
= 0 J Part B
Calculate the kinetic energy of the cat when the cat is at its highest point.
ANSWER:
K
= 0 J Part C
Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at
its highest point.
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ANSWER:
Ugrav
= 3.92 J Part D
Calculate the sum of these three energies when the cat is at its highest point.
ANSWER:
E
= 3.92 J Part E
Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at
its lowest point.
ANSWER:
Uspring
= 3.92 J Part F
Calculate the kinetic energy of the cat when the cat is at its lowest point.
ANSWER:
K
= 0 J Part G
Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at
its lowest point.
ANSWER:
Ugrav
= 0 J Part H
Calculate the sum of these three energies when the cat is at its lowest point.
ANSWER:
E
= 3.92 J https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3435779
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Part I
Calculate the elastic potential energy of the spring (take it to be zero for the unstretched spring) when the cat is at
its equilibrium position.
ANSWER:
Uspring
= 0.980 J Part J
Calculate the kinetic energy of the cat when the cat is at its equilibrium position.
ANSWER:
K
= 0.980 J Part K
Calculate the gravitational potential energy of the system relative to the lowest point of the motion when the cat is at
its equilibrium position.
ANSWER:
Ugrav
= 1.96 J Part L
Calculate the sum of these three energies when the cat is at its equilibrium position.
ANSWER:
E
= 3.92 J Exercise 14.48
Description: A certain simple pendulum has a period on the earth of T. (a) What is its period on the surface of
Mars, where g = 3.71 ( m/s)^2?
A certain simple pendulum has a period on the earth of 1.40s .
Part A
What is its period on the surface of Mars, where g
= 3.71 m/s
2
?
ANSWER:
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TM
= = 2.28 s Exercise 14.54
Description: A 1.80kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a
physical pendulum. The period for smallangle oscillations is 0.940 s. (a) What is the moment of inertia of the
wrench about an axis through the pivot? (b)...
A 1.80kg monkey wrench is pivoted 0.250 m from its center of mass and allowed to swing as a physical pendulum.
The period for smallangle oscillations is 0.940 s .
Part A
What is the moment of inertia of the wrench about an axis through the pivot?
ANSWER:
I
= 9.87×10−2 kg ⋅ m2 Part B
If the wrench is initially displaced 0.400 rad from its equilibrium position, what is the angular speed of the wrench as
it passes through the equilibrium position?
ANSWER:
Ω max
= 2.66 rad/s Exercise 14.56
Description: A holiday ornament in the shape of a hollow sphere with mass ## kg and radius ## m is hung from a
tree limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance
and released, it swings back and...
A holiday ornament in the shape of a hollow sphere with mass 1.5×10−2kg and radius 5.5×10−2m is hung from a tree
limb by a small loop of wire attached to the surface of the sphere. If the ornament is displaced a small distance and
released, it swings back and forth as a physical pendulum.
Part A
Calculate its period. (You can ignore friction at the pivot. The moment of inertia of the sphere about the pivot at the
tree limb is 5M R2 /3 .)
Take the free fall acceleration to be 9.80m/s 2 . Express your answer using two significant figures.
ANSWER:
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T
= = 0.61 s Exercise 14.59
Description: An unhappy rodent of mass ## kg, moving on the end of a spring with force constant ## N/m, is acted
on by a damping force F_x= b * v_x. (a) If the constant b has the value ## kg/s, what is the frequency of
oscillation of the mouse? (b) For what...
An unhappy rodent of mass 0.287kg , moving on the end of a spring with force constant 2.54N/m , is acted on by a
damping force Fx = −b ⋅ v x .
Part A
If the constant b has the value 0.891kg/s , what is the frequency of oscillation of the mouse?
ANSWER:
f
= 0.404 Hz = Part B
For what value of the constant b will the motion be critically damped?
ANSWER:
b
= = 1.71 kg/s Exercise 14.64
Description: A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant k and
mass m. If the damping constant has a value b_1, the amplitude is A_1 when the driving angular frequency equals
sqrt (k/m). (a) In terms of A_1,...
A sinusoidally varying driving force is applied to a damped harmonic oscillator of force constant k and mass m. If the
−
−
−
−
damping constant has a value b1 , the amplitude is A1 when the driving angular frequency equals √ k/m .
Part A
In terms of A1 , what is the amplitude for the same driving frequency and the same driving force amplitude Fmax , if
the damping constant is 3b1 ?
Express your answer in terms of the given quantities.
ANSWER:
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A
= Part B
In terms of A1 , what is the amplitude for the same driving frequency and the same driving force amplitude Fmax , if
the damping constant is b1 /2?
Express your answer in terms of the given quantities.
ANSWER:
A
= Problem 14.72
Description: A block with mass M rests on a frictionless surface and is connected to a horizontal spring of force
constant k. The other end of the spring is attached to a wall (the figure ). A second block with mass m rests on top
of the first block. The...
A block with mass M rests on a frictionless surface and is connected to a horizontal spring of force constant k . The
other end of the spring is attached to a wall (the figure ). A second block with mass m rests on top of the first block. The
coefficient of static friction between the blocks is μ s .
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Part A
Find the maximum amplitude of oscillation such that the top block will not slip on the bottom block.
Express your answer in terms of the variables m, M , k , μ s , and appropriate constants.
ANSWER:
Problem 14.80
Description: An object with mass ## kg is acted on by an elastic restoring force with force constant ## N/m. The
object is set into oscillation with an initial potential energy of ## J and an initial kinetic energy of ## J. (a) What is
the amplitude of...
An object with mass 0.190kg is acted on by an elastic restoring force with force constant 11.0N/m . The object is set
into oscillation with an initial potential energy of 0.150J and an initial kinetic energy of 7.00×10−2J .
Part A
What is the amplitude of oscillation?
ANSWER:
A
= = 0.200 m Part B
What is the potential energy when the displacement is onehalf the amplitude?
ANSWER:
U
= = 5.50×10−2 J Part C
At what displacement are the kinetic and potential energies equal?
ANSWER:
x
= = 0.141 m https://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=3435779
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Part D
What is the value of the phase angle ϕ if the initial velocity is positive and the initial displacement is negative?
ANSWER:
ϕ
= 3.74 rad = Also accepted: = 2.54
Problem 14.83
Description: A partridge of mass ## kg is suspended from a pear tree by an ideal spring of negligible mass. When
the partridge is pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of ## s. (a)
What is its speed as it...
A partridge of mass 5.09kg is suspended from a pear tree by an ideal spring of negligible mass. When the partridge is
pulled down 0.100 m below its equilibrium position and released, it vibrates with a period of 4.24s .
Part A
What is its speed as it passes through the equilibrium position?
ANSWER:
v
= = 0.148 m/s Part B
What is its acceleration when it is 0.050 m above the equilibrium position?
ANSWER:
a
= = 0.110 m/s 2 Part C
When it is moving upward, how much time is required for it to move from a point 0.050 m below its equilibrium
position to a point 0.050 m above it?
ANSWER:
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t
= 0.707 s = Part D
The motion of the partridge is stopped, and then it is removed from the spring. How much does the spring shorten?
ANSWER:
|Δl|
= = 4.46 m Problem 14.95
Description: In the figure the upper ball is released from rest, collides with and the stationary lower ball, and sticks
to it. The strings are both 50.0 cm long. The upper ball has mass is 2.00 kg and it is initially 10.0 cm higher than the
lower ball, which has...
In the figure the upper ball is released from rest, collides with and the stationary lower ball, and sticks to it. The strings
are both 50.0 cm long. The upper ball has mass is 2.00 kg and it is initially 10.0 cm higher than the lower ball, which has
mass 3.00 kg.
Part A
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Find the frequency of the motion after the collision.
ANSWER:
f
= 0.705 Hz Part B
Find the maximum angular displacement of the motion after the collision.
ANSWER:
θ
= 14.5 ∘ Copyright © 2015 Pearson. All rights reserved.
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