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14.1. Solve: The frequency generated by a guitar string is 440 Hz. The period is the inverse of the frequency, hence
Tf
= = = × =−1 12 27 10 2 273
440 Hz s ms. .
14.2. Solve: Your pulse or heart beat is 75 beats per minute. The frequency of your heart’s oscillations is thus
f = = =75 beats
60 s beats s Hz1 25 1 25. .
The period is the inverse of the frequency, hence
Tf
= = =1 10 80
1.25 Hz s.
14.3. Model: The air-track glider oscillating on a spring is in simple harmonic motion.Solve: The glider completes 10 oscillations in 33 s, and it oscillates between the 10 cm mark and the 60 cm mark.
(a) T = = =33 s
10 oscillations s oscillation s3 3 3 3. .
(b) fT
= = =1 10 303
3.3 s Hz.
(c) ω π π= = ( ) =2 2 1 904f 0.303 Hz rad s.
(d) The oscillation from one side to the other is equal to 60 cm – 10 cm = 50 cm = 0.50 m. Thus, the amplitude isA = ( ) =1
2 0 250.50 m m..(e) The maximum speed is
v AT
Amax 1.904 rad s 0.25 m m s= =
= ( )( ) =ω π2
0 476.
14.4. Model: The air-track glider attached to a spring is in simple harmonic motion.Visualize: The position of the glider can be represented as x(t) = A cos ωt.Solve: The glider is pulled to the right and released from rest at t = 0 s . It then oscillates with a period T = 2 0. sand a maximum speed vmax cm s m s= =40 0 40. .
(a) v AT
Av
maxmax and
s rad s
m s
rad s m cm= = = = ⇒ = = = =ω ω π π π
ω π2 2
2 0
0 400 127 12 7
.
.. .
(b) The glider’s position at t = 0.25 s is
x0.25 s 0.127 m rad s 0.25 s m cm= ( ) ( )( )[ ] = =cos . .π 0 090 9 0
14.5. Model: The oscillation is the result of simple harmonic motion.Visualize: Please refer to Figure Ex14.5.Solve: (a) The amplitude A = 10 cm.(b) The time to complete one cycle is the period, hence T = 2 0. s and
fT
= = =1 1
2 00 50
..
s Hz
(c) The position of an object undergoing simple harmonic motion is x t A t( ) = +( )cos ω φ0 . At s, cm,t x= =0 50 thus
5 cm cm s
5 cm
10 cm rad or 60
= ( ) ( ) +[ ]⇒ = = ⇒ =
= °−
10 0
1
2
1
2 3
0
0 01
cos
cos cos
ω φ
φ φ π
14.6. Model: The oscillation is the result of simple harmonic motion.Visualize: Please refer to Figure Ex14.6.Solve: (a) The amplitude A = 20 cm.(b) The period T = 4.0 s, thus
fT
= = =1 10 25
4.0 s Hz.
(c) The position of an object undergoing simple harmonic motion is x t A t( ) = +( )cos ω φ0 . At t x= = −0 100 s, cm.Thus,
− = ( ) ⇒ = −
= −
= ± = ± °− −10
1
2
2
31200 0
1 1 cm 20 cm10 cm
20 cm radcos cos cosφ φ π
Because the object is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram and thusmust have a phase constant between π and 2π radians. Therefore, φ π0
23 120= − = − ° rad .
14.7. Visualize: The phase constant 23 π has a plus sign, which implies that the object undergoing simple
harmonic motion is in the second quadrant of the circular motion diagram. That is, the object is moving to the left.Solve: The position of the object is
x t A t A ft t( ) = +( ) = +( ) = ( ) ( ) +[ ]cos cos cosω φ π φ π π0 0232 44.0 cm rad s rad
The amplitude is A = 4 cm and the period is T f= =1 0 50. s. A phase constant φ π0 2 3 1= = ° rad 20 (secondquadrant) means that x starts at − 1
2 A and is moving to the left (getting more negative).
Assess: We can see from the graph that the object starts out moving to the left.
14.8. Visualize: A phase constant of − 23 π implies that the object that undergoes simple harmonic motion is in
the third quadrant of the circular motion diagram. That is, the object is moving to the right.Solve: The position of the object is given by the equation
x t A t A ft t( ) = +( ) = +( ) = ( ) ( ) −[ ]cos cos cosω φ π φ π π0 0232 6.0 cm rad s rad
The amplitude is A = 6 cm and the period is T f= =1 2 0. s. With φ π0 2 3= − rad, x starts at − 12 A and is moving
to the right (getting more positive).
Assess: As we see from the graph, the object starts out moving to the right.
14.9. Solve: The position of the object is given by the equation
x t A t A ft( ) = +( ) = +( )cos cosω φ π φ0 02
We can find the phase constant φ0 from the initial condition:
0 0 00 0 01 1
2 cm 4.0 cm rad= ( ) ⇒ = ⇒ = ( ) = ±−cos cos cosφ φ φ π
Since the object is moving to the right, the object is in the lower half of the circular motion diagram. Hence,φ π0
12= − rad. The final result, with f = 4.0 Hz, is
x t t( ) = ( ) ( ) −[ ]4.0 cm 8.0 rad s radcos π π12
14.10. Solve: The position of the object is given by the equation
x t A t A ft( ) = +( ) = +( )cos cosω φ π φ0 02
We can find the phase constant φ0 from the initial condition:
− = ⇒ = −( ) =−A A radcos cosφ φ π0 01 1
The final result, with f = 0.50 Hz, is
x t t( ) = ( ) ( ) +[ ]10.0 cm rad scos π π
14.11. Model: The air-track glider is in simple harmonic motion.Solve: (a) We can find the phase constant from the initial conditions for position and velocity:
x0 = Acosφ0 v0x = –ωAsinφ0
Dividing the second by the first, we see that
sin
costan
φφ
φω
0
00
0
0
= = − v
xx
The glider starts to the left (x0 = –5.0 cm) and is moving to the right (v0x = +36.3 cm/s). With a period of 1.5 s =32 s, the angular frequency is ω = 2π/T = 4
3 π rad/s. Thus
φπ
π π01 1
3
36 3= −−
= ° °−tan. cm / s
(4 / 3 rad / s) ( 5.0 cm) rad (60 ) or – rad (–120 )2
3
The tangent function repeats every 180°, so there are always two possible values when evaluating the arctanfunction. We can distinguish between them because an object with a negative position but moving to the right is inthe third quadrant of the corresponding circular motion. Thus φ0 = − 2
3 π rad or –120°.(b) At time t, the phase is φ ω φ π π= + = −t t0
43
23( rad / s) rad . This gives φ = − 2
3 π rad, 0 rad, 23 π rad, and
43 π rad at, respectively, t = 0 s, 0.5 s, 1.0 s, and 1.5 s. This is one period of the motion.
14.12. Model: The block attached to the spring is in simple harmonic motion.Solve: The period of an object attached to a spring is
Tm
kT 2 s= = =π 0 2 0.
where m is the mass and k is the spring constant.(a) For mass = 2m,
Tm
kT= = ( ) =2
22 2 830π . s
(b) For mass 12 m ,
Tm
kT= = =2 2 1 41
12
0π . s
(c) The period is independent of amplitude. Thus T T= =0 2 00. s(d) For a spring constant = 2k,
Tm
kT= = =2
22 1 410π . s
14.13. Model: The air-track glider attached to a spring is in simple harmonic motion.Solve: Experimentally, the period is T = ( ) ( ) =12 0 1 20. . s 10 oscillations s. Using the formula for the period,
Tm
kk
Tm= ⇒ =
=
( ) =2
2 25 48
2 2
π π π1.20 s
0.200 kg N m.
14.14. Model: The oscillating mass is in simple harmonic motion.Solve: (a) The amplitude A = 2.0 cm.(b) The period is calculated as follows:
ω π π= = ⇒ = =210
20 628
TT rad s
10 rad s s.
(c) The spring constant is calculated as follows:
ω ω= ⇒ = = ( )( ) =k
mk m 2 2
5 00.050 kg 10 rad s N m.
(d) The phase constant φ π014= − rad.
(e) The initial conditions are obtained from the equations
x t t v t tx( ) = ( ) −( ) ( ) = −( ) −( )2.0 cm and 20.0 cm scos sin10 1014
14π π
At t = 0 s, these equations become
x v x014 0
141 414 14 14= ( ) −( ) = = −( ) −( ) =2.0 cm cm and 20.0 cm s cm scos . sin .π π
In other words, the mass is at +1.414 cm and moving to the right with a velocity of 14.14 cm/s.(f) The maximum speed is v Amax 2.0 cm 10 rad s cm s= = ( )( ) =ω 20 0. .
(g) The total energy E kA= = ( )( ) = × −12
2 12
2 31 0 105.0 N m 0.02 m J. .(h) At t = 0.40 s, the velocity is
v x014 1 46= −( ) ( )( ) −[ ] =20.0 cm s 10 rad s 0.40 s cm ssin .π
14.15. Model: The mass attached to the spring oscillates in simple harmonic motion.Solve: (a) The period T f= = =1 1 2 0 0 50. . Hz s .(b) The angular frequency ω π π π= = =2 2 2 4f ( ) Hz rad/s .(c) Using energy conservation
12
2 12 0
2 12 0
2kA kx mv x= +
Using x0 = 5.0 cm, v0x = −30 cm/s and k = mω2 = (0.2 kg) (4 rad sπ ) ,2 we get A = 5.54 cm.(d) To calculate the phase constant φ0 ,
cm
5.0 cm
5.54 cm rad
A xcos .
cos .
φ
φ
0 0
01
5 0
0 445
= =
⇒ =
=−
(e) The maximum speed is v Amax 4 rad s 5.54 cm cm s= = ( )( ) =ω π 69 6. .(f) The maximum acceleration is
a A Amax24 rad s 69.6 cm s cm s= = ( ) = ( )( ) =ω ω ω π2 875
(g) The total energy is E mv= = ( )( ) =12
2 12
20 0484max 0.200 kg 0.696 m s J. .
(h) The position at t = 0.40 s is
x0 4 0 445 3 81. cos . . s 5.54 cm 4 rad s 0.40 s rad cm= ( ) ( )( ) +[ ] = +π
14.16. Model: The mass attached to the spring is in simple harmonic motion.Solve: (a) The period is
Tm
k= = ( )
( ) =2 2 1 00π π0.507 kg
20 N m s.
(b) The angular frequency is ω π π π= = =2 2 1 0 2T . s rad s.(c) To calculate the phase constant φ0 ,
x A0 0 01 1
2130 05 0= ⇒ = ( ) ⇒ = ( ) = ±−cos . cos cosφ φ φ π m .10 m rad0
Because the mass is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram. Hence,φ π0
13= − rad.
(d) At t = 0 s, v t tx ( ) = −( )( ) +( )0.10 m rad s2 2 0π π φsin becomes
−( )( ) −( ) =0.10 m 2 rad s rad m sπ πsin .13 0 544
(e) The maximum speed v Amax 2 rad s 0.10 m m s= = ( )( ) =ω π 0 628. .
(g) At t = 1.3 s, x1 3132 0 0669. cos . s 0.10 m 1.3 s rad m= ( ) ( ) −[ ] =π π .
(h) At t = 1.3 s, v x1 313 0 467. sin . s 0.10 m 2 rad s 2 rad s 1.3 s rad m s( ) = −( )( ) ( )( ) −[ ] = −π π π .
14.17. Model: The block attached to the spring is in simple harmonic motion.Visualize:
Solve: (a) The conservation of mechanical energy equation Kf + Usf = Ki + Usi is12 1
2 12
2 12 0
2 12
2 12 0
2
0
0 0 0
0 10
mv k x mv kA mv
Am
kv
+ ( ) = + ⇒ + = +
⇒ = = ( ) =
∆ J J J
1.0 kg
16 N m0.40 m s m = 10.0 cm.
(b) We have to find the velocity at a point where x = A/2. The conservation of mechanical energy equationK U K U2 + = +s2 i si is
1
2
1
2 2
1
20
1
2
1
2
1
4
1
2
1
2
1
4
1
2
3
4
1
2
3
4
3
40 346 34 6
22
2
02
22
02 2
02
02
02
2 0
mv kA
mv mv mv kA mv mv mv
v v
+
= + ⇒ = −
= −
=
⇒ = = ( ) = =
J
0.40 m s m s cm s. .
14.18. Model: The vertical oscillations constitute simple harmonic motion.Visualize:
Solve: (a) At equilibrium, Newton’s first law applied to the physics book is
F mg k y mg
k mg y
ysp
2
N N
0.500 kg 9.8 m s 0.20 m N m
( ) − = ⇒ − − =
⇒ = − = −( )( ) −( ) =
0 0
24 5
∆
∆ .
(b) To calculate the period:
ω πω
π= = = = = =k
mT
24.5 N m
0.50 kg rad s and
2 rad
7.0 rad s s7 0
20 898. .
(c) The maximum speed is
v Amax 0.10 m 7.0 rad s m s= = ( )( ) =ω 0 70.
Maximum speed occurs as the book passes through the equilibrium position.
14.19. Model: The vertical oscillations constitute simple harmonic motion.Visualize:
Solve: The period and angular frequency are
TT
= = = = =20 s
30 oscillations s and
s rad s0 6667
2 2
0 66679 425.
..ω π π
(a) The mass can be found as follows:
ωω
= ⇒ = =( )
=k
mm
k2 2 0 169
15.0 N / m
9.425 rad s kg.
(b) The maximum speed v Amax 9.425 rad s 0.060 m= = ( )( )ω = 0.565 m/s.
14.20. Model: The vertical oscillations constitute simple harmonic motion.Solve: To find the oscillation frequency using ω π= =2 f k m , we first need to find the spring constant k. In
equilibrium, the weight mg of the block and the spring force k∆L are equal and opposite. That is,mg k L k mg L= ⇒ =∆ ∆ . The frequency of oscillation f is thus given as
fk
m
mg L
m
g
L= = = = =1
2
1
2
1
2
1
2
9 83 52
π π π π∆
∆.
. m s
0.020 m Hz
2
14.21. Solve: (a) and (b)
θ (°) θ (rad) sin θ cos θ
0
2
4
6
8
10
12
0
.0349
.0698
.1047
.1396
.1745
.2094
0
.0349
.0698
.1045
.1392
.1736
.2079
1.0000
.9994
.9976
.9945
.9903
.9848
.9781
(c) 12°(d) 10°(e) The small-angle approximation is good to three significant figures for θ up to 10°.
14.22. Model: Assume a small angle of oscillation so there is simple harmonic motion.Solve: The period of the pendulum is
TL
g002 4 0= =π . s
(a) The period is independent of the mass and depends only on the length. Thus T T= =0 4 0. s.(b) For a new length L = 2L0,
TL
gT= = =2
22 5 660
0π . s
(c) For a new length L = L0/2,
TL
gT= = =2
2 1
22 830
0π . s
(d) The period is independent of the amplitude as long as there is simple harmonic motion. Thus T = 4 0. s.
14.23. Model: Assume the small-angle approximation so there is simple harmonic motion.Solve: The period is T = =12.0 s 10 oscillations s1 20. and is given by the formula
TL
gL
Tg= ⇒ =
=
( ) =2
2 235 7
2 2
ππ π
1.20 s9.8 m s cm2 .
14.24. Model: Assume a small angle of oscillation so there is simple harmonic motion.Solve: (a) On the earth the period is
TL
gearth 2
1.0 m
9.80 m s s= = =2 2 2 01π π .
(b) On Venus the acceleration due to gravity is
gGM
R
TL
g
VenusVenus
Venus
2 2
2
VenusVenus
2
N m kg kg
m m s
1.0 m
8.86 m s s
= =× ⋅( ) ×( )
×( )=
⇒ = = =
−
2
11 24
6 2
6 67 10 4 88 10
6 06 108 86
2 2 2 11
. .
..
.π π
14.25. Model: Assume the pendulum to have small-angle oscillations. In this case, the pendulum undergoessimple harmonic motion.Solve: Using the formula g GM R= 2 , the periods of the pendulums on the moon and on the earth are
TL
g
L R
GMT
L R
GMearthearth earth
earthmoon
moon moon
moon
and = = =2 2 22 2
π π π
Because Tearth = Tmoon,
2 2
7 36 10
5 98 10
6 37 1033
2 2 2
22
24
6 2
π πL R
GM
L R
GML
M
M
R
RLearth earth
earth
moon moon
moonmoon
moon
earth
earth
moonearth
6
kg
kg
m
1.74 10 m2.0 m cm
= ⇒ =
= ××
××
( ) =.
.
.
14.26. Model: The pendulum undergoes simple harmonic motion.Solve: (a) The amplitude is 0.10 rad.(b) The frequency of oscillations is
f = = =ωπ π2
5
2 Hz 0.796 Hz
(c) The phase constant φ = π rad.(d) The length can be obtained from the period:
ω ππ π
= = ⇒ =
=( )
( ) =21
2
1
20 392
2 2
fg
LL
fg
0.796 Hz9.8 m s m2 .
(e) At t = 0 s, θ π0 0 10= ( ) ( ) = −0.10 rad radcos . . To find the initial condition for the angular velocity we take thederivative of the angular position:
θ π θ πt td t
dtt( ) = ( ) +( ) ⇒ ( ) = −( )( ) +( )0.10 rad 0.10 radcos sin5 5 5
At t = 0 s, d dtθ π( ) = −( ) ( ) =0
0 50 0. sin rad rad s.
(f) At s, 0.10 rad 2.0 s rad.t = = ( ) ( ) +( ) =2 0 5 0 0842 0. cos ..θ π
14.27. Model: The motion is a damped oscillation.Solve: The amplitude of the oscillation at time t is given by Equation 14.55: A t A e t( ) = −
02τ , where τ = m b is
the time constant. Using x = 0.368 A and t = 10.0 s, we get
0 368 0 36810
5 0010 0. ln . ..A Ae s
2
10.0 s
2ln 0.368 s s 2= ⇒ ( ) = − ⇒ = −
( )=− τ
ττ
14.28. Model: The motion is a damped oscillation.
Solve: The position of a damped oscillator is x t Ae tt( ) = +( )−( )2τ ω φcos 0 . The frequency is 1.0 Hz and the
damping time constant τ is 4.0 s. Let us assume φ0 0= rad and A = 1 with arbitrary units. Thus,
x t e t x t e t( ) = ( )[ ] ⇒ ( ) = ( )− ( ) −t 8.0 s t1.0 Hzcos cos.2 20 125π π
where t is in s. Values of x(t) at selected values of t are displayed in the following table:
t(s) x(t) t(s) x(t) t(s) x(t)
0
0.25
0.50
0.75
1.00
1.25
1.50
1.75
1
0
– 0.939
0
0.882
0
– 0.829
0
2.00
2.50
3.00
3.50
4.00
4.50
5.00
5.50
0.779
– 0.732
0.687
– 0.646
0.607
– 0.570
0.535
– 0.503
6.00
6.50
7.00
7.50
8.00
8.50
9.00
9.50
10.00
0.472
– 0.444
0.417
– 0.392
0.368
– 0.346
0.325
– 0.305
0.286
14.29. Model: The motion is a damped oscillation.
Solve: The position of the air-track glider is x t Ae tt( ) = +( )−( )20
τ ω φcos , where τ = m b and
ω = −k
m
b
m
2
24Using A = 0.20 m, φ0 0= rad , and b = 0.015 kg/s,
ω = − ( )( )
= − × =−4 0
416 9 10 4 00
2
24.
. N m
0.250 kg
0.015 kg s
0.250 kg rad s rad s
Thus the period is
T = = =21 57
πω
π2 rad
4.0 rad s s.
The amplitude at t = 0 s is x0 = A and the amplitude will be equal to e A−1 at a time given by1
2 2 33 3e
A Ae tm
bt= ⇒ = = =−( )2 sτ τ .
The number of oscillations in a time of 33.3 s is (33.3 s)/(1.57 s) = 21.
14.30. Model: The vertical oscillations are damped and follow simple harmonic motion.
Solve: The position of the ball is given by x t Ae tt( ) = +( )−( )2τ ω φcos 0 . The amplitude A t Ae t( ) = −( )2τ is a functionof time. The angular frequency is
ω πω
= = ( ) = ⇒ = =k
mT
15.0 N m
0.500 kg rad s s5 477
21 147. .
Because the ball’s amplitude decreases to 3.0 cm from 6.0 cm after 30 oscillations, that is, after 30 × 1.147 s =34.41 s, we have
3 0 6 0 0 5 0 534 41
. . . ln ..
cm cm s
2 24.8 s 34.414 s 2 34.41 s 2= ( ) ⇒ = ⇒ ( ) = − ⇒ =−( ) −( )e eτ τ
ττ
14.31. Model: The object is undergoing simple harmonic motion.Visualize:
Solve: The formula for the period is
Tf
m
k= =1
2π
(a) When the frequency f is halved, the period is doubled. That is, the period increases from 4.0 s to 8.0 s.(b) When the mass m is quadrupled, the period doubles.
14.32. Model: The object is undergoing simple harmonic motion.Visualize:
Solve: The velocity of an object oscillating on a spring is
v t A tx ( ) = − +( )ω ω φsin 0
(a) For ′ =A A2 and ′ =ω ω 2, we have
′ ( ) = −( )( ) ( ) +[ ] = − +( )v t A t A tx 2 2 2 20 0ω ω φ ω ω φsin sin
That is, the maximum velocity ′ ′A ω remains the same, but the frequency of oscillation is halved.(b) For ′ =m m4 ,
and max max′ =′
= = ′ = ⇒ ′ = ′ ′ = =ω ω ω ωk
m
k
mA A v A A v
4 22 2
Quadrupling of the mass halves the frequency or doubles the time period, and halves the maximum velocity.
14.33. Visualize: Please refer to Figure P14.33.Solve: The position and the velocity of a particle in simple harmonic motion are
x t A t v t A t v tx( ) = +( ) ( ) = − +( ) = − +( )cos sin sinω φ ω ω φ ω φ0 0 0 and max
(a) At t = 0 s, the equation for x yields
5.0 cm 10.0 cm rad( ) = ( ) ( ) ⇒ = ( ) = ±−cos cos .φ φ π0 01 1
30 5
Because the particle is moving to the right at t = 0 s, it is in the lower half of the circular motion diagram, and thephase constant is between π and 2π radians. Thus, φ π0
13= − rad.
(b) At t = 0 s,
v ATx0 0
2
36 80= − = −( )
−
= +ω φ π π
sin sin .10.0 cm cm
(c) The maximum speed is
v Amax 8.0 s10.0 cm cm s= =
( ) =ω π2
7 85.
Assess: The positive velocity at t = 0 s is consistent with the position-versus-time graph and the negative sign ofthe phase constant.
14.34. Visualize: Please refer to Figure P14.34.Solve: The position and the velocity of a particle in simple harmonic motion are
x t A t v t A t v tx( ) = +( ) ( ) = − +( ) = − +( )cos sin sinω φ ω ω φ ω φ0 0 0 and max
From the graph, T = 12 s and the angular frequency is
ω π π π= = =2 2
6T 12 s rad s
(a) Because v Amax cm s= =ω 60 , we have
A = = =60 cm s 60 cm s
rad s cm
ω π 6114 6.
(b) At t = 0 s,
v Ax0 cm 60 cm s cm
0.5 rad rad 30 or rad 150
= − = − ⇒ −( ) = −
⇒ = ( ) = °( ) °( )−
ω φ φ
φ π π
sin sin
sin
0 0
01 1
656
30 30
Because the velocity at t = 0 s is negative and the particle is slowing down, the particle is in the second quadrant ofthe circular motion diagram. Thus φ π0
56= rad.
(c) At t = 0 s, x056 99 2= ( ) ( ) = −114.6 cm rad cmcos .π .
14.35. Model: The vertical mass/spring systems are in simple harmonic motion.Visualize: Please refer to Figure P14.35.Solve: (a) For system A, the maximum speed while traveling in the upward direction corresponds to themaximum positive slope, which is at t = 3.0 s. The frequency of oscillation is 0.25 Hz.(b) For system B, all the energy is potential energy when the position is at maximum amplitude, which for the first timeis at t = 1.5 s. The time period of system B is thus 6.0 s.(c) Spring/mass A undergoes three oscillations in 12 s, giving it a period TA s.= 4 0. Spring/mass B undergoes 2oscillations in 12 s, giving it a period TB s.= 6 0. We have
Tm
kT
m
k
T
T
m
m
k
kAA
AB
B
B
A
B
A
B
B
A
and 4.0 s
6.0 s= = ⇒ =
= =2 22
3π π
If mA = mB, thenk
k
k
kB
A
A
B
= ⇒ = =4
9
9
42 25.
14.36. Model: The astronaut attached to the spring is in simple harmonic motion.Visualize: Please refer to Figure P14.36.Solve: (a) From the graph, T = 3.0 s, so we have
Tm
km
Tk= ⇒ =
=
( ) =2 54 7
2 2
ππ π2
3.0 s
2240 N m kg.
(b) Oscillations occur about an equilibrium position of 1.0 m. From the graph, A = ( ) = =12 00 80 0 40. . m m, φ
0 rad, and
ω π π= = =2 22 094
T 3.0 s rad s.
The equation for the position of the astronaut is
x t A t t
t t t
( ) = + = ( ) ( )[ ] +
⇒ ( ) ( )[ ] + ⇒ ( )[ ] = ⇒ =
cos . . .
. cos . cos . .
ω 1 0 0 4 1 0
1 2 1 0 0 5 0 50
m m cos 2.094 rad s m
m = 0.4 m 2.094 rad s m 2.094 rad s s
The equation for the velocity of the astronaut is
v t A t
v
x ( ) = − ( )⇒ = −( )( ) ( )( )[ ] = −
ω ωsin
sin ..0 5 0 725 s 0.4 m 2.094 rad s 2.094 rad s 0.50 s m s
Thus her speed is 0.725 m/s.
14.37. Model: The block attached to the spring is in simple harmonic motion.Visualize: The position and the velocity of the block are given by the equations
x t A t v t A tx( ) = +( ) ( ) = − +( )cos sinω φ ω ω φ0 0 and
Solve: To graph x(t) we need to determine ω, φ0 , and A. These quantities will be found by using the initial (t = 0 s)conditions on x(t) and vx(t). The period is
Tm
k T= = = ⇒ = = =2 2 1 405
24 472π π ω π π1.0 kg
20 N m s
2 rad
1.405 s rad s. .
At t = 0 s, x A v Ax0 0 0 0= = −cos sinφ ω φ and . Dividing these equations,
tan.
. .φω
φ00
00
1 01 1181 0 841= − = −
−( )( )( )
= ⇒ =v
xx m s
4.472 rad s 0.20 m rad
From the initial conditions,
A xv x= +
= ( ) + −
=02 0
22
21 0
0 300ω
0.20 m m s
4.472 rad s m
..
The position-versus-time graph can now be plotted using the equation
x t t( ) = ( ) ( ) +[ ]0.300 m 4.472 rad s radcos .0 841
14.38. Solve: The object’s position as a function of time is x t A t( ) = +( )cos ω φ0 . Letting x = 0 m at t = 0 s, gives
0 0 012= ⇒ = ±Acosφ φ π
Since the object is traveling to the right, it is in the lower half of the circular motion diagram, giving a phaseconstant between −π and 0 radians. Thus, φ π0
12= − and
x t A t x t A t t( ) = −( ) ⇒ ( ) = = ( ) ( )cos sin . sinω π ω π12
120 10 m
where we have used A = 0.10 m and
ω π π π= = =2
2T
2 rad
4.0 s rad s
Let us now find t where x = 0.060 m:
0 0602
20 4101. sin sin . m 0.10 m
0.060 m
0.10 m s= ( )
⇒ =
=−π
πt t
Assess: The answer is reasonable because it is approximately 18 of the period.
14.39. Model: The particle is in simple harmonic motion.Solve: The equation for the velocity of the particle is
v tx ( ) = −( )( ) ( )25 cm 10 rad s 10 tsin
Substituting into K = 2U gives
1
22
1
2
1
210
2 21
100
10 21
100
2 2 2 2
2
2
2
22
2 2
mv t kx t m k
k
m
x ( ) = ( )
⇒ −( ) ( )[ ] = ( ) ( )[ ]
⇒ ( )( )
=
( )( )
=
⇒ ( ) = ( )
250 cm s t 25 cm 10 t
10 t
10 t
25 cm
250 cm s s
t 10 rad s s
2
sin cos
sin
cos
tan
ω
22 s= ⇒ = =−2 01
102 0 0 09551. tan . .t
14.40. Model: The spring undergoes simple harmonic motion.Solve: (a) Total energy is E = 1
2 kA2. When the displacement is x = 12 A, the potential energy is
U kx k A kA E K E U E= = = ( ) = ⇒ = − =12
2 12
12
2 14
12
2 14
34( )
One quarter of the energy is potential and three-quarters is kinetic.(b) To have U = 1
2 E requires
U kx E kA xA= = = ( ) ⇒ =1
22 1
212
12
2
2
14.41. Solve: Average speed is vavg = ∆x/∆t. During half a period (∆t = 12 T), the particle moves from x = –A to
x = +A (∆x = 2A). Thus
vx
t
A
T
A
T
AA v v vavg max max avg= ∆
∆= = = = = ⇒ =2
2
4 4
2
2 2
2/ /( )
π ω πω
ππ
14.42. Model: The block on a spring is in simple harmonic motion.Solve: (a) The position of the block is given by x t A t( ) = +( )cos ω φ0 . Because x t A( ) = at t = 0 s, we have
φ0 0= rad , and the position equation becomes x t A t( ) = cosω . At t = 0.685 s, 3 0. cos cm 0.685= ( )A ω and at t =0.886 s, − = ( )3 0. cos cm 0.886A ω . These two equations give
cos cos cos .
. . .
0.685 0.886
rad s
ω ω π ωω π ω ω
( ) = − ( ) = −( )⇒ = − ⇒ =
0 886
0 685 0 886 2 00
(b) Substituting into the position equation,
3 0 1 370 0 20 15 0. cos cos . . . cm 2.00 rad s 0.685 s3.0 cm
0.20 cm= ( )( )( ) = ( ) = ⇒ = =A A A A
14.43. Model: The ball attached to a spring is in simple harmonic motion.Solve: (a) Let t = 0 s be the instant when x0 = −5 cm and v0 = 20 cm/s. The oscillation frequency is
ω = = =k
m
2.5 N m
0.10 kg rad / s5 0.
Using Equation 14.27, the amplitude of the oscillation is
A xv= +
= −( ) +
=02 0
22
2
520
6 40ω
cm cm / s
5 rad / s cm.
(b) The maximum acceleration is a Amax2 cm s= =ω 2 160 .
(c) For an oscillator, the acceleration is most positive a a=( )max when the displacement is most negative
x x A= − = −( )max . So the acceleration is maximum when x = −6 40. cm .(d) We can use the conservation of energy between x0 5= − cm and x1 = 3 cm:
12 0
2 12 0
2 12 1
2 12 1
21 0
202
12 0 283 28 3mv kx mv kx v v
k
mx x+ = + ⇒ = + −( ) = =. . m s cm s
Because k is known in SI units of N/m, the energy calculation must be done using SI units of m, m/s, and kg.
14.44. Solve: (a) The velocity is –0.25 m/s at times that satisfy the equation
− = − + ⇒ + = −−
=0 25 0 35 20 200 25
0 350 7143. ( . sin( ) sin( )
.
.. m / s m / s)
m / s
m / st tπ π
The time is
20 0 7143 0 79560 7956
0 1171t t+ = = ⇒ = − = −−π π rad rad
rad rad
20 rad / s ssin ( . ) .
..
This is a time at which v = –0.25 m/s, but it’s not a time after t = 0 s. But the sine function is periodic with period2π rad, so another angle whose sine is 0.7143 is
20 0 7143 0 7956 20 7956
0 1971t t+ = = + ⇒ = + =−π π π rad rad rad
rad rad
20 rad / s ssin ( . ) .
..
This is a time t > 0 s at which v = –0.25 m/s.(b) The amplitude is A = vmax/ω = (0.35 m/s)(20 rad/s) = 0.0175 m = 1.75 cm. Thus the position at t = 0.197 s is
x A t= + = + =cos( ) ( . cos(( .ω φ π0 1 75 20 1 22 cm) rad / s) (0 / 197 s) rad) cm
14.45. Model: The oscillator is in simple harmonic motion. Energy is conserved.Solve: The energy conservation equation E E1 2= is
12 1
2 12 1
2 12 2
2 12 2
2
2 2 2 21
2
1
2
1
2
1
244 48
mv kx mv kx
k k
k
+ = +
( )( ) + ( ) = ( )( ) + ( )
⇒ =
0.30 kg 0.954 m s 0.030 m 0.30 kg 0.714 m s 0.060 m
N m.
The total energy of the oscillator is
E mv kxtotal 0.30 kg 0.954 m s 44.48 N m 0.030 m J= + = ( )( ) + ( )( ) =1
2
1
2
1
2
1
20 15651
212 2 2 .
Because E mvtotal max2= 1
2 ,
0 15651
20 300 1 02. . . J kg m smax
2 max= ( ) ⇒ =v v
Assess: A maximum speed of 1.02 m/s is reasonable.
14.46. Model: The transducer undergoes simple harmonic motion.Solve: Newton’s second law for the transducer is
F ma a arestoring max max max2 N = 0.10 10 kg m s= ⇒ ×( ) ⇒ = ×−40 000 4 0 103 8, .
Because a Amax = ω 2 ,
Aa= = ×
×( )[ ]= × −max
2
2 m s
Hz m = 10.1 m
ω πµ4 0 10
2 1 0 101 01 10
8
62
5.
..
(b) The maximum velocity is
v Amax Hz m m s= = ×( ) ×( ) =−ω π2 1 0 10 1 01 10 63 66 5. . .
14.47. Model: The block attached to the spring is in simple harmonic motion.Solve: (a) The frequency is
fk
m= = =1
2
1
23 183
π π2000 N m
5.0 kg Hz.
(b) From energy conservation,
A xv= +
= +
⋅
=0
2 02
22
0 0501 0
2 3 1830 0707
ω π( . )
. /
.. m
m s
Hz m
(c) The total mechanical energy is
E kA= = ( )( ) =12
2 12
2 5 02000 N m 0.0707 m J.
14.48. Model: The spider is in simple harmonic motion.Solve: Your tapping is a driving frequency. Largest amplitude at fext = 1.0 Hz means that this is the resonancefrequency, so f0 = fext = 1.0 Hz. That is, the spider’s natural frequency of oscillation f0 is 1.0 Hz and ω π0 02= =f2π rad s . We have
ω ω π0 02 2
0 079= ⇒ = = ( )( ) =k
mk m 0.002 kg 2 rad s N m.
14.49. Model: The block undergoes simple harmonic motion.Visualize:
Solve: (a) The frequency of oscillation is
fk
m= = =1
2
1
2
101 125
π π N / m
0.20 kg Hz.
(b) Using conservation of energy, 12 1
2 12 1
2 12 0
2 12 0
2mv kx mv kx+ = + , we find
x xm
kv v1 0
202
12 0 20
0 201 00 0 50
0 235
= + − = − + −
=
( ) ( ..
(( . ( . )
.
m) kg
10.0 N / m m / s) m / s)
m or 23.5 cm
2 2 2
(c) At time t, the displacement is x = Acos(ω t + φ0). The angular frequency is ω = 2π f = 7.071 rad/s. The amplitudeis
A xv= +
= − +
=02 0
2 2
0 201 00
0 245ω
( ..
. m) m / s
7.071 rad / s m2
The phase constant is
φ01 0 1 0 200
2 526=
= −
= ± ± °− −cos cos
..
x
A
m
0.245 m rad or 145
A negative displacement (below the equilibrium point) and positive velocity (upward motion) indicate that thecorresponding circular motion is in the third quadrant, so φ0 = –2.526 rad. Thus at t = 1.0 s,
x = −( ) = − = −( . cos ( . . . .0 245 7 071 2 526 0 409 4 09 m) rad / s) (1.0 s) rad m cm
The block is 4.09 cm below the equilibrium point.
14.50. Model: The mass is in simple harmonic motion.Visualize:
The high point of the oscillation is at the point of release. This conclusion is based on energy conservation.Gravitational potential energy is converted to the spring’s elastic potential energy as the mass falls and stretches thespring, then the elastic potential energy is converted 100% back into gravitational potential energy as the massrises, bringing the mass back to exactly its starting height. The total displacement of the oscillation – high point tolow point – is 20 cm. Because the oscillations are symmetrical about the equilibrium point, we can deduce that theequilibrium point of the spring is 10 cm below the point where the mass is released. The mass oscillates about thisequilibrium point with an amplitude of 10 cm, that is, the mass oscillates between 10 cm above and 10 cm belowthe equilibrium point.Solve: The equilibrium point is the point where the mass would hang at rest, with F w mgsp = = . At the equilibrium
point, the spring is stretched by ∆y = 10 cm = 0.1 m. Hooke’s law is F k ysp = ∆ , so the equilibrium condition is
F k y w mgk
m
g
ysp
229.8 m s
0.1 m s=[ ] = =[ ] ⇒ = = = −∆
∆98
The ratio k m is all we need to find the oscillation frequency:
fk
m= = =−1
2
1
298 1 582
π π s Hz.
14.51. Model: The compact car is in simple harmonic motion.Solve: (a) The mass on each spring is 1200 kg 4 kg( ) = 300 . The spring constant can be calculated as follows:
ω ω π π2 2 2 2 42 300 2 2 0 4 74 10= ⇒ = = ( ) = ( ) ( )[ ] = ×k
mk m m f kg Hz N m. .
(b) The car carrying four persons means that each spring has, on the average, an additional mass of 70 kg. That is,m = + =300 kg kg kg70 370 . Thus,
fk
m= = = × =ω
π π π2
1
2
1
21 80
4.74 10 N m
370 kg Hz
4
.
Assess: A small frequency change from the additional mass is reasonable because frequency is inverselyproportional to the square root of the mass.
14.52. Model: The blocks undergo simple harmonic motion.Visualize:
The length of the stretched spring due to a block of mass m is ∆L1 . In the case of the two block system, the spring isfurther stretched by an amount ∆L2 .Solve: The equilibrium equations from Newton’s second law for the single block and double block systems are
∆ ∆ ∆L k mg L L k m g1 1 2 2( ) = +( ) = ( ) and
Using ∆L2 5 0= . cm , and subtracting these two equations, gives us
∆ ∆ ∆L L k L k m g mg k mg
k
mf
1 2 1
2
2 0 05
14 02
2 23
+( ) − = ( ) − ⇒ ( ) =
⇒ = = ⇒ = ⇒ = =
.
. .
m
9.8 m s
0.05 m rad s Hz
2
ω ω ωπ
14.53. Model: Hooke’s law for the spring. The spring’s compression and decompression constitutes simpleharmonic motion.Visualize:
Solve: (a) The spring’s compression or decompression is one-half of the oscillation cycle. This means the contacttime is ∆t T= 1
2 , where T is the period. The period is calculated as follows:
ω πω
π= = = ⇒ = = = =
⇒ = =
k
mT
f
tT
50 N m
0.50 kg rad s
rad s s
s
101 2 2
100 628
20 314
.
.∆
(b) There is no change in contact time, because period of oscillation is independent of the amplitude or themaximum speed.
14.54. Model: The two blocks are in simple harmonic motion, without the upper block slipping. We will alsoapply the model of static friction between the two blocks.Visualize:
Solve: The net force acting on the upper block m1 is the force of friction due to the lower block m2. The model ofstatic friction gives the maximum force of static friction as
f n m g m a a gs max s s max max s= = ( ) = ⇒ =µ µ µ1 1
Using µs = 0 5. , a gmax S2 29.8 m s m s= = ( )( ) =µ 0 5 4 9. . . That is, the two blocks will ride together if the maximum
acceleration of the system is equal to or less than amax. We can calculate the maximum value of A as follows:
a Ak
m mA A
a m m
kmax max max maxmax
24.9 m s 1.0 kg 5.0 kg
50 N m m= =
+⇒ =
+( )=
( ) +( )=ω 2
1 2
1 2 0 588.
14.55. Model: Assume simple harmonic motion for the two-block system without the upper block slipping. Wewill also use the model of static friction between the two blocks.Visualize:
Solve: The net force on the upper block m1 is the force of static friction due to the lower block m2. The two blocksride together as long as the static friction doesn’t exceed its maximum possible value. The model of static frictiongives the maximum force of static friction as
f n m g m a a g
a
g
A
g T
A
g
s max s s max max s
smax max max
2 s
0.40 m
9.8 m s
= = ( ) = ⇒ =
⇒ = = =
=
=
µ µ µ
µ ω π π
1 1
2 2 22 2
1 50 716
..
Assess: Because the period is given, we did not need to use the block masses or the spring constant in our calculation.
14.56. Model: Assume that the swinging lamp makes a small angle with the vertical so that there is simpleharmonic motion.Visualize:
Solve: (a) Using the formula for the period of a pendulum,
TL
gL g
T= ⇒ =
= ( )
=2
27 51
2 2
ππ π
9.8 m s5.5 s
2 m2 .
(b) The conservation of mechanical energy equation K U K U0 1+ = +g0 g1 for the swinging lamp is
12 0
20
12 1
21
12
20 0
2 2 3
2 1 3 0 449
mv mgy mv mgy mgh mv
v gh g L L
+ = + ⇒ + = +
⇒ = = − °( )
= ( )( ) − °( ) =
J J
9.8 m s 7.51 m m s
max
max
2
cos
cos .
14.57. Model: Assume that the pendulum makes a small angle with the vertical so that there is simple harmonicmotion.Solve: The angle θ made by the string with the vertical as a function of time is
θ θ ω φt t( ) = +( )max cos 0
The pendulum starts from maximum displacement, thus φ0 0= . Thus, θ θ ωt t( ) = max cos . To find the time t whenthe pendulum reaches 4.0° on the opposite side:
− °( ) = °( ) ⇒ = −( ) =−4 0 8 0 0 5 2 0941. . cos cos . .ω ωt t rad
Using the formula for the angular frequency,
ωω
= = = ⇒ = = =g
Lt
9.8 m s
1.0 m rad s
2.0944 rad 2.094 rad
3.130 rad s s
2
3 130 0 669. .
Assess: Because T = =2 2 00π ω . s , a value of 0.669 s for the pendulum to cover a little less than half theoscillation is reasonable.
14.58. Model: Assume a small angle of oscillation so that the pendulum has simple harmonic motion.Solve: The time periods of the pendulums on the earth and on Mars are
TL
gT
L
gearthearth
MarsMars
and = =2 2π π
Dividing these two equations,
T
T
g
gg g
T
Tearth
Mars
Mars
earthMars earth
earth
Mars
2 29.8 m s1.50 s
2.45 s m s= ⇒ =
= ( ) =
2 2
3 67.
14.59. Model: Assume a small angle oscillation of the pendulum so that it has simple harmonic motion.Solve: (a) At the equator, the period of the pendulum is
Tequator 2
1.000 m
9.78 m s s= =2 2 009π .
The time for 100 oscillations is 200.9 s.(b) At the north pole, the period is
Tpole 2
1.000 m
9.83 m s s= =2 2 004π .
The time for 100 oscillations is 200.4 s.(c) The difference between the two answers is 0.5 s, and this difference is quite measurable with a hand-operatedstopwatch.(d) The period on the top of the mountain is 2.010 s. The acceleration due to gravity can be calculated byrearranging the formula for the period:
g LTmountain
mountain
21.000 m2.010 s
m s=
= ( ) =2 2
9 77
2 2π π.
Assess: This last result is reasonable because g decreases with altitude.
14.60. Model: The pendulum is a damped oscillator.Solve: The period of the pendulum and the number of oscillations in 4 hours are calculated as follows:
TL
gN= = = ⇒ = ( ) =2 2 7 773
41853π π 15.0 m
9.8 m s s
3600 s
7.773 s2 osc.
The amplitude of the pendulum as a function of time is A t Ae bt m( ) = − /2 . The exponent of this expression can becalculated to be
− = − ( ) ×( )( ) = −bt
m2
4 3600
20 6545
0.010 kg s s
110 kg.
We have A t e( ) = ( ) =−1.5 m m.0 6545 0 780. .
14.61. Model: A completely inelastic collision between the two gliders resulting in simple harmonic motion.Visualize:
Let us denote the 250 g and 500 g masses as m1 and m2, which have initial velocities vi1 and vi2. After m1 collideswith and sticks to m2, the two masses move together with velocity vf.Solve: The momentum conservation equation p pf i= for the completely inelastic collision is m m v1 2+( ) =f
m v m v1 2i1 i2+ . Substituting the given values,
0.750 kg 0.25 kg 1.20 m s 0.50 kg 0 m s m sf f( ) = ( )( ) + ( )( ) ⇒ =v v 0 40.
We now use the conservation of mechanical energy equation:
K U K U kA m m v
Am m
kv
+( ) = +( ) ⇒ + = +( ) +
⇒ = + = ( ) =
s compressed s equilibrium f
f
J J
0.750 kg
10 N m0.40 m s m
0 0
0 110
12
2 12 1 2
2
1 2 .
The period is
Tm m
k= + = =2 2 1 721 2π π 0.750 kg
10 N m s.
14.62. Model: The block attached to the spring is oscillating in simple harmonic motion.Solve: (a) Because the frequency of an object in simple harmonic motion is independent of the amplitude and/orthe maximum velocity, the new frequency is equal to the old frequency of 2.0 Hz.(b) The speed v0 of the block just before it is given a blow can be obtained by using the conservation of mechanicalenergy equation as follows:
12
2 12
2 12 0
2
0 2 2 0 25
kA mv mv
vk
mA A f A
= =
⇒ = = = ( ) = ( )( )( ) =
max
2.0 Hz 0.02 m m sω π π .
The blow to the block provides an impulse that changes the velocity of the block:
J F t p mv mv
v v
x x= = = −
−( ) ×( ) = ( ) − ( )( ) ⇒ =−
∆ ∆ f
f f N s 0.200 kg 0.200 kg 0.25 m s m s
0
320 1 0 10 0 15. .
Since vf is the new maximum velocity of the block at the equilibrium position, it is equal to Aω. Thus,
A = =( )
= =0 15 0 15
20 012 1 2
. .. .
m s m s
2.0 Hz m cm
ω π
Assess: Because vf is positive, the block continues to move to the right even after the blow.
14.63. Model: A completely inelastic collision between the bullet and the block resulting in simple harmonicmotion.Visualize:
Solve: (a) The equation for conservation of energy after the collision is
1
2
1
24 9752 2kA m m v v
k
m mA= +( ) ⇒ =
+= ( ) =b B f f
b B
2500 N m
1.010 kg0.10 m m s.
The momentum conservation equation for the perfectly inelastic collision p pafter before= is
m m v m v m v
v v
b B f b b B B
b b1.010 kg 4.975 m s 0.010 kg 1.00 kg 0 m s m s
+( ) = +
( )( ) = ( ) + ( )( ) ⇒ = 502
(b) No. The oscillation frequency k m mb B+( ) depends on the masses but not on the speeds.
14.64. Model: The pendulum falls, then undergoes small-amplitude oscillations in simple harmonic motion.Visualize:
We placed the origin of the coordinate system at the bottom of the arc.Solve: We need to find the length of the pendulum. The conservation of mechanical energy equation for thependulum’s fall is K U K U+( ) = +( )g top g bottom
:
12 0
20
12 1
21
12
2
2
0 2 0
1
40 6377
mv mgy mv mgy mg L m
Lg
+ = + ⇒ + ( ) = ( ) +
⇒ = ( ) =
J 5.0 m s J
5.0 m s m.
Using L = 0.6377 m, we can find the frequency f as
fg
L= = =1
2
1
20 624
π π9.8 m s
0.6377 m Hz
2
.
14.65. Model: Assume the small-angle approximation.Visualize:
Solve: The tension in the two strings pulls downward at angle θ. Thus Newton’s second law is
F T may y∑ = − =2 sinθ
From the geometry of the figure we can see that
sinθ =+y
L y2 2
If the oscillation is small, then y << L and we can approximate sinθ ≈ y/L. Since y/L is tanθ, this approximation isequivalent to the small-angle approximation sinθ ≈ tanθ if θ << 1 rad. With this approximation, Newton’s secondlaw becomes
− ≈ − = = ⇒ = −22 22
2
2
2TT
Ly ma m
d y
dt
d y
dt
T
mLyysinθ
This is the equation of motion for simple harmonic motion (see Eqs. 14.33 and 14.47). The constants 2T/mL areequivalent to k/m in the spring equation or g/L in the pendulum equation. Thus the oscillation frequency is
fT
mL= 1
2
2
π
14.66. Visualize: Please refer to Figure P14.66.Solve: The potential energy curve of a simple harmonic oscillator is described by U k x= ( )1
22∆ , where
∆x x x= − 0 is the displacement from equilibrium. From the graph, we see that the equilibrium bond length is x0 =0.13 nm. We can find the bond’s spring constant by reading the value of the potential energy U at a displacement∆x and using the potential energy formula to calculate k.
x (nm) ∆x (nm) U (J) k (N/m)
0.11
0.10
0.09
0.02
0.03
0.04
0.8 × 10−19 J
1.9 × 10−19 J
3.4 × 10−19 J
400
422
425
The three values of k are all very similar, as they should be, with an average value of 416 N/m. Knowing the springconstant, we can now calculate the oscillation frequency of a hydrogen atom on this “spring” to be
fk
m= =
×= ×−
1
2
1
27 9 1013
π π416 N m
1.67 10 kg Hz27 .
14.67. Model: Assume that r Rmarble hoop<< and that θ is a small angle.
Visualize:
Solve: The marble is like an object on an inclined plane. The net force on the marble in the tangential direction is
− = = = ⇒ − =w ma mR mRd
dtmg mR
d
dtsin sinθ α θ θ θ2
2
2
2
where α is the angular acceleration. With the small-angle approximation sinθ θ≈ , this becomes
d
dt
g
R
2
22θ θ ω θ= − = −
This is the equation of motion of an object in simple harmonic motion with a period of
TR
g= =2
2π
ωπ
14.68. Visualize:
Solve: (a) Newton’s second law applied to the penny along the y-axis is
F n mg maynet = − =rFnet is upward at the bottom of the cycle (positive ay), so n > mg. The speed is maximum when passing through
equilibrium, but ay = 0 so n = mg. The critical point is the highest point. rFnet points down and ay is negative. If ay
becomes sufficiently negative, n drops to zero and the penny is no longer in contact with the surface.(b) When the penny loses contact (n = 0), the equation for Newton’s law becomes a gmax = . For simple harmonicmotion,
a Ag
A
f
max
29.8 m s
0.04 m rad
15.65 rad s Hz
= ⇒ = = =
⇒ = = =
ω ω
ωπ π
2 15 65
2 22 49
.
.
14.69. Model: The vertical oscillations constitute simple harmonic motion.Visualize:
Solve: At the equilibrium position, the net force on mass m on Planet X is:
F k L mgk
m
g
Lnet XX N= − = ⇒ =∆
∆0
For simple harmonic motion k m = ω 2 , thus
ω ω π π π22 2
2 2 25 86= ⇒ = = ⇒ =
=
( ) =g
L
g
L Tg
TLX X
X2
14.5 s 100.312 m m s
∆ ∆∆ .
14.70. Model: The doll’s head is in simple harmonic motion and is damped.Solve: (a) The oscillation frequency is
fk
mk m f= ⇒ = ( ) = ( )( ) ( ) =1
22 0 015 2 9 475
2 2 2
ππ π. . kg 4.0 Hz N m
(b) The maximum speed is
v Ak
mAmax
9.475 N m
0.015 kg0.020 m m s= = = ( ) =ω 0 503.
(c) Using A t A e bt m( ) = −0
2/ , we get
0.5 cm 2.0 cm
133.33 s kg kg / s
4.0 s 2 0.015 kg 133.33 s kg( ) = ( ) ⇒ =
⇒ −( ) = ⇒ =
− ( ) ×( ) −( )e e
b b
b b0 25
0 25 0 0104
.
ln . .
14.71. Model: The oscillator is in simple harmonic motion.Solve: (a) The maximum displacement at time t of a damped oscillator is
x t Aet x t
At
max2 max( ) = ⇒ − = ( )
− τ
τ2ln
Using xmax = 0.98A at t = 0.5 s, we can find the time constant τ to be
τ = −( )
=0.5 s
2ln 0.98 s12 375.
25 oscillations will be completed at t = 25T = 12.5 s. At that time, the amplitude will be
x emax, 12.5 s s 2 s10.0 cm cm= ( ) =− ( )( )12 5 12 375 6 03. . .
(b) The energy of a damped oscillator decays more rapidly than the amplitude: E(t) = E0e–t/τ. When the energy is
60% of its initial value, E(t)/E0 = 0.60. We can find the time this occurs as follows:
− = ( )
⇒ = − ( )
= −( ) ( ) =t E t
Et
E t
Eττln ln ln . .
0 0
0 60 6 3212.375 s s
14.72. Model: The oscillator is in simple harmonic motion.Solve: The maximum displacement, or amplitude, of a damped oscillator decreases as xmax(t) = −Ae t 2τ , where τ isthe time constant. We know x Amax = 0 60. at t = 50 s, so we can find τ as follows:
− = ( )
⇒ = −( )
=t x t
A2 2 0 6048 9
ττln
ln ..max 50 s
s
Now we can find the time t30 at which xmax A 0.30= :
tx t
A30 2 2 48 9 0 30 118= − ( )
= − ( ) ( ) =τ ln . ln .max s s
The undamped oscillator has a frequency f = 2 Hz = 2 oscillations per second. Damping changes the oscillationfrequency slightly, but the text notes that the change is negligible for “light damping.” Damping by air, whichallows the oscillations to continue for well over 100 s, is certainly light damping, so we will use f = 2.0 Hz. Thenthe number of oscillations before the spring decays to 30% of its initial amplitude is
N f t= ⋅ = ( ) ⋅ ( ) =30 2362 oscillations s 118 s oscillations
14.73. Model: Samson on the trapeze is in simple harmonic motion.Visualize:
The time t t1D 0D−( ) for Delilah to fall 4.0 m must be equal to the time t t1S 0S−( ) Samson takes to swing from the
angle θ to the lowest point of the arc.Solve: The time of Delilah’s fall can be found from kinematics as follows:
y y v t t a t t
t t t t
1 012
2
12
20 4 0 0 9 8 0 0 9035
− = −( ) + −( )− = ( ) −( ) + −( ) −( ) ⇒ =
0y 1D 0D y 1D 0D
1D 0D2
1D 1D m m m s m s s s. . .
Treating Samson as a pendulum, his period is
TL
g= = =2 2 4 916π π 6.0 m
9.8 m s s2 .
If Samson starts his swing at t = 0 s, his motion is θ = θmaxcos(ω t) = (8.0°)cos[(2π/4.916 s)t)]. He reaches the lowestpoint (θ = 0°) at t1S = 1
4 T = 1.229 s. In order for Samson and Delilah to meet at this point, she had to jump 0.9035 searlier, at t0S = 0.3255 s. Samson’s angle at that instant was
θ π0 8 0
2
4 9160 3255 7 3= °
= °( . )cos.
( . . s
s)
Delilah should jump when Samson makes a 7.3° angle on his downward swing.
14.74. Model: The astronaut will be attached to a spring and will oscillate in simple harmonic motion. She candetermine her mass by measuring her period.Solve: (a)
(b) One end of a spring is attached to the wall of the space ship. The spring was calibrated prior to leaving earthand its spring constant was measured. An astronaut is strapped to the other end of the spring. Another astronautpulls her back and releases her. She will oscillate back and forth on the spring with a period T m k= 2π . It iseasy to measure the period accurately with a stopwatch. Her mass is determined from the above equation.(c) The only required component is the spring. The design value of the spring constant is 300 N/m. The actual valuewill be determined by calibration measurements before launch.(d) The period needs to be slow enough to be easily measured (and also slow enough not to make the astronautsick!) But it can’t be so slow the measurements take too long. A period of approximately 3 s is reasonable, allowingthe astronaut to oscillate 10 times in approximately 30 s. Astronaut masses are most likely to be in the range 50 kg(100 pounds) to 100 kg (220 pounds). Sample oscillation periods for three different astronauts are
Mass m Period T for k = 300 N/m
50 kg
75 kg
100 kg
2.57 s
3.14 s
3.63 s
A target value of k ≈ 300 N/m gives acceptable oscillation periods for the full range of possible astronaut masses.Assess: Spring constants ranging from k ≈ 30 N/m (T = 10 s for 75 kg) to k ≈ 700 N/m (T = 2 s for 75 kg) wouldbe acceptable designs.
14.75. Model: The two springs obey Hooke’s law.Visualize:
Solve: There are two restoring forces on the block. If the block’s displacement x is positive, both restoring forces– one pushing, the other pulling – are directed to the left and have negative values:
F F F k x k x k k x k xx x xnet sp 1 sp 2 eff( ) = ( ) + ( ) = − − = − +( ) = −1 2 1 2
where k k keff = +1 2 is the effective spring constant. This means the oscillatory motion of the block under theinfluence of the two springs will be the same as if the block was attached to a single spring with spring constant keff.The frequency of the blocks, therefore, is
fk
m
k k
m
k
m
k
mf f= = + = + = +1
2
1
2 4 41 2 1
222 1
222
π π π πeff
14.76. Model: The two springs obey Hooke’s law. Assume massless springs.Visualize: Each spring is shown separately. Note that ∆x = ∆x1 + ∆x2.
Solve: Only spring 2 touches the mass, so the net force on the mass is Fm = F2 on m. Newton’s third law tells us thatF2 on m = Fm on 2 and that F2 on 1 = F1 on 2. From Fnet = ma, the net force on a massless spring is zero. Thus Fw on 1 = F2 on 1 =k1∆x1 and Fm on 2 = F1 on 2 = k2∆x2. Combining these pieces of information,
Fm = k1∆x1 = k2∆x2
The net displacement of the mass is ∆x = ∆x1 + ∆x2, so
∆ = ∆ + ∆ = + = +
= +x x x
F
k
F
k k kF
k k
k kFm m
m m1 21 2 1 2
1 2
1 2
1 1
Turning this around, the net force on the mass is
Fk k
k kx k x k
k k
k km =+
∆ = ∆ =+
1 2
1 2
1 2
1 2eff eff where
keff, the proportionality constant between the force on the mass and the mass’s displacement, is the effective springconstant. Thus the mass’s angular frequency of oscillation is
ω = =+
k
m m
k k
k keff 1 1 2
1 2
Using ω12
1= k m/ and ω 22
2= k m/ for the angular frequencies of either spring acting alone on m, we have
ω ω ωω ω
=+
=+
( / ) ( / )
( / ) ( / )
k m k m
k m k m1 2
1 2
12
22
12
22
Since the actual frequency f is simply a multiple of ω, this same relationship holds for f:
ff f
f f=
+12
22
12
22
14.77. Model: The block undergoes SHM after sticking to the spring. Energy is conserved throughout the motion.Visualize:
It’s essential to carefully visualize the motion. At the highest point of the oscillation the spring is stretched upward.Solve: We’ve placed the origin of the coordinate system at the equilibrium position, where the block would sit onthe spring at rest. The spring is compressed by ∆L at this point. Balancing the forces requires k∆L = mg. Theangular frequency is w2 = k/m = g/∆L, so we can find the oscillation frequency by finding ∆L. The block hits thespring (1) with kinetic energy. At the lowest point (3), kinetic energy and gravitational potential energy have beentransformed into the spring’s elastic energy. Equate the energies at these points:
K U U U mv mg L k L A mg A1 1g 3s 3g+ = + ⇒ + ∆ = ∆ + + −12 1
2 12
2( ) ( )
We’ve used y1 = ∆L as the block hits and y3 = –A at the bottom. The spring has been compressed by ∆y = ∆L + A.Speed v1 is the speed after falling distance h, which from free-fall kinematics is v1
2 = 2gh. Substitute this expressionfor v1
2 and mg/∆L for k, giving
mgh mg Lmg
LL A mg A+ ∆ =
∆∆ + + −
22
( )( ) ( )
The mg term cancels, and the equation can be rearranged into the quadratic equation
( ) ( )∆ + ∆ − =L h L A2 22 0
The positive solution is
∆ = + − = + − =L h A h2 2 0 030 0 100 0 030 0 0744( . ( . . . m) m) m m2 2
Now that ∆L is known, we can find
ω ωπ
=∆
= = ⇒ = =g
Lf
9 80
0 074411 48
21 83
.
.. .
m / s
m rad / s Hz
2
14.78. Model: Model the bungee cord as a spring. The motion is damped SHM.Visualize:
Solve: (a) For light damping, the oscillation period is
Tm
kk m
T= ⇒ =
=
=2
275 185
2 2
π π π( kg)
2
4.0 s N/m
(b) The maximum speed is vmax = ωA = (2π/4.0 s)(11.0 m) = 17.3 m/s.(c) Jose oscillates about the equilibrium position at which he would hang at rest. Balancing the forces, ∆L = mg/k =(75 kg)(9.80 m/s2)/(185 N/m) = 3.97 m. Jose’s lowest point is 11.0 m below this point, so the bungee cord isstretched by ∆ymax = ∆L + A = 14.97 m. Choose this lowest point as y = 0. Because Jose is instantaneously at rest atthis point, his energy is entirely the elastic potential energy of the stretched bungee cord. Initially, his energy wasentirely gravitational potential energy. Equating his initial energy to his energy at the lowest point,
U U k y mgh
hk y
mg
lowest point highest point max
max2
2
N / m)(14.97 m)
kg)(9.80 m / s m
= ⇒ ∆ =
= ∆ = =
12
2
2
2
185
2 7528 2
( )
( ) (
( ).
Jose jumped 28.2 m above the lowest point.(d) The amplitude decreases due to damping as A(t) = Ae–bt/2m. At the time when the amplitude has decreased from11.0 m to 2.0 m,
2 0 2 2
11
2 751 705 42 62.
ln(
( . ) ./ m
11.0 m
kg)
6.0 kg / s s= ⇒ = −
= − − =−e t
m
bbt m
With a period of 4.0 s, the number of oscillations is Nosc = (42.6 s)/(4.0 s) = 10.7 oscillations.
14.79. Model: The vertical movement of the car is simple harmonic motion.Visualize:
The fact that the car has a maximum oscillation amplitude at 5 m/s implies a resonance. The bumps in the roadprovide a periodic external force to the car’s suspension system, and a resonance will occur when the “bumpfrequency” fext matches the car’s natural oscillation frequency f0.Solve: Now the 5.0 m/s is not a frequency, but we can convert it to a frequency because we know the bumps arespaced every 3.0 meters. The time to drive 3.0 m at 5.0 m/s is the period:
Tx
v= = =∆ 3.0 m
5.0 m s s0 60.
The external frequency due to the bumps is thus f Text Hz= =1 1 667. . This matches the car’s natural frequency f0,which is the frequency the car oscillates up and down with if you push the car down and release it. This is enoughinformation to deduce the spring constant of the car’s suspension:
1.667 Hz N
mext= ⇒ = ( ) =1
22 131 600
2
ππk
mk m f ,
where we used m = mtotal = mcar + 2mpassenger = 1200 kg. When at rest, the car is in static equilibrium with Fnet N.= 0The downward weight mtotalg of the car and passengers is balanced by the upward spring force k∆y of thesuspension. Thus the compression ∆y of the suspension is
∆ym g
k= total
Initially mtotal = mcar + 2mpassenger = 1200 kg, causing an initial compression ∆yi = 0.0894 m = 8.94 cm. When threeadditional passengers get in, the mass increases to mtotal = mcar + 5mpassenger = 1500 kg. The final compression is ∆yf =0.1117 m = 11.17 cm. Thus the three new passengers cause the suspension to “sag” by 11.17 cm – 8.94 cm = 2.23 cm.
14.80. Solve: The solution of the equation
d x
dt
b
m
dx
dt
k
mx
2
2 0+ + =
is x t Ae tbt m( ) = +( )− 2 cos .ω φ0 The first and second derivatives of x(t) are
dx
dt
Ab
me t A e t
d x
dt
Ab
mA t
Ab
mt e
bt m bt m
b m
= − +( ) − +( )
= −
+( ) + +( )
− −
−
2
4
0 0
2
2
22
0 0
2 2
2t 2
cos sin
cos sin
ω φ ω ω φ
ω ω φ ω ω φ
Substituting these expressions into the differential equation, the terms involving sin(ωt + φ0) cancel and we obtainthe simplified result
− − +
+( ) =b
m
k
mt
22
0 04 2 ω ω φcos
Because cos ω φt +( )0 is not equal to zero in general,
− − + = ⇒ = −b
m
k
m
k
m
b
m
22
2
04 42 2ω ω
