Baixe Tipler chap 14 solution e outras Exercícios em PDF para Física, somente na Docsity!
Chapter 14
Oscillations
Conceptual Problems
1 • True or false:
( a ) For a simple harmonic oscillator, the period is proportional to the square of the amplitude. ( b ) For a simple harmonic oscillator, the frequency does not depend on the amplitude. ( c ) If the net force on a particle undergoing one-dimensional motion is proportional to, and oppositely directed from, the displacement from equilibrium, the motion is simple harmonic.
( a ) False. In simple harmonic motion, the period is independent of the amplitude.
( b ) True. In simple harmonic motion, the frequency is the reciprocal of the period which, in turn, is independent of the amplitude.
( c ) True. This is the condition for simple harmonic motion
2 • If the amplitude of a simple harmonic oscillator is tripled, by what factor is the energy changed?
Determine the Concept The energy of a simple harmonic oscillator varies as the square of the amplitude of its motion. Hence, tripling the amplitude increases the energy by a factor of 9.
3 •• [SSM] An object attached to a spring exhibits simple harmonic motion with an amplitude of 4.0 cm. When the object is 2.0 cm from the equilibrium position, what percentage of its total mechanical energy is in the form of potential energy? ( a ) One-quarter. ( b ) One-third. ( c ) One-half. ( d ) Two-thirds. ( e ) Three-quarters.
Picture the Problem The total energy of an object undergoing simple harmonic motion is given by E (^) tot = 21 kA^2 , where k is the force constant and A is the amplitude of the motion. The potential energy of the oscillator when it is a distance x from its equilibrium position is U ( x ) = 21 kx^2.
Express the ratio of the potential energy of the object when it is 2.0 cm from the equilibrium position to its total energy:
( ) 2
2 2 2 1
2 2 1
tot A
x kA
kx E
U x = =
1348 Chapter 14
Evaluate this ratio for x = 2.0 cm and
A = 4.0 cm:
4.0cm
2 cm 2.0cm 2
2
tot
E
U
and ( a ) is correct.
4 • An object attached to a spring exhibits simple harmonic motion with an amplitude of 10.0 cm. How far from equilibrium will the object be when the system’s potential energy is equal to its kinetic energy? ( a ) 5.00 cm. ( b ) 7.07 cm. ( c ) 9.00 cm. ( d ) The distance can’t be determined from the data given.
Determine the Concept Because the object’s total energy is the sum of its kinetic
and potential energies, when its potential energy equals its kinetic energy, its potential energy (and its kinetic energy) equals one-half its total energy.
Equate the object’s potential energy
to one-half its total energy:
2 total
1 U (^) s = E
Substituting for U s and E total yields: ( 2 )
2 1 2
(^21) 2 (^1) kx = kA ⇒ 2
A
x =
Substitute the numerical value of A
and evaluate x to obtain:
07 cm 2
0 cm x = =
and (^ b )^ is correct.
5 • Two identical systems each consist of a spring with one end attached to a block and the other end attached to a wall. The springs are horizontal, and the blocks are supported from below by a frictionless horizontal table. The blocks are oscillating in simple harmonic motions such that the amplitude of the motion of block A is four times as large as the amplitude of the motion of block B. How do their maximum speeds compare? ( a ) v (^) A max= v Bmax, ( b ) v (^) A max= 2 v Bmax,
( c ) v (^) A max= 4 v Bmax, ( d ) This comparison cannot be done by using the data given.
Determine the Concept The maximum speed of a simple harmonic oscillator is
the product of its angular frequency and its amplitude. The angular frequency of a
simple harmonic oscillator is the square root of the quotient of the force constant of the spring and the mass of the oscillator.
Relate the maximum speed of system
A to the amplitude of its motion:
v A max= ωA A A
Relate the maximum speed of system B to the amplitude of its motion:
v B max= ωB A B
1350 Chapter 14
Substituting for k A and simplifying
yields:
B
B Bmax
A max= = k
k v
v ⇒ v A (^) max= 2 v Bmax
( b ) is correct.
7 •• [SSM] Two systems each consist of a spring with one end attached to a block and the other end attached to a wall. The identical springs are horizontal, and the blocks are supported from below by a frictionless horizontal table. The blocks are oscillating in simple harmonic motions with equal amplitudes. However, the mass of block A is four times as large as the mass of block B. How do their maximum speeds compare? ( a ) v (^) A max= v Bmax,
( b ) v (^) A max= 2 v Bmax, ( c ) v (^) A max= 21 v Bmax, ( d ) This comparison cannot be done by
using the data given.
Determine the Concept The maximum speed of a simple harmonic oscillator is
the product of its angular frequency and its amplitude. The angular frequency of a simple harmonic oscillator is the square root of the quotient of the force constant
of the spring and the mass of the oscillator.
Relate the maximum speed of system
A to its force constant: (^) A A
A A max A A A m
k v =ω A =
Relate the maximum speed of system B to its force constant: (^) B B
B B max B B m A
k v =ω A =
Divide the first of these equations by
the second and simplify to obtain:
B
A B
A A
B
B B
B
A A
A
Bmax
Amax A
A
k
k m
m
A m
k
A
m
k
v
v = =
Because the systems differ only in
the masses attached to the springs: (^) A
B Bmax
Amax m
m v
v
Substituting for m A and simplifying yields: 2
1 B
B Bmax
Amax 4
m
m v
v ⇒ v A (^) max= 21 v Bmax
( c ) is correct.
8 •• Two systems each consist of a spring with one end attached to a block and the other end attached to a wall. The identical springs are horizontal, and the blocks are supported from below by a frictionless horizontal table. The blocks are
Oscillations 1351
oscillating in simple harmonic motions with equal amplitudes. However, the mass of block A is four times as large as the mass of block B. How do the magnitudes of their maximum acceleration compare? ( a ) a (^) A max= a Bmax, ( b ) a (^) A max= 2 a Bmax,
( c ) a (^) A max= 21 a Bmax, ( d ) a (^) A max= 41 a Bmax, ( e ) This comparison cannot be done by
using the data given.
Determine the Concept The maximum acceleration of a simple harmonic oscillator is the product of the square of its angular frequency and its amplitude.
The angular frequency of a simple harmonic oscillator is the square root of the quotient of the force constant of the spring and the mass of the oscillator.
Relate the maximum acceleration of
system A to its force constant: (^) A A
A A
2 A max A m A
k
a =ω A =
Relate the maximum acceleration of system B to its force constant: (^) B B
B B
2 B max B m A
k
a =ω A =
Divide the first of these equations by the second and simplify to obtain:
B
A B
A A
B
B B
B
A A
A
,max
Amax A
A
k
k m
m
A m
k
A
m
k
a
a = =
Because the systems differ only in
the masses attached to the springs: (^) A
B Bmax
Amax m
m a
a
Substituting for m A and simplifying yields: 4
1 B
B Bmax
Amax 4
m
m a
a ⇒ a A (^) max=^14 a Bmax
( d ) is correct.
9 •• [SSM] In general physics courses, the mass of the spring in simple harmonic motion is usually neglected because its mass is usually much smaller than the mass of the object attached to it. However, this is not always the case. If you neglect the mass of the spring when it is not negligible, how will your calculation of the system’s period, frequency and total energy compare to the actual values of these parameters? Explain.
Determine the Concept Neglecting the mass of the spring, the period of a simple
harmonic oscillator is given by T = 2 π ω= 2 π mk where m is the mass of the
oscillating system (spring plus object) and its total energy is given by E (^) total = 21 kA^2.
Oscillations 1353
Substituting for T A and simplifying
yields: 4
2
B
B B
A =
T
T
m
m ⇒ m A (^) = 4 m B
( a ) is correct.
11 •• Two mass–spring systems oscillate at frequencies f A and f B. If f A = 2 f (^) B and the systems’ springs have identical force constants, it follows that the
systems’ masses are related by ( a ) m A = 4 m B , ( b ) m A (^) = m B 2 , ( c ) m A (^) = m B 2 ,
( d ) m (^) A = m B 4.
Picture the Problem We can use m
k f
= to express the frequencies of the
two mass-spring systems in terms of their masses. Dividing one of the equations
by the other will allow us to express m A in terms of m B.
Express the frequency of mass-spring system A as a function of its mass: A^2 A
m
k f
Express the frequency of mass-
spring system B as a function of its mass:
B
B 2
m
k f
Divide the second of these equations by the first to obtain: (^) B
A A
B m
m f
f
Solve for m A: 4 B 1 B
2
B
B B
2
A
B A (^2) f m m
f m f
f m (^) ⎟⎟ = ⎠
( d ) is correct.
12 •• Two mass–spring systems A and B oscillate so that their total mechanical energies are equal. If m A = 2 m B , which expression best relates their
amplitudes? ( a ) A A = A B /4, ( b ) A A (^) = A B 2 , ( c ) A A = A B , ( d ) Not enough
information is given to determine the ratio of the amplitudes.
Picture the Problem We can relate the energies of the two mass-spring systems
through either E = 21 kA^2 or E = 21 m ω^2 A^2 and investigate the relationship between
their amplitudes by equating the expressions, substituting for m A, and expressing A A in terms of A B.
1354 Chapter 14
Express the energy of mass-spring system A:
2 A
2 2 A A
(^21) 2 A A 1
E A = k A = m ω A
Express the energy of mass-spring system B:
2 B
2 2 B B
(^21) 2 B B
1
E B = k A = m ω A
Divide the first of these equations by the second to obtain: (^21) B B^2 B^2
2 A
2 2 A A 1
B
A 1
m A
m A E
E
Substitute for m A and simplify: 2 B
2 B
2 A
2 A 2 B
2 B B
2 A
2 1 2 B^ A^2 A
A
m A
m A
Solve for A A: B A
B A 2
A A
Without knowing how ωA and ωB , or k A and k B , are related, we cannot simplify
this expression further. ( d ) is correct.
13 •• [SSM] Two mass–spring systems A and B oscillate so that their total mechanical energies are equal. If the force constant of spring A is two times the force constant of spring B, then which expression best relates their amplitudes? ( a ) A A = A B /4, ( b ) A A (^) = A B 2 , ( c ) A A = A B , ( d ) Not enough information is given to determine the ratio of the amplitudes.
Picture the Problem We can express the energy of each system using 2 2 E = 1 kA and, because the energies are equal, equate them and solve for A A.
Express the energy of mass-spring
system A in terms of the amplitude
of its motion:
2 2 A A 1 E A (^) = k A
Express the energy of mass-spring system B in terms of the amplitude
of its motion:
2 2 B B
1 E B (^) = k A
Because the energies of the two systems are equal we can equate
them to obtain:
2 2 B B
(^21) 2 A A
(^1) k A = k A ⇒ B A
B A (^) k A
k A =
1356 Chapter 14
( b ) L A = 4 L B and m A = m B , ( c ) L A = 4 L B whatever the ratio m A/ m B , ( d ) L A (^) = 2 L B whatever the ratio m A/ m B.
Picture the Problem The period of a simple pendulum is independent of the mass
of its bob and is given by T = 2 π Lg.
Express the period of pendulum A:
g
L
T A = 2 π A
Express the period of pendulum B:
g
L
T B = 2 π B
Divide the first of these equations by
the second and solve for L A/ L B :
2
B
A B
A ⎟⎟ ⎠
T
T
L
L
Substitute for T A and solve for L B to obtain: B B
2
B
B A 4
L L
T
T
L ⎟⎟ =
( c ) is correct.
17 •• [SSM] Two simple pendulums are related as follows. Pendulum A has a length L A and a bob of mass m A; pendulum B has a length L B and a bob of mass m B. If the frequency of A is one-third that of B, then ( a ) L A = 3 L B and m A = 3 m B , ( b ) L A = 9 L B and m A = m B , ( c ) L A = 9 L B regardless of the ratio m A/ m B , ( d ) L A (^) = 3 L Bregardless of the ratio m A/ m B.
Picture the Problem The frequency of a simple pendulum is independent of the
mass of its bob and is given by. 2
f g L
Express the frequency of pendulum A: A^2 A
L
g f
A
A (^4 2) f
g L
Similarly, the length of pendulum B
is given by: (^2) B^2 B 4 f
g L
Oscillations 1357
Divide the first of these equations by the second and simplify to obtain:
2
A
B 2 A
2 B
2 B
2
2 A
2
B
A
f
f f
f
f
g
f
g
L
L
Substitute for f A to obtain: 9
2
3 B
1
B B
A =
f
f L
L
⇒ ( c ) is correct.
18 •• Two simple pendulums are related as follows. Pendulum A has a length L A and a bob of mass m A; pendulum B has a length L B a bob of mass m B. They have the same period. The only thing different between their motions is that the amplitude of A’s motion is twice that of B’s motion, then ( a ) L A = L B and m A = m B , ( b ) L A = 2 L B and m A = m B , ( c ) L A = L B whatever the ratio m A/ m B , ( d ) L A (^) = 21 L Bwhatever the ratio m A/ m B.
Picture the Problem The period of a simple pendulum is independent of the mass
of its bob and is given by T = 2 π Lg .For small amplitudes, the period is
independent of the amplitude.
Express the period of pendulum A: g
L
T A = 2 π A
Express the period of pendulum B: g
L
T B = 2 π B
Divide the first of these equations by the second and solve for L A/ L B :
2
B
A B
A ⎟⎟ ⎠
T
T
L
L
Because their periods are the same: 1
2
B
B B
A =
T
T
L
L
⇒ ( c ) is correct.
19 •• True or false:
( a ) The mechanical energy of a damped, undriven oscillator decreases exponentially with time. ( b) Resonance for a damped, driven oscillator occurs when the driving frequency exactly equals the natural frequency. ( c ) If the Q factor of a damped oscillator is high, then its resonance curve will be narrow.
Oscillations 1359
Because their damping constants are the same: B
A B
A m
m
Substituting for m A yields: 4
B
B B
A = =
m
m
⇒ τ A = 4 τB
( a ) is correct.
21 •• Two damped spring-mass oscillating systems have identical spring constants and decay times. However, system A’s mass m A is twice system B’s mass m B. How do their damping constants, b , compare? ( a ) b A (^) = 4 b B,
( b ) b A (^) = 2 b B,( c ) b A (^) = b B, ( d ) b A (^) = 21 b B, ( e ) Their decay times cannot be
compared, given the information provided.
Picture the Problem The decay time τ of a damped oscillator is related to the
mass m of the oscillator and the damping constant b for the motion according to
τ= mb.
Express the damping constant of System A: A
A A
m b =
The damping constant for System B is given by: (^) B
B
B τ
m b =
Dividing the first of these equations by the second and simplifying yields:
A
B B
A
B
B
A
A
B
A
m
m m
m
b
b = =
Because their decay times are the same: B
A B
A m
m b
b
Substituting for m A yields: 2
B
B B
A = =
m
m b
b ⇒ b A (^) = 2 b B
( b ) is correct.
22 •• Two damped, driven spring-mass oscillating systems have identical driving forces as well as identical spring and damping constants. However, the mass of system A is four times the mass of system B. Assume both systems are very weakly damped. How do their resonant frequencies compare?
( a ) ω A= ωB , ( b ) ω A = 2 ωB, ( c ) ωA = 21 ωB, ( d ) ωA = 41 ωB , ( e ) Their resonant
frequencies cannot be compared, given the information provided.
1360 Chapter 14
Picture the Problem For very weak damping, the resonant frequency of a spring- mass oscillator is the same as its natural frequency and is given by
ω 0 = k m ,where m is the oscillator’s mass and k is the force constant of the
spring.
Express the resonant frequency of System A: A
A A (^) m
k
The resonant frequency of System B is given by: B
B B (^) m
k
Dividing the first of these equations by the second and simplifying yields:
A
B B
A
B
B
A
A
B
A m
m k
k
m
k
m
k
= =
Because their force constants are the same: A
B B
A m
m
Substituting for m A yields: 2
1 B
B B
A 4
m
m
⇒ ( c ) is correct.
23 •• [SSM] Two damped, driven spring-mass oscillating systems have identical masses, driving forces, and damping constants. However, system A’s force constant k A is four times system B’s force constant k B. Assume they are both very weakly damped. How do their resonant frequencies compare?
( a ) ω A= ωB , ( b ) ω A = 2 ωB, ( c ) ωA = 21 ωB, ( d ) ωA = 41 ωB , ( e ) Their resonant
frequencies cannot be compared, given the information provided.
Picture the Problem For very weak damping, the resonant frequency of a spring- mass oscillator is the same as its natural frequency and is given by
ω 0 = k m ,where m is the oscillator’s mass and k is the force constant of the
spring.
Express the resonant frequency of System A: A
A A (^) m
k
The resonant frequency of System B is given by: B
B B (^) m
k
1362 Chapter 14
Estimation and Approximation
25 • [SSM] Estimate the width of a typical grandfather clocks’ cabinet relative to the width of the pendulum bob, presuming the desired motion of the pendulum is simple harmonic.
Picture the Problem If the motion of
the pendulum in a grandfather clock is to be simple harmonic motion, then its
period must be independent of the angular amplitude of its oscillations.
The period of the motion for large-
amplitude oscillations is given by Equation 14-30 and we can use this
expression to obtain a maximum value for the amplitude of swinging
pendulum in the clock. We can then use this value and an assumed value for the
length of the pendulum to estimate the width of the grandfather clocks’
cabinet.
L
w
θ
w bob w amplitude
Referring to the diagram, we see that
the minimum width of the cabinet is determined by the width of the bob
and the width required to
accommodate the swinging pendulum:
w = w bob + w amplitude and
bob
amplitude bob
w
w w
w = + (1)
Express w amplitude in terms of the
angular amplitude θ and the length
of the pendulum L :
w amplitude = 2 L sin θ
Substituting for w amplitude in equation
(1) yields: bob bob
2 sin 1 w
L
w
w θ
Equation 14-30 gives us the period
of a simple pendulum as a function of its angular amplitude:
sin 2
T T 0 1 2 2 θ
Oscillations 1363
If T is to be approximately equal to
T 0 , the second term in the brackets must be small compared to the first
term. Suppose that:
sin 4
Solving for θ yields: θ≤ 2 sin−^1 ( 0. 0632 ) ≈ 7. 25 °
If we assume that the length of a
grandfather clock’s pendulum is about 1.5 m and that the width of the
bob is about 10 cm, then equation (2) yields:
( ) 5
10 m
5 msin 7. 25 1 bob
w
w
26 • A small punching bag for boxing workouts is approximately the size and weight of a person’s head and is suspended from a very short rope or chain. Estimate the natural frequency of oscillations of such a punching bag.
Picture the Problem For the purposes of this estimation, model the punching bag
as a sphere of radius R and assume that the spindle about which it rotates to be 1. times the radius of the sphere. The natural frequency of oscillations of this
physical pendulum is given by I
MgD f
0 =^0 = where^ M^ is the mass of the
pendulum, D is the distance from the point of support to the center of mass of the
punching bag, and I is its moment of inertia about an axis through the spindle from which it is supported and about which it swivels.
Express the natural frequency of
oscillation of the punching bag: (^0 2) spindle
I
MgD f
From the parallel-axis theorem we have:
2 I spindle (^) = I cm+ Mh where h = 1. 5 R + 0. 5 R = 2 R
Substituting for I cm and h yields: (^) I spindle (^) = 52 MR^2 + M ( 2 R )^2 = 4. 4 MR^2
Substitute for I spindle in equation (1) to obtain:
( ) R
g MR
Mg R f 2 2. 2
Assume that the radius of the
punching bag is 10 cm, substitute numerical values and evaluate f 0 :
( )
1 Hz
- 10 m
81 m/s 2
f
Oscillations 1365
( a ) Express the period of a uniform rod pivoted at one end: MgD
I
T = 2 π
where I is the moment of inertia of the stick about an axis through one end, M is the mass of the stick, and D (= L /2) is the distance from the end of the stick to its center of mass.
Express the moment of inertia of a rod about an axis through its end:
2 3
I =^1 ML
Substitute the values for I and D in the expression for T and simplify to
obtain: (^ )^ g
L
Mg L
ML
T
2 1
2 3 1
Substitute numerical values and evaluate T :
5 s
81 m/s
80 m
T = 2 π 2 =
( b ) Express the period of a simple pendulum: (^) g
L'
T '= 2 π
where L ′ is slightly longer than the arm length due to the size of the briefcase.
Assuming L ′ = 1.0 m, evaluate the period of the simple pendulum:
0 s
81 m/s
0 m
T' = 2 π 2 =
( c ) From observation of people as they walk, these estimates seem reasonable.
Simple Harmonic Motion
29 • The position of a particle is given by x = (7.0 cm) cos 6 π t , where t is
in seconds. What are ( a ) the frequency, ( b ) the period, and ( c ) the amplitude of the particle’s motion? ( d ) What is the first time after t = 0 that the particle is at its equilibrium position? In what direction is it moving at that time?
Picture the Problem The position of the particle is given by x = A cos(ω + t δ)
where A is the amplitude of the motion, ω is the angular frequency, and δ is a
phase constant. The frequency of the motion is given by f = ω 2 πand the period
of the motion is the reciprocal of its frequency.
( a ) Use the definition of ω to
determine f :
- 00 Hz 2
6 s 2
1 = = =
−
f
1366 Chapter 14
( b ) Evaluate the reciprocal of the
frequency:
- 333 s
- 00 Hz
f
T
( c ) Compare x = (7.0 cm) cos 6 π t to
x = A cos ( ω t +δ)to conclude that:
A = 7. 0 cm
( d ) Express the condition that must
be satisfied when the particle is at its equilibrium position:
cos ω t = 0 ⇒
ω t = ⇒
t =
Substituting for ω yields:
( )
- 0833 s 2 6
t
Differentiate x to find v ( t ): [( ) ]
( ) t
t dt
d v
42 cm/ssin 6
- 0 cmcos 6
= −
Evaluate v (0.0833 s):
v ( 0. 0833 s) =−( 42 πcm/s) sin 6 π( 0. 0833 s) < 0
Because v < 0, the particle is moving in the − x direction at t = 0.0833 s.
30 • What is the phase constant δ in x = A cos( ω + t δ)(Equation 14-4) if the
position of the oscillating particle at time t = 0 is ( a ) 0, ( b ) – A , ( c ) A , ( d ) A /2?
Picture the Problem The initial position of the oscillating particle is related to
the amplitude and phase constant of the motion by x 0 = A cos δ where 0 ≤ δ < 2 π.
( a ) For x 0 = 0: cos δ = 0 ⇒ ( ) 2
cos 10
( b ) For x 0 = − A : − A = A cos δ⇒ δ = cos −^1 ( − 1 )= π
( c ) For x 0 = A : A = A cos δ⇒ δ= cos −^1 ( ) 1 = 0
( d ) When x = A /2:
cos δ
A
A
cos 1
