**UFRJ**

# Respostas e resolução Física Halliday 1 2 3 4 8 edição - ch15

(Parte **1** de 5)

Chapter 15

1. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2pif since there are 2pi radians in one cycle). Therefore, in this circumstance, we obtain

2. (a) The angular frequency ω is given by ω = 2pif = 2pi/T, where f is the frequency and T is the period. The relationship f = 1/T was used to obtain the last form. Thus ω = 2pi/(1.0 × 10–5 s) = 6.28 × 105 rad/s.

(b) The maximum speed vm and maximum displacement xm are related by vm = ωxm, so

rad/s m

3. (a) The amplitude is half the range of the displacement, or xm = 1.0 m.

(b) The maximum speed vm is related to the amplitude xm by vm = ωxm, where ω is the angular frequency. Since ω = 2pif, where f is the frequency,

4. (a) The acceleration amplitude is related to the maximum force by Newton’s second law: Fmax = mam. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency (ω = 2pif since there are 2pi radians in one cycle). The frequency is the reciprocal of the period: f = 1/T =

1/0.20 = 5.0 Hz, so the angular frequency is ω = 10pi (understood to be valid to two significant figures). Therefore,

CHAPTER 15

()()2 | 0.12kg10 rad/s1.210N/m.km |

5. (a) During simple harmonic motion, the speed is (momentarily) zero when the object is at a “turning point” (that is, when x = +xm or x = –xm). Consider that it starts at x = +xm and we are told that t = 0.25 second elapses until the object reaches x = –xm. To execute a full cycle of the motion (which takes a period T to complete), the object which started at x = +xm must return to x = +xm (which, by symmetry, will occur 0.25 second after it was at x = –xm). Thus, T = 2t = 0.50 s.

(b) Frequency is simply the reciprocal of the period: f = 1/T = 2.0 Hz.

(c) The 36 cm distance between x = +xm and x = –xm is 2xm. Thus, xm = 36/2 = 18 cm.

6. (a) Since the problem gives the frequency f = 3.0 Hz, we have ω = 2pif = 6pi rad/s (understood to be valid to three significant figures). Each spring is considered to support one fourth of the mass mcar so that Eq. 15-12 leads to

1 | 1450kg6 rad/s1.2910N/m. |

(b) If the new mass being supported by the four springs is mtotal = [1450 + 5(73)] kg = 1815 kg, then Eq. 15-12 leads to

new new total

1.2910 N/m | 2.68Hz. |

7. (a) The motion repeats every 0.500 s so the period must be T = 0.500 s. (b) The frequency is the reciprocal of the period: f = 1/T = 1/(0.50 s) = 2.0 Hz.

(c) The angular frequency ω is ω = 2pif = 2pi(2.0 Hz) = 12.6 rad/s.

(d) The angular frequency is related to the spring constant k and the mass m by ω=km. We solve for k and obtain k = mω2 = (0.500 kg)(12.6 rad/s)2 = 79.0 N/m.

(e) Let xm be the amplitude. The maximum speed is vm = ωxm = (12.6 rad/s)(0.350 m) = 4.40 m/s.

(f) The maximum force is exerted when the displacement is a maximum and its magnitude is given by Fm = kxm = (79.0 N/m)(0.350 m) = 27.6 N.

8. (a) The problem describes the time taken to execute one cycle of the motion. The period is T = 0.75 s.

(b) Frequency is simply the reciprocal of the period: f = 1/T ≈ 1.3 Hz, where the SI unit abbreviation Hz stands for Hertz, which means a cycle-per-second.

(c) Since 2pi radians are equivalent to a cycle, the angular frequency ω (in radians-per-second) is related to frequency f by ω = 2pif so that ω ≈ 8.4 rad/s.

9. The magnitude of the maximum acceleration is given by am = ω2xm, where ω is the angular frequency and xm is the amplitude.

(a) The angular frequency for which the maximum acceleration is g is given by ω=gxm/, and the corresponding frequency is given by gf xω pi pi pi −

(b) For frequencies greater than 498 Hz, the acceleration exceeds g for some part of the motion.

10. We note (from the graph) that xm = 6.0 cm. Also the value at t = 0 is xo = − 2.0 cm. Then Eq. 15-3 leads to

The other “root” (+4.37 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0.

1. (a) Making sure our calculator is in radians mode, we find

(b) Differentiating with respect to time and evaluating at t = 2.0 s, we find dt − FHG IKJ −pi pib g b gsin m/spi

CHAPTER 15 (c) Differentiating again, we obtain dt − FHG IKJ − ×pi pi pibgb g b gcos m/s

(d) In the second paragraph after Eq. 15-3, the textbook defines the phase of the motion. In this case (with t = 2.0 s) the phase is 3pi(2.0) + pi/3 ≈ 20 rad.

(e) Comparing with Eq. 15-3, we see that ω = 3pi rad/s. Therefore, f = ω/2pi = 1.5 Hz. (f) The period is the reciprocal of the frequency: T = 1/f ≈ 0.67 s.

4.0 cm/s | Then Eq. 15-6 leads to |

12. We note (from the graph) that vm = ωxm = 5.0 cm/s. Also the value at t = 0 is vo = φ = sin−1(− 4.0/5.0) = – 0.927 rad or +5.36 rad.

The other “root” (+4.07 rad) can be rejected on the grounds that it would lead to a positive slope at t = 0.

13. When displaced from equilibrium, the net force exerted by the springs is –2kx acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration 2/adxdt=, Newton’s second law yields

Substituting x = xm cos(ωt + φ) and simplifying, we find

where ω is in radians per unit time. Since there are 2pi radians in a cycle, and frequency f measures cycles per second, we obtain

14. The statement that “the spring does not affect the collision” justifies the use of elastic collision formulas in section 10-5. We are told the period of SHM so that we can find the mass of block 2:

2 | 0.600 kg.4 |

m kTT mkpi pi

At this point, the rebound speed of block 1 can be found from Eq. 10-30:

This becomes the initial speed v0 of the projectile motion of block 1. A variety of choices for the positive axis directions are possible, and we choose left as the +x direction and down as the +y direction, in this instance. With the “launch” angle being zero, Eq. 4-21 and Eq. 4-2 (with –g replaced with +g) lead to

9.8 m/s hx x v t v g

Since x – x0 = d, we arrive at d = 4.0 m.

15. (a) Eq. 15-8 leads to

Therefore, f = ω/2pi = 5.58 Hz. (b) Eq. 15-12 provides a relation between ω (found in the previous part) and the mass:

400 N/m= | 0.325kg. |

(c) By energy conservation, 122kxm (the energy of the system at a turning point) is equal to the sum of kinetic and potential energies at the time t described in the problem.

v xm m⇒

16. From highest level to lowest level is twice the amplitude xm of the motion. The period is related to the angular frequency by Eq. 15-5. Thus, xdm=12 and ω = 0.503 rad/h. The phase constant φ in Eq. 15-3 is zero since we start our clock when xo = xm (at the highest point). We solve for t when x is one-fourth of the total distance from highest to lowest level, or (which is the same) half the distance from highest level to middle level (where we locate the origin of coordinates). Thus, we seek t when the ocean surface is at

CHAPTER 15

which has t = 2.08 h as the smallest positive root. The calculator is in radians mode during this calculation.

17. The maximum force that can be exerted by the surface must be less than µsFN or else the block will not follow the surface in its motion. Here, µs is the coefficient of static friction and FN is the normal force exerted by the surface on the block. Since the block does not accelerate vertically, we know that FN = mg, where m is the mass of the block. If the block follows the table and moves in simple harmonic motion, the magnitude of the maximum force exerted on it is given by

F = mam = mω2xm = m(2pif)2xm, where am is the magnitude of the maximum acceleration, ω is the angular frequency, and f is the frequency. The relationship ω = 2pif was used to obtain the last form. We substitute F = m(2pif)2xm and FN = mg into F < µsFN to obtain m(2pif)2xm < µsmg. The largest amplitude for which the block does not slip is

pi pib g b gc h b g

= | m |

A larger amplitude requires a larger force at the end points of the motion. The surface cannot supply the larger force and the block slips.

18. They pass each other at time t, at xxxm121 2== where

From this, we conclude that cos()cos()ωφωφt+=+=1212, and therefore that the phases (the arguments of the cosines) are either both equal to pi/3 or one is pi/3 while the other is

–pi/3. Also at this instant, we have v1 = –v2≠0 where

This leads to sin(ωt + φ1) = – sin(ωt + φ 2). This leads us to conclude that the phases have opposite sign. Thus, one phase is pi/3 and the other phase is –pi /3; the wt term cancels if we take the phase difference, which is seen to be pi /3 – (–pi /3) = 2pi /3.

19. (a) Let cos piFHG IKJ

be the coordinate as a function of time for particle 1 and cos pi piFHG IKJ

be the coordinate as a function of time for particle 2. Here T is the period. Note that since the range of the motion is A, the amplitudes are both A/2. The arguments of the cosine functions are in radians. Particle 1 is at one end of its path (x1 = A/2) when t = 0. Particle 2 is at A/2 when 2pit/T + pi/6 = 0 or t = –T/12. That is, particle 1 lags particle 2 by one-twelfth a period. We want the coordinates of the particles 0.50 s later; that is, at t = 0.50 s,

and

Their separation at that time is x1 – x2 = –0.25A + 0.43A = 0.18A. (b) The velocities of the particles are given by

==2 11vdxdtATtT pi pisinFH IK and

==2+6. 22vdxdtATtT pi pi pisinFH IK

We evaluate these expressions for t = 0.50 s and find they are both negative-valued, indicating that the particles are moving in the same direction.

20. We note that the ratio of Eq. 15-6 and Eq. 15-3 is v/x = –ωtan(ωt + φ) where ω = 1.20 rad/s in this problem. Evaluating this at t = 0 and using the values from the graphs shown in the problem, we find

One can check that the other “root” (4.17 rad) is unacceptable since it would give the wrong signs for the individual values of vo and xo.

21. Both parts of this problem deal with the critical case when the maximum acceleration becomes equal to that of free fall. The textbook notes (in the discussion immediately after

CHAPTER 15

Eq. 15-7) that the acceleration amplitude is am = ω2xm, where ω is the angular frequency; this is the expression we set equal to g = 9.8 m/s2.

(a) Using Eq. 15-5 and T = 1.0 s, we have

piT x g x gT mmFHGIKJ⇒ m

(b) Since ω = 2pif, and xm = 0.050 m is given, we find

()212= | =2.2 Hz.2m |

m gf x g f xpi pi

2. Eq. 15-12 gives the angular velocity:

Energy methods (discussed in §15-4) provide one method of solution. Here, we use trigonometric techniques based on Eq. 15-3 and Eq. 15-6.

(a) Dividing Eq. 15-6 by Eq. 15-3, we obtain

=+ vx t−ω ω φtanb g so that the phase (ωt + φ) is found from

back into Eq. 15-3 leads to0.129mcos(1.31) | 0.500m.mx=−⇒= |

With the calculator in radians mode, this gives the phase equal to –1.31 rad. Plugging this

(b) Since ωt + φ = –1.31 rad at t = 1.0 s, we can use the above value of ω to solve for the phase constant φ. We obtain φ = –8.38 rad (though this, as well as the previous result, can have 2pi or 4pi (and so on) added to it without changing the physics of the situation). With this value of φ, we find xo = xm cos φ = – 0.251 m.

(c) And we obtain vo = –xmω sinφ = 3.06 m/s.

23. Let the spring constants be k1 and k2. When displaced from equilibrium, the magnitude of the net force exerted by the springs is |k1x + k2 x| acting in a direction so as to return the block to its equilibrium position (x = 0). Since the acceleration a = d2x/d2,

Newton’s second law yields

Substituting x = xm cos(ωt + φ) and simplifying, we find

where ω is in radians per unit time. Since there are 2pi radians in a cycle, and frequency f measures cycles per second, we obtain

The single springs each acting alone would produce simple harmonic motions of frequency

1=30 Hz, | =45 Hz, |

respectively. Comparing these expressions, it is clear that

24. To be on the verge of slipping means that the force exerted on the smaller block (at the point of maximum acceleration) is fmax = µs mg. The textbook notes (in the discussion immediately after Eq. 15-7) that the acceleration amplitude is am =ω2xm, where ω=+kmM/() is the angular frequency (from Eq. 15-12). Therefore, using Newton’s second law, we have

m M x gm s m sµ µ⇒ which leads to

200 N/msm g m Mx k

25. (a) We interpret the problem as asking for the equilibrium position; that is, the block is gently lowered until forces balance (as opposed to being suddenly released and allowed to oscillate). If the amount the spring is stretched is x, then we examine force-components along the incline surface and find

sin(14.0N)sin40.0sin | 0.0750 m |

CHAPTER 15

120 N/m mgkx mg x k at equilibrium. The calculator is in degrees mode in the above calculation. The distance from the top of the incline is therefore (0.450 + 0.75) m = 0.525 m.

(b) Just as with a vertical spring, the effect of gravity (or one of its components) is simply to shift the equilibrium position; it does not change the characteristics (such as the period) of simple harmonic motion. Thus, Eq. 15-13 applies, and we obtain

26. We wish to find the effective spring constant for the combination of springs shown in the figure. We do this by finding the magnitude F of the force exerted on the mass when the total elongation of the springs is ∆x. Then keff = F/∆x. Suppose the left-hand spring is elongated by ∆x and the right-hand spring is elongated by ∆xr. The left-hand spring exerts a force of magnitude kx∆ on the right-hand spring and the right-hand spring exerts a force of magnitude k∆xr on the left-hand spring. By Newton’s third law these must be equal, so ∆∆xxr =. The two elongations must be the same and the total elongation is twice the elongation of either spring: ∆∆x=2 . The left-hand spring exerts a force on the block and its magnitude is Fkx=∆ . Thus kkxxkreff==∆∆ //2. The block behaves as if it were subject to the force of a single spring, with spring constant k/2. To find the frequency of its motion replace keff in fkm=12//piafeff with k/2 to obtain

With m = 0.245 kg and k = 6430 N/m, the frequency is f = 18.2 Hz.

27. When the block is at the end of its path and is momentarily stopped, its displacement is equal to the amplitude and all the energy is potential in nature. If the spring potential energy is taken to be zero when the block is at its equilibrium position, then

28. (a) The energy at the turning point is all potential energy: Ekxm=122 where E = 1.0 J and xm = 0.100 m. Thus,

(b) The energy as the block passes through the equilibrium position (with speed vm = 1.20 m/s) is purely kinetic:

vm m

⇒ kg

(c) Eq. 15-12 (divided by 2pi) yields

= | Hz |

29. The total energy is given by Ekxm=122, where k is the spring constant and xm is the amplitude. We use the answer from part (b) to do part (a), so it is best to look at the solution for part (b) first.

(a) The fraction of the energy that is kinetic is

where the result from part (b) has been used.

(b) When xxm=12 the potential energy is Ukxkxm==12 2182. The ratio is kxU

(c) Since Ekxm=122 and Ukx=122, U/E = xxm 2. We solve xxm

2 = 1/2 for x. We should

30. The total mechanical energy is equal to the (maximum) kinetic energy as it passes through the equilibrium position (x = 0):

Looking at the graph in the problem, we see that U(x=10)=0.5 J. Since the potential function has the form 2()Uxbx=, the constant is 325.010J/cmb−=×. Thus, U(x) = 0.72 J when x = 12 cm.

(a) Thus, the mass does turn back before reaching x = 15 cm. (b) It turns back at x = 12 cm.

31. (a) Eq. 15-12 (divided by 2pi) yields

CHAPTER 15

(c) With vo = 10.0 m/s, the initial kinetic energy is Kmv01 2 02 250= = J.

(Parte **1** de 5)