SM chapter 15 - Solucionario capitulo 15 Serway 7ma edición PDF

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15 Oscillatory Motion CHAPTER OUTLINE 15 15 15 15 15 15 15 Motion of an Object Attached to a Spring The Particle in Simple Harmonic Motion Energy of the Simple Harmonic Oscillator Comparing Simple Harmonic Motion with Uniform Circular Motion The Pendulum Damped Oscillations Forced Oscillations Q15 A...

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15 Oscillatory Motion ANSWERS TO QUESTIONS

CHAPTER OUTLINE 15.1 15.2 15.3 15.4

15.5 15.6 15.7

Motion of an Object Attached to a Spring The Particle in Simple Harmonic Motion Energy of the Simple Harmonic Oscillator Comparing Simple Harmonic Motion with Uniform Circular Motion The Pendulum Damped Oscillations Forced Oscillations

Q15.3

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.

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 proportional 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 12 1 2k the oscillation frequency in space is ⎛ ⎞ ⎛ ⎞ = 2 f . The absence of a force required to ⎝ 2π ⎠ ⎝ m ⎠ 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 symmetric 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 ( ωt + 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 − x 2

2

)

12

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)

(ii) *Q15.10 (i) (ii)

Answer (e). We have Ti = times, to become 5 s.

Li and Tf = g

Lf g

=

4 Li = 2 Ti. The period gets larger by 2 g

Answer (c). Changing the mass has no effect on the period of a simple pendulum. Answer (b). The upward acceleration has the same effect as an increased gravitational field. 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.

Oscillatory Motion

397

Q15.12 The motion will be periodic—that is, it will repeat. The period is nearly constant as the angular amplitude increases through small values; then the period becomes noticeably larger asθ increases farther. *Q15.13 The mechanical energy of a damped oscillator changes back and forth between kinetic and potential while it gradually and permanently changes into internal energy. *Q15.14 The oscillation of an atom in a crystal at constant temperature is not damped but keeps constant amplitude forever. Q15.15 No. If the resistive force is greater than the restoring force of the spring (in particular, if b2 > 4 mk), the system will be overdamped and will not oscillate. Q15.16 Yes. An oscillator with damping can vibrate at resonance with amplitude that remains constant in time. Without damping, the amplitude would increase without limit at resonance. Q15.17 Higher frequency. When it supports your weight, the center of the diving board flexes down less than the end does when it supports your weight. Thus the stiffness constant describing the center ⎛ 1 ⎞ k is of the board is greater than the stiffness constant describing the end. And then f = ⎝ 2π ⎠ m greater for you bouncing on the center of the board. Q15.18 An imperceptibly slight breeze may be blowing past the leaves in tiny puffs. As a leaf twists in the wind, the fibers in its stem provide a restoring torque. If the frequency of the breeze matches the natural frequency of vibration of one particular leaf as a torsional pendulum, that leaf can be driven into a large-amplitude resonance vibration. Note that it is not the size of the driving force that sets the leaf into resonance, but the frequency of the driving force. If the frequency changes, another leaf will be set into resonant oscillation. Q15.19 We assume the diameter of the bob is not very small compared to the length of the cord supporting it. As the water leaks out, the center of mass of the bob moves down, increasing the effective length of the pendulum and slightly lowering its frequency. As the last drops of water dribble out, the center of mass of the bob hops back up to the center of the sphere, and the pendulum frequency quickly increases to its original value.

SOLUTIONS TO PROBLEMS Section 15.1 P15.1

Motion of an Object Attached to a Spring

(a)

Since the collision is perfectly elastic, the ball will rebound to the height of 4.00 m and then repeat the motion over and over again. Thus, the motion is periodic .

(b)

To determine the period, we use: x =

1 2 gt . 2

The time for the ball to hit the ground is t =

2x = g

2 ( 4. 00 m ) = 0 .904 s . 9 .80 m s 2

This equals one-half the period, so T = 2( 0.904 s) = 1.81 s . (c)

The motion is not simple harmonic. The net force acting on the ball is a constant given by F = −mg (except when it is in contact with the ground), which is not in the form of Hooke’s law

398

Chapter 15

Section 15.2 P15.2

P15.3

The Particle in Simple Harmonic Motion

(a)

π⎞ x = ( 5. 00 cm ) cos ⎛ 2t + ⎝ 6⎠

(b)

v=

(c)

a=

(d)

A = 5.00 cm

x = ( 5.00 cm) cos

dx π ⎛ = − (10.0 cm s )sin 2t + ⎞ ⎝ dt 6⎠

At t = 0,

v = −5.00 cm s

π⎞ dv = −( 20 .0 cm s2 ) cos ⎛ 2t + ⎝ 6⎠ dt

At t = 0,

a = −17 .3 cm s 2

and

T=

⎛ π ⎞ = 4. 33 cm ⎝ 6⎠

2π 2π = = 3. 14 s 2 ω

x = ( 4.00 m) cos ( 3. 00 π t + π ) Compare this with x = A cos (ω t + φ ) to find (a)

ω = 2π f = 3.00π or

P15.4

At t = 0,

f = 1.50 Hz

T=

1 = 0 .667 s f

(b)

A = 4.00 m

(c)

φ = π rad

(d)

x ( t = 0.250 s ) = (4.00 m )cos (1.75π ) = 2.83 m

(a)

The spring constant of this spring is k=

F 0 .45 kg 9 .8 m s 2 = = 12. 6 N m x 0. 35 m

we take the x-axis pointing downward, so φ = 0 x = A cos ωt = 18 .0 cm cos

12 .6 kg 84 .4 s = 18.0 cm cos 446.6 rad = 15.8 cm 0.45 kg ⋅ s2

We choose to solve the parts in a different order. (d)

Now 446.6 rad = 71 × 2π + 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 = x = A cos

(e)

0 .44 kg 9 .8 m s2 = 12 .1 N m 0 .355 m

k 12. 1 84 .4 = 18 .0 cm cos 443.5 rad = −15. 9 cm t = 18. 0 cm cos 0. 44 m

443.5 rad = 70 ( 2π ) + 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

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 ωt

Since f = 1.50 Hz,

ω = 2 π f = 3.00 π x = ( 2.00 cm) sin 3.00π t

Also, A = 2.00 cm, so that (b)

v max = vi = Aω = 2 .00 (3 .00 π ) = 6 .00π cm s = 18.8 cm s The particle has this speed at t = 0 and next at

(c)

2 a max = Aω 2 = 2 .00 ( 3 .00π ) = 18 .0π

2

P15.6

(a) (b) (c)

(a)

t=

3 T = 0 .500 s 4

2 s and A = 2.00 cm, the particle will travel 8.00 cm in this time. 3 3 8. 00 cm + 4 .00 cm = 12. 0 cm . Hence, in 1. 00 s ⎛ = T ⎞ , the particle will travel ⎝ 2 ⎠ 12 .0 s = 2 .40 s 5 1 1 f = = = 0. 417 Hz T 2. 40

T=

ω = 2π f = 2π ( 0 .417 ) = 2.62 rad s 1 ω = 2π 2π

k m

or

Solving for k, *P15.8

1 T = s 2 3

Since T =

f =

P15.7

t=

cm s2 = 178 cm s2

This positive value of acceleration first occurs at (d)

399

T=

m 1 = 2π k f

k=

4π 2 m 4π 2 ( 7 .00 kg) = 40 .9 N m = T2 (2 .60 s )2

For constant acceleration position is given as a function of time by

x

1 x = xi + vxi t + ax t 2 2

4.5 t

1 2 = 0. 27 m +( 0. 14 m s ) (4.5 s ) + ( − 0. 32 m s2 ) ( 4. 5 s ) 2 = −2. 34 m FIG. P15.8(a)

(b) (c)

v x = v xi + ax t = 0.14 m s − ( 0.32 m s2 )( 4.5 s ) = −1.30 m s For simple harmonic motion we have instead x = Acos (ω t + φ) and v = − Aω sin (ω t + φ ) 2 where a = −ω x , so that −0. 32 m s 2 = −ω 2 ( 0. 27 m ), ω = 1.09 rad s. At t = 0, 0.27 m = 0 .14 m s = − (1 .09 s) tanφ , A cos φ and 0.14 m s = − A (1. 09 s ) sin φ . Dividing gives 0 .27 m tan φ = −0.476, φ = −25. 5°. Still at t = 0, 0.27 m = Acos (− 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. 299 m) cos ( 4. 90 rad − 25. 5°) = ( 0 299 m) cos 2 55° = −0.076 3 m

400

Chapter 15

(d)

v = − ( 0.299 m ) (1.09 s) sin 255° = +0 .315 m s

x 4.5

P15.9

x = A cosω t

A = 0.05 m

v = −Aω sin ωt

t, s

a = − Aω 2 cos ωt

If f = 3 600 rev min = 60 Hz, then ω = 120π s −1 vmax = 0.05 (120π ) m s = 18 .8 m s P15.10

amax = 0 .05 (120π )

2

m s2 = 7 .11 km s 2

m = 1.00 kg, k = 25. 0 N m, and A = 3.00 cm. At t = 0, x = −3.00 cm (a)

ω=

k = m

so that, (b)

25 .0 = 5 .00 rad s 1 .00 2π 2π T= = = 1. 26 s ω 5. 00

vmax = Aω = 3 .00 ×10 −2 m (5 .00 rad s) = 0 .150 m s amax = Aω 2 = 3. 00 × 10− 2 m ( 5. 00 rad s) = 0. 750 m s2 2

(c)

Because x = −3.00 cm and v = 0 at t = 0, the required solution is x = − A cosω t x = −3 .00 cos (5. 00 t ) cm

or

dx = 15.0 sin (5.00t ) cm s dt dv a= = 75. 0 cos( 5.00t ) cm s2 dt v=

P15.11

(a)

ω=

k 8 .00 N m 1 = = 4.00 s − 0. 500 kg m

From this we find that

(b)

so position is given by

x = 10 .0 sin( 4.00t cm

v = 40. 0 cos ( 4 .00t ) cm s

vmax = 40.0 cm s

a = − 160 sin (4 .00t ) cm s2

amax = 160 cm s 2

)

x ⎞ ⎛ 1 ⎞ sin −1 ⎛ and when t= ⎝ 4. 00⎠ ⎝ 10. 0 ⎠

x = 6.00 cm, t = 0.161 s.

We find

v = 40.0 cos[ 4. 00 ( 0.161)] = 32 .0 cm s a = −160 sin[ 4.00 ( 0.161)] = − 96.0 cm s2

(c)

Using t =

⎛ 1 ⎞ ⎛ x ⎞ sin −1 ⎝ 4 .00 ⎠ ⎝ 10 .0 ⎠

when x = 0, t = 0 and when

x = 8.00 cm, t = 0. 232 s

Therefore,

∆t = 0. 232 s

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 ΣFy = 0

−Fg + kx = 0

mg = k(0.183 m)

mⲐk = (0.183 m)Ⲑ(9.8 NⲐkg)

The object’s bouncing is described by T = 2π ( mⲐk)1Ⲑ2 = 2π [(0.183 m)Ⲑ(9.8 m /s2)]1Ⲑ2 = 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 /s2). 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=

2π = 6.28 s T

and vmax = ω A = (6 .28 s) ( 0 .100 m ) = 0 .628 m s .

Section 15.3 P15.14

P15.15

Energy of the Simple Harmonic Oscillator

m = 200 g, T = 0. 250 s, E = 2.00 J; ω =

2π 2π = = 25. 1 rad s T 0. 250

(a)

k = mω 2 = 0.200 kg (25.1 rad s ) = 126 N m

(b)

E=

2

kA2 ⇒ A= 2

2E = k

2 (2 .00 ) = 0. 178 m 126

Choose the car with its shock-absorbing bumper as the system; by conservation of energy, 1 1 mv 2 = kx 2 : 2 2

5. 00 × 10 6 k = 2 .23 m s = ( 3. 16 × 10−2 m ) m 10 3

v=x

−2 kA2 250 N m (3 .50 × 10 m ) = = 0 .153 J 2 2 2

P15.16

P15.17

(a)

E=

(b)

vmax = Aω

(c)

amax = Aω 2 = 3. 50 × 10− 2 m ( 22. 4 s− 1) = 17. 5 m s2

(a)

E=

(b)

where

ω=

k 250 = = 22. 4 s −1 0. 500 m

vmax = 0.784 m s

2

2 1 2 1 kA = (35 .0 N m ) ( 4 .00 × 10 −2 m ) = 28 .0 mJ 2 2 k 2 2 2 2 v =ω A − x = A −x m

v =

35 .0 50. 0 × 10− 3

( 4. 00 × 10 ) − (1.00 × 10 ) −2 2

−2 2

(c)

1 2 1 2 1 2 1 2 m v = kA − kx = ( 35 .0) ⎡⎣( 4 .00 × 10 − 2 2 2 2

(d)

1 2 1 2 kx = E − mv = 15. 8 mJ 2 2

02 m s = 1.0

) − (3.00 × 10 ) 2

−2 2

⎤ ⎦ = 12. 2 mJ

402

P15.18

P15.19

Chapter 15

20 .0 N F = = 100 N m x 0.200 m

(a)

k=

(b)

ω=

(c)

vmax = ω A = 50. 0 ( 0. 200) = 1. 41 m s at

(d)

2 2 amax = ω A = 50. 0( 0. 200) = 10. 0 m s at

(e)

E=

(f )

v = ω A − x = 50. 0

(g)

a = ω 2 x = 50. 0

k = m

f =

so

50. 0 rad s

x=0 x = ±A

1 2 1 kA = (100 ) (0 .200)2 = 2 .00 J 2 2 2

2

8 2 (0.200 ) = 1.33 m s 9

⎛ 0. 200 ⎞ = 3 .33 m s2 ⎝ 3 ⎠

Model the oscillator as a block-spring system. v 2 + ω 2x 2 = ω 2A 2

From energy considerations, vmax = ω A and v =

ωA 2

From this we find x2 = P15.20

ω = 1.13 Hz 2π

(a)

⎛ ωA ⎞ + ω 2x 2 = ω 2A 2 ⎝ 2⎠

and

x=

2

3 2 A 4

3 A = ±2.60 cm where A = 3.00 cm 2

1 2 at 2 y 1 − 11 m = 0 + 0 + ( −9. 8 m s 2 ) t 2 2 y f = yi + v yi t +

t= (b)

so

22 m ⋅ s2 = 1 .50 s 9 .8 m

Take the initial point where she steps off the bridge and the final point at the bottom of her motion.

(K +U

g

) (

+ U s = K + Ug + Us i

)

f

1 2 0 + mgy + 0 = 0 + 0 + kx 2 1 2 65 kg 9. 8 m s 36 m = k (25 m )2 2 k = 73 .4 N m (c)

(d)

2 F 65 kg 9.8 m s = = 8 .68 m , so this point is k 73.4 N m 11 + 8. 68 m = 19. 7 m below the bridge and the amplitude of her oscillation is 36 − 19.7 = 16.3 m.

The spring extension at equilibrium is x =

ω=

k = m

73 .4 N m = 1 .06 rad s 65 kg

Oscillatory Motion

(e)

Take the phase as zero at maximum downward extension. We find what the phase was 25 m higher, where x = −8.68 m: In x = A cosω t

16.3 m = 16 .3 m cos 0

t⎞ ⎛ − 8 .68 m = 16 .3 m cos 1.06 ⎝ s⎠

t 1.06 = − 122° = − 2 .13 rad s

t = −2.01 s Then +2 .01 s is the time over which the spring stretches. (f ) P15.21

total time = 1.50 s + 2. 01 s = 3. 50 s

The potential energy is Us =

1 2 1 2 kx = kA cos2 ( ω t ) 2 2

The rate of change of potential energy is

(a)

1 2 dU s 1 2 = kA 2 cos (ωt )[ − ω sin(ω t )] = − kA ω sin2ω t 2 dt 2 This rate of change is maximal and negative at 2ωt =

π π π , 2ω t = 2π + , or in general, 2ω t = 2nπ + for integer n 2 2 2

Then, t =

π π ( 4 n + 1) (4n + 1 ) = 4ω 4 (3.60 s −1 )
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