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Oscillatory Motion
CHAPTER OUTLINE 15.1 Motion of an Object Attached to a Spring 15.2 The Particle in Simple Harmonic Motion 15.3 Energy of the Simple Harmonic Oscillator 15.4 Comparing Simple Harmonic Motion with Uniform Circular Motion 15.5 The Pendulum 15.6 Damped Oscillations 15.7 Forced Oscillations
ANSWERS TO QUESTIONS
Q15.1 Neither are examples of simple harmonic motion, although they are both periodic motion. In neither case is the acceleration proportional to the position. Neither motion is so smooth as SHM. The ball’s acceleration is very large when it is in contact with the floor, and the student’s when the dismissal bell rings.
*Q15.2 (i) Answer (c). At 120 cm we have the midpoint between the turning points, so it is the equilibrium position and the point of maximum speed.
(ii) Answer (a). In simple harmonic motion the acceleration is a maximum when the excursion from equilibrium is a maximum.
(iii) Answer (a), by the same logic as in part (ii).
(iv) Answer (c), by the same logic as in part (i).
(v) Answer (c), by the same logic as in part (i).
(vi) Answer (e). The total energy is a constant.
395
Q15.3 You can take φ = π , or equally well, φ = −π. At t = 0, the particle is at its turning point on the negative side of equilibrium, at x = − A.
*Q15.4 The amplitude does not affect the period in simple harmonic motion; neither do constant forces that offset the equilibrium position. Thus a, b, e, and f all have equal periods. The period is pro- portional to the square root of mass divided by spring constant. So c, with larger mass, has larger period than a. And d with greater stiffness has smaller period. In situation g the motion is not quite simple harmonic, but has slightly smaller angular frequency and so slightly longer period. Thus the ranking is c > g > a = b = e = f > d.
*Q15.5 (a) Yes. In simple harmonic motion, one-half of the time, the velocity is in the same direction as the displacement away from equilibrium.
(b) Yes. Velocity and acceleration are in the same direction half the time.
(c) No. Acceleration is always opposite to the position vector, and never in the same direction.
396 Chapter 15
*Q15.6 Answer (e). We assume that the coils of the spring do not hit one another. The frequency will be higher than f by the factor 2. When the spring with two blocks is set into oscillation in space, the coil in the center of the spring does not move. We can imagine clamping the center coil in place without affecting the motion. We can effectively duplicate the motion of each individual block in space by hanging a single block on a half-spring here on Earth. The half-spring with its center coil clamped—or its other half cut off—has twice the spring constant as the original uncut spring, because an applied force of the same size would produce only one-half the extension distance. Thus the oscillation frequency in space is
1 2 π
k m f. The absence of a force required to support the vibrating system in orbital free fall has no effect on the frequency of its vibration.
*Q15.7 Answer (c). The equilibrium position is 15 cm below the starting point. The motion is symmet- ric about the equilibrium position, so the two turning points are 30 cm apart.
Q15.8 Since the acceleration is not constant in simple harmonic motion, none of the equations in Table 2.2 are valid.
Equation Information given by equation x t ( ) = A cos( ω t +φ ) position as a function of time v ( ) = − t ω A sin ( ω t +φ) velocity as a function of time v ( x ) = ± ω (^) ( A^2 − x^2 ) 1 2 velocity as a function of position a t ( ) = − ω 2 A cos (^) ( ω t +φ) acceleration as a function of time
a t ( ) = − ω 2 x t ( ) acceleration as a function of position
The angular frequency ω appears in every equation. It is a good idea to figure out the value of angular frequency early in the solution to a problem about vibration, and to store it in calculator memory.
*Q15.9 (i) Answer (e). We have T L i g = i and T L g
L
g f =^ f^ =^4 i =^2 T^ i. The period gets larger by 2 times, to become 5 s.
(ii) Answer (c). Changing the mass has no effect on the period of a simple pendulum.
*Q15.10 (i) Answer (b). The upward acceleration has the same effect as an increased gravitational field.
(ii) Answer (a). The restoring force is smaller for the same displacement.
(iii) Answer (c).
Q15.11 (a) No force is exerted on the particle. The particle moves with constant velocity.
(b) The particle feels a constant force toward the left. It moves with constant acceleration toward the left. If its initial push is toward the right, it will slow down, turn around, and speed up in motion toward the left. If its initial push is toward the left, it will just speed up.
(c) A constant force towards the right acts on the particle to produce constant acceleration toward the right.
(d) The particle moves in simple harmonic motion about the lowest point of the potential energy curve.
398 Chapter 15
Section 15.2 The Particle in Simple Harmonic Motion
P15.2 (a) x = ( 5 00 ) ⎛⎝ 2 t + ⎞⎠
. cm cos π (^) At t = 0,
x = ( 5 00 ) ⎛⎝ ⎞⎠ =
. cm cos 4 33. cm π
(b) v = = − ( ) ⎛⎝ + ⎞⎠ dx dt 10 0 2 t 6
. cm ssin π At t = 0, v = −5 00. cm s
(c) a d dt = = − ( ) ⎛⎝ t + ⎞⎠ v 20 0 2 6
. cm s 2 cos π At t = 0, a^ =^ −17 3.^ cm s^2
(d) A = 5 00. cm and T = = =
π ω
π
. s
P15.3 x = ( 4 00. m) cos ( 3 00. π t +π )Compare this with x = A cos( ω t +φ )to fi nd
(a)^ ω^ =^2 π^ f^ =3 00. π or f^ =^ 1 50.^ Hz^ T f
0 667. s
(b) A = 4 00. m
(c) φ = πrad
(d) x t ( =0 250. s ) = ( 4 00. m ) cos ( 1 75. π) = 2 83. m
P15.4 (a) The spring constant of this spring is
k
F
x
kg m s m N m
2
we take the x -axis pointing downward, so φ = 0
x = A t = ⋅
cos ω 18 0. cm cos 12 6.^ kg 84 4. = 0.45 kg s 2 s^1 18 0.^ cm^ cos^ 446 6.^ rad^ = 15 8. cm We choose to solve the parts in a different order.
(d) Now 446 6. rad = 71 × 2 π +0 497. rad. In each cycle the object moves 4 18( ) = 72 cm, so it
has moved 71 72( cm ) + ( 18 −15 8. ) cm = 51 1. m.
(b) By the same steps, k = 0 44^ 9 8 = 0 355
kg m s. m
N m
2
x A k m = cos t =. cos =
18 0.. co
cm 84 4 18 0cm ss 443 5. rad = −15 9. cm
(e) 443 5.^ rad^ =^ 70 2(^ π) +3 62. rad
Distance moved = 70 72( cm ) + 18 + 15 9. cm = 50 7. m
(c) The answers to (d) and (e) are not very different given the difference in the data about the two vibrating systems. But when we ask about details of the future, the imprecision in our knowledge about the present makes it impossible to make precise predictions. The two oscillations start out in phase but get completely out of phase.
Oscillatory Motion 399
P15.5 (a) At t = 0, x = 0 and v is positive (to the right). Therefore, this situation corresponds to x = A sin ω t and v = vi cos ω t Since f = 1 50. Hz, ω = 2 π f =3 00. π
Also, A = 2 00. cm, so that x = ( 2 00. cm)sin 3 00. π t
(b) v max = vi = A ω= 2 00 3 00. (. π ) = 6 00. π cm s = 18 8. cm s
The particle has this speed at t = 0 and next at t = T = 2
s
(c) a max = A ω^2 = 2 00 3 00. (. π ) 2 = 18 0.π 2 cm s 2 = 178 cm s^2
This positive value of acceleration first occurs at t = 3 T = 4
0 500. s
(d) Since T = 2 3
s and A = 2 00. cm, the particle will travel 8.00 cm in this time.
Hence, in 1 00
. s ⎛⎝= T ⎞⎠, the particle will travel 8 00. cm + 4 00. cm = 12 0. cm.
P15.6 (a) T = 12 0.^ s= 2 40. 5
s
(b) f T
. Hz
(c) ω = 2 π f = 2 π ( 0 417. ) = 2 62. rad s
P15.7 f^
k m
ω 2 π π
2 or^ T^ f
m k
2 π
Solving for k , k m T
( (^) )
( )^
2 2
2 2
π π^. .
kg s
N m
*P15.8 (a) For constant acceleration position is given as a function of time by
x = x (^) i + (^) xit + a tx
= + ( )( ) +
v^1 2 0 27 0 14 4 5 1 2
2
. m. m s. s (^) ( −− )( )
= −
..^2
m s s
m
2
(b) v^ x =^ vxi +^ a tx =^ 0 14.^ m s^ − ( 0 32.^ m s^2 )(^ 4 5.^ s) =^ −1 30. mm s
(c) For simple harmonic motion we have instead x = A cos( ω t +φ )and v = − A ω sin( ω t +φ)
where a = −ω 2 x , so that − 0 32. m s 2 = − ω 2 ( 0 27. m), ω = 1 09. rad s. At t = 0, 0.27 m =
A cos φ and 0 14. m s = − A (1 09. s)sin φ. Dividing gives 0 14 0 27
m s. tan m
= − ( s) φ ,
tan φ = −0 476. , φ = −25 5. °. Still at t = 0 , 0 27. m = A cos ( −25 5.°) , A = 0 299. m. Now
at t = 4 5. s,
x = ( 0 299. m) cos ⎡⎣( 1 09. rad s )( 4 5. s ) −25 5.°⎤⎦ = 0 0 299 4 90 25 5 0 299 2
. cos.. . cos
m rad m
555 ° = −0 076 3. m
t
x
FIG. P15.8(a)
continued on next page
Oscillatory Motion 401
*P15.12 We assume that the mass of the spring is negligible and that we are on Earth. Let m represent the mass of the object. Its hanging at rest is described by Σ Fy = 0 − Fg + kx = 0 mg = k (0.183 m) m k = (0.183 m)(9.8 Nkg) The object’s bouncing is described by T = 2 π( m k ) 1 ^2 = 2 π[(0.183 m)(9.8 m /s 2 )]^1 ^2 = 0.859 s
We do have enough information to find the period. Whether the object has small or large mass, the ratio m / k must be equal to 0.183 m/(9.80 m /s^2 ). The period is 0.859 s.
P15.13 The 0.500 s must elapse between one turning point and the other. Thus the period is 1.00 s.
ω π = =
T
. s
and v max =^ ω^ A^ = (^ 6 28.^ s^ )(^ 0 100.^ m^ ) = 0 628. m s^.
Section 15.3 Energy of the Simple Harmonic Oscillator
P15.14 m = 200 g, T = 0 250. s, E = 2 00. J; ω π π = = =
T.
. rad s
(a) k = m ω 2 = 0 200. kg 25.1 rad s( ) 2 = 126 N m
(b) E kA A
E
k
2 2
. m
P15.15 Choose the car with its shock-absorbing bumper as the system; by conservation of energy,
1 2
m v^2 = kx^2 : v = x k = (^) ( × − ) ×^ = m
6
. (^3)
m. m s
P15.16 (a) E^ = kA^2 = (^ × −^2 )^ =
2
2
N m m J
(b) v max = A ω where ω = = = − k m
. s v max = 0 784 m s.
(c) a (^) max = A ω 2 =. × −^2 (. −^1 ) =. 2 3 50 10 m 22 4 s 17 5m s 2
P15.17 (a) E^ =^ kA =^ (^ )^ ( × − ) =
2
. N m. m. mJ
(b) v = ω A − x = − k m
(^2 2) A (^2) x 2
v = × − (^ × −^ ) −^ (^ × −)^ =
3 4 00^10 2 1 00^10
.. .002 m s
(c)
2 m v = kA − kx = (. ) ⎡⎣(. × −) −(. ×× 10 −^2 )⎤⎦ = 12 2 2
. mJ
(d)
kx^2 = E − mv^2 = 15 8. mJ
402 Chapter 15
P15.18 (a) (^) k F x
= = 20 0.^ N = 100
0.200 m
N m
(b) ω =^ k = m 50 0. rad s so^ f = ω = 2 π
1 13. Hz
(c) v max = ω A = 50 0 0 200. (. ) = 1 41 m s. at x = 0
(d) a max = ω 2 A = 50 0 0 200. (. ) = 10 0 m s. 2 at x = ± A
(e) E = 1 kA = ( )( ) =
2 100 0 200. 2 2 00. J
(f) v = ω A^2 − x^2 = 50 0 8 ( ) 2 =
. 0 200. 1 33. m s
(g) a^ =^ ω^2 x =^ 50 0^ ⎛⎝ ⎞⎠ =
..^ 3 33. m s 2
P15.19 Model the oscillator as a block-spring system.
From energy considerations, v^2 + ω 2 x^2^ =ω^2 A^2
v max =ω A and v =ω^ A 2
so^ ω^^ A^ ω x ω A 2
2 ⎛ 2 2 2 2 ⎝
⎠ +^ =
From this we fi nd x^2 3 A^2 4
= and x = 3 A = ± 2
2 60. cm where A = 3 00. cm
P15.20 (a) y (^) f = y (^) i + vyi t +^1 a ty 2
2
− = + + (^) ( − )
= ⋅^ =
m m s^2
m s m
2
2
t
t s
(b) Take the initial point where she steps off the bridge and the final point at the bottom of her motion. K U U K U U
mgy kx
( +^ g + s ) = i^ ( +^ g + s ) f
2
kg m s 36 m m
N m
.^2
k
k
(c) The spring extension at equilibrium is x F k
= = 65 kg 9.8 m s =8 68 73.4 N m
m
2
. , so this point is 11 + 8 68. m = 19 7. m below the bridge and the amplitude of her oscillation is 36 − 19.7 = 16.3 m.
(d) ω =^ k = = m
. N m 1 06. kg
rad s
continued on next page
404 Chapter 15
P15.23 (a)^ The motion is simple harmonic because the tire is rotating with constant velocity and you are looking at the motion of the bump projected in a plane perpendicular to the tire. (b) Since the car is moving with speed v = 3 00. m s, and its radius is 0.300 m, we have
ω = 3 00 = 0 300
m s. m
rad s
Therefore, the period of the motion is
T = = ( )
π ω
π .
rad s s
Section 15.5 The Pendulum
P15.24 The period in Tokyo is T
L
T g T T
= 2 π
and the period in Cambridge is T^
L
C g C C
= 2 π
We know T (^) T = TC = 2 00. s
For which, we see L g
L
g
T T
C C
or g g
L
L
C T
C T
P15.25 Using the simple harmonic motion model:
A r
g L
θ π
ω
m 15 m
m s m
2
. (^) .113 rad s
(a) v max =^ A^ ω^ =^ 0 262.^ m 3.13 s^ = 0 820. m s
(b) a^ max =^ A ω^2 =^.^ (^ )^ =. 0 262 m 3.13 s 2 2 57m s 2
a (^) tan = r α α = = = a r
tan m s^22 m rad s
(c) F^ =^ ma =^ 0 25.^ kg 2.57 m s^2 = 0 641. N
More precisely,
(a) mgh = m
v^2 and h = L ( 1 −cos θ)
∴ v max = 2 gL ( 1 −cos θ ) = 0 817. m s
g FIG. P15.
continued on next page
Oscillatory Motion 405
(b) I α = mgL sinθ
αmax=^ mgL sinθ^ = sinθ =. mL
g (^2) L i 2 54 rad s 2
(c) F max = mg sin θ i = 0 250 9 80. (. )( sin 15 0. °) = 0 634. N
The answers agree to two digits. The answers computed from conservation of energy and from Newton’s second law are more precisely correct. With this amplitude the motion of the pendu- lum is approximately simple harmonic.
*P15.26 Note that the angular amplitude 0.032 rad = 1.83 degree is small, as required for the SHM model of a pendulum.
ω = 2 π T
: T^ =^ =^ =
π 1 42 ω
π .
. s
ω = g L
: L = g =
ω 2 2
. m
P15.27 Referring to the sketch we have
F = − mg sin θ and tan θ = x R For small displacements, tan θ ≈sinθ
and F mg R
= − x = − kx
Since the restoring force is proportional to the displacement from equilibrium, the motion is simple harmonic motion.
Comparing to F^ = − m ω^^2 x shows^ ω =^ = k m
g R
P15.28 (a) The string tension must support the weight of the bob, accelerate it upward, and also provide
the restoring force, just as if the elevator were at rest in a gravity field of (9 80. +5 00. ) m s 2.
Thus the period is
T L g T
π π.
.
m 14.8 m s s
2
(b) T = ( − )
π. 6 41 .
m. 9.80 m s m s 2 2 s
(c) geff = ( 9 80 ) + ( 5 00 ) =11 0 2 2
. m s 2. m s 2. m s^2
T = 2 π = 5 00.^ m = 4 24. 11.0 m s 2 s
P15.29 f = 0 450. Hz, d = 0 350. m, and m = 2 20. kg
T
f T
I
mgd
T
I
mgd
I T mgd f
⎝⎜^
2 2
2 2
π ; π
π
22 4 2 2 1 2
mgd π (^) π
( )
. s
..944 kg m ⋅ 2
FIG. P15.
FIG. P15.
Oscillatory Motion 407
P15.32 (a) The parallel-axis theorem:
I I Md ML Md
M M
CM
m m
2 2 2
2
( )
M
T
I
Mgd
M
Mg
m
m
2
2 π π
. m m 12 9.80 m s 2 s
( )
π 2 09.
(b) For the simple pendulum
T = 2 π 1 00.^ m =2 01. 9.80 m s 2 s difference^ =^ 2 09.^ s^ −^ 2 01.^ s= 4 08.^ % 2.01 s
P15.33 T = 0 250. s, I = mr^2 = ( × −^3 ) ( × −^3 ) 2 20 0. 10 kg 5 00. 10 m
(a) I = 5 00. × 10 −^7 kg m⋅ 2
(b) I d dt
2 2
θ = −κθ ;^ κ^ ω π I T
κ = I ω^2 = (^) ( × −^7 )⎛⎝ π ⎞⎠ = × − ⋅
2 5 00 10 2 4 0 250
. N m rrad
Section 15.6 Damped Oscillations
P15.34 The total energy is E = 1 m + kx 2
v^2
Taking the time derivative, dE dt
m d x dt
= v + kxv
2 2
Use Equation 15.31: md x dt kx b
2 2 = −^ −^ v dE dt
= v ( − kx − b v ) + k xv
Thus, dE dt = − bv^2 < 0
We have proved that the mechanical energy of a damped oscillator is always decreasing.
P15.35 θ i = 15 0. ° θ ( t =1 000 ) =5 50. °
x = Ae −^ bt^^2 m
x x
Ae A
e i
bt m 1 000 b m 2 5 50 1 000 2 15 0
−
. − ( ) .
ln. .
⎠ = −^ =
− ( )
∴ = ×
b m b m
−− (^3) s− 1
θ
FIG. P15.
FIG. P15.
408 Chapter 15
P15.36 To show that x = Ae − bt^^^2 m cos( ω t +φ)
is a solution of − kx − b = dx dt m d x dt
2 2 (1)
where ω =^ k − ⎛⎝ ⎞⎠ m
b 2 m
2 (2)
We take x = Ae − bt^^^2 m cos( ω t +φ )and compute (3) dx dt
Ae b m
= −^ bt^^2 m^^ ⎛⎝− ⎞⎠ ( t + ) − Ae − bt^^2 m t 2
cos ω φ ω sin(ω + +φ ) (4)
d x dt
b m Ae b m
bt m (^) t Aebt m 2 2
2 2 2 2 = − ⎡ −^ ⎛⎝− ⎞⎠ cos( ω +φ ) − − ωω sin (ω t +φ) ⎣⎢^
− Ae −^ ⎛⎝− b ⎞⎠ ( + ) + − + m
bt 2 m (^) t Ae bt 2 m (^2) t 2
⎡ ω sin ω φ ω cos((ω φ) ⎣⎢^
⎦⎥^
We substitute (3), (4) into the left side of (1) and (5) into the right side of (1);
− kAe −^ ( t + ) + b − ( + ) + m
bt 2 m (^) Ae bt m t b A 2 2 2
cos ω φ cosω φ ω ee t
b Ae b m t
bt m
bt m
−
−
( + )
= − ⎛⎝− ⎞⎠ ( +
2
2 2 2
sin
cos
ω φ
⎡ ω φ)) − ( + ) ⎣⎢^
−
−
Ae t
b Ae t
bt m
bt m
2
2 2
ω ω φ
ω ω
sin
sin( φφ ) − m ω^2 Ae − bt^^2 m^ cos(ω t +φ)
Compare the coefficients of Ae − bt^^^2 m^ cos( ω t +φ )and Ae − bt^^^2 m sin( ω t +φ ):
cosine-term: − + = − ⎛⎝− ⎞⎠ − = − −
⎝⎜^
k b m
b b m
m b m
m k m
b m
2 2 2 2 2 2 2 ω^4 42 ⎟⎟ = −^ k^ +
b m
2 2
sine-term: b ω = + b^ (ω ) + b^ (ω ) = b ω
Since the coefficients are equal, x = Ae − bt^^^2 m^ cos( ω t +φ )is a solution of the equation.
P15.37 The frequency if undamped would be ω 0
×
k m
N m kg s.
(a) With damping
ω = ω − ⎛⎝ ⎞⎠ = ⎛⎝ ⎞⎠ −
2 2 2 2
b 3 m
s
kg ⎜⎜ (^) s 2 10.6 kg
s
s
f^ ω π π
..00 Hz
(b) In x = A e 0 − bt^^2 m cos( ω t +φ )over one cycle, a time T = 2 π ω
, the amplitude changes from A 0 to A e 0 − b^^2 π^2 m ω^ for a fractional decrease of A A e A
e e
b m 0 0 0
− −π^ ω^ = 1 − − π (^3) (10 6 44 0. ⋅. ) (^) = 1 − −0 020 2. (^) = 11 − 0 979 98. = 0 020 0. = 2 00. %
continued on next page
410 Chapter 15
P15.40 (^) F t kx m d x (^0) dt
2 sin ω − = 2 ω 0 = k m
x = A cos( ω t +φ ) (2) dx dt = − A ω sin ( ω t +φ) (3)
d x dt A t
2 2 = − ω 2 cos ( ω +φ) (4)
Substitute (2) and (4) into (1): F 0 sin ω t − kA cos (^) (ω t +φ (^) ) = m (^) ( − A ω (^2) ) cos(ω t +φ)
Solve for the amplitude: (^) ( kA − mA ω (^2) ) cos (^) ( ω t +φ (^) ) = F 0 (^) sin ω t = F 0 (^) cos ( ω t − 90 °)
These will be equal, provided only that φ must be − 90 ° and kA − mA ω 2 = F 0
Thus, A F m k m
( ) −
0 2
ω
P15.41 From the equation for the amplitude of a driven oscillator with no damping,
A F m F m
f
( − )
= = ( −)
0 2 0 2 2
0 2 0 2
1 2 20 0 0
ω ω ω^ ω
ω π. π s ω^22
0 2 0 2
= (^) ( − )
k − m F mA
F
. s
ω ω
00
= ⎛⎝. ⎞⎠ ( 2 00 × 10 −) (3 950 −49 0 ) = 318 .
.. N
P15.42 A
F m b m
( − ) + ( )
ext ω 2 ω 02 2 ω^2
With b = 0 , A F m F m F m = ( − )
± (^) ( − )
ext ext ext ω 2 ω 02 2 ω^ ω^ ω^ ω
2 0 2 2 0 22
Thus, ω 2 ω 02 6 30 0 150
= ±^ F^ m = ± = ± 1 7 A
k m
F
mA
ext ext N m kg
N
( 0.150 kg )(. m) This yields ω = 8 23. rad s or ω = 4 03. rad s
Then, f = ω 2 π
gives either f = 1 31. Hz or f = 0 641. Hz
P15.43 The beeper must resonate at the frequency of a simple pendulum of length 8.21 cm:
f g L
π π
m s. m
Hz
2
Oscillatory Motion 411
Additional Problems
*P15.44 (a) Consider the first process of spring compression. It continues as long as glider 1 is moving faster than glider 2. The spring instantaneously has maximum compression when both glid- ers are moving with the same speed va.
FIG. P15.44(a) Momentum conservation: m (^) 1 1 i m (^) 2 2 i m (^) 1 1 f m 2 2 f 0 24 0 74 0
v + v = v + v (. kg)(. m s) + .. 36 0 12. 0 24. 0 36. 0
( kg )( m s ) = ( kg ) v (^) a + ( kg) va .. .
kg m s 0 368 kg
⋅ (^) = va v a = m s i
(b) Energy conservation: K K U K K U
m m
s (^) i s (^) f
i i
1 2 1 2
1 1 2 2 2
( + + ) = ( + + )
v + v + =^11 2
1 2 2 2
2
(^ m^ + m^ ) (^) a + kx
( )( ) +
v
. kg. m s 0 0 36 0 12 1 2
2
2
kg m s
kg m s
( )( )
= ( )( ) ++ ( )
= + ( )
2
2
N m
J J N m
x
x
x
⎛ ( ) ⎝⎜^
⎠⎟^
1 2 . .
J
N m m
(c) Conservation of momentum guarantees that the center of mass moves with constant velocity. Imagine viewing the gliders from a reference frame moving with the center of mass. We see the two gliders approach each other with momenta in opposite directions of equal magnitude. Upon colliding they compress the ideal spring and then together bounce, extending and compressing it cyclically.
(d) 1 2
m tot v CM^2 = ( 0 60. kg )( 0 368. m s )^2 = 0 040 6. J
(e)
kA^2^ = ( 45 N m )( 0 035 1. m )^2 = 0 027 7. J
Oscillatory Motion 413
P15.48 (a) Total energy = 1 = ( )( ) = 2
kA^2^100 N m 0 200. m 2 2 00. J
At equilibrium, the total energy is: 1 2
(^ m 1^ + m 2 ) v^2^^ =^ (16 0^.^ kg^ )^ v^2^^ = (8 00^. kg) v^2
Therefore,
( 8 00. kg ) v^2 =2 00. J, and v = 0 500. m s
This is the speed of m 1 and m 2 at the equilibrium point. Beyond this point, the mass m 2 moves with the constant speed of 0.500 m /s while mass m 1 starts to slow down due to the restoring force of the spring.
(b) The energy of the m 1 -spring system at equilibrium is: 1 2
m 1 v^2^ = (9 00. kg)( 0 500. m s) 2 =1 125. J
This is also equal to 1 2
k A ( ′)^2 , where A ′is the amplitude of the m
1 -spring system. Therefore, 1 2
( 100 ) ( A ′)^2 =1 125. or A ′ = 0 150. m
The period of the m 1 -spring system is T m k = 2 π 1 =1 885. s
and it takes 1 4
T = 0 471. s after it passes the equilibrium point for the spring to become fully stretched the first time. The distance separating m 1 and m 2 at this time is:
D = v ⎛⎝ T^ ⎞⎠ − A ′ = ( ) − =
0 500. m s 0 471. s 0 150. m 0 08. 5 5 6 = 8 56. cm
P15.49 The maximum acceleration of the oscillating system is a (^) max = A ω 2 = 4 π 2 Af^2. The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip,
f = f (^) max= μ s n = μ s mg = m (^) ( 4 π 2 Af^2 )
A
g f
= μ s = = 4 π π
cm/s s)
c
2
mm
P15.50 Refer to the diagram in the previous problem. The maximum acceleration of the oscillating sys- tem is a (^) max = A ω 2 = 4 π 2 Af^2. The friction force exerted between the two blocks must be capable of accelerating block B at this rate. Thus, if Block B is about to slip,
f = f (^) max= μ s (^) n = μ s mg = m (^) ( 4 π 2 Af^2 ) or A g f = μ s 4 π 2 2
f
n
m g
B
P
B
s
FIG. P15.
414 Chapter 15
P15.51 Deuterium is the isotope of the element hydrogen with atoms having nuclei consisting of one proton and one neutron. For brevity we refer to the molecule formed by two deuterium atoms as D and to the diatomic molecule of hydrogen-1 as H.
M (^) D = 2 MH ω ω
D H
D H
H D
k M k M
M
M
f f D^ =^ H 2 =^ 0 919^ ×^10
. 14 Hz
*P15.52 (a) A time interval. If the interaction occupied no time, each ball would move with infinite acceleration. The force exerted by each ball on the other would be infinite, and that cannot happen.
(b) k = ⏐ F ⏐⏐ x ⏐ = 16 000 N0.000 2 m = 80 MN m
(c) We assume that steel has the density of its main constituent, iron, shown in Table 14.1. Then its mass is ρ V = ρ (43) π r^3 = (4 π3)(7860 kgm 3 )(0.0254 m2) 3 = 0.0674 kg and K = (12) mv^2 = (12) (0.0674 kg)(5 m /s) 2 = 0 843. J
(d) Imagine one ball running into an infinitely hard wall and bouncing off elastically. The original kinetic energy becomes elastic potential energy 0.843 J = (12) (8 × 107 Nm) x^2 x = 0 145. mm (e) The half-cycle is from the equilibrium position of the model spring to maximum compres- sion and back to equilibrium again. The time is one-half the period, (12) T = (12)2 π( m k ) 1/2^ = π(0.0674 kg 80 × 106 kgs 2 )1/2^ = 9 12. × 10 −^5 s
P15.53 (a)
a a
L
h
Li
FIG. P15.53(a)
(b) T
L
g = 2 π dT dt g L
dL dt
= π^1 (1)
We need to find L ( t ) and dL dt
. From the diagram in (a),
L L a h = (^) i + 2 − 2 and dL dt
dh dt
But dM dt
dV dt
A
dh dt = ρ = −ρ. Therefore,
dh dt A
dM dt
dL dt A
dM dt
ρ 2 ρ
Also, dL A
dM dt t L L L
L i i
∫ =
⎠ =^ −
2 ρ
Substituting Equation (2) and Equation (3) into Equation (1): dT dt g a
dM dt (^) L (^) i t a dM dt
⎠ (^) + ( )
π ρ ρ
(^2 ) (( ) continued on next page
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