# Solucionario Capitulo 14 - Paul E. Tippens

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Chapter 14. Simple Harmonic Motion

Physics, 6th Edition

Chapter 14. Simple Harmonic MotionPeriodic Motion and the Reference Circle14-1. A rock swings in a circle at constant speed on the end of a string, making 50 revolutions in 30 s. What is the frequency and the period for this motion? f = 50 rev = 1.67 rev/s ; f = 1.67 Hz 30 s T= 1 1 = ; f 1.67 hz T = 0.600 s

14-2. A child sits at the edge of a platform rotating at 30 rpm. The platform is 10 m in diameter. What is the period of the motion and what is the childs speed? [ R = (D/2) = 5 m ] 30 rev 1 min = 0.500 rev/s; f = 0.500 Hz; min 60 s v= 2 R 2 (5 m) = ; T 2.00 s T= 1 1 = f 0.500 hz T = 2.00 s

v = 15.7 m/s

14-3. A rubber ball swings in a horizontal circle 2 m in diameter and makes 20 revolutions in one minute. The shadow of the ball is projected on a wall by a distant light. What are the amplitude, frequency, and period for the motion of the shadow? A = R = 1.00 m f = 20 rev 1 min = 0.333 rev/s; min 60 s T = 3.00 s R

f = 0.333 Hz

T=

1 1 = ; f 0.333 hz

-R

0

x +R

14-4. Assume a ball makes 300 rpm while moving in a circle of radius 12 cm. What are the amplitude, frequency, and period for the motion of its shadow projected on a wall? A = R = 12 cm f = 300 rev 1 min = 5.00 rev/s; min 60 s T = 0.200 s f = 5.00 Hz

T = 1/f = 1/5 s

187

Chapter 14. Simple Harmonic Motion

Physics, 6th Edition

14-5. A mass oscillates at a frequency of 3 Hz and an amplitude of 6 cm. What are its positions at time t = 0 and t = 2.5 s? At t = 0: x = A cos (2ft) = (6 cm) cos [2(3 Hz)(0)]; x = 6.00 cm x = 1.85 cm

At t = 2.5 s: x = A cos (2ft) = (6 cm) cos [2(3 Hz)(2.5 s)];

14-6. A 50-g mass oscillates with SHM of frequency 0.25 Hz. Assume the t = 0 when the mass is at its maximum displacement. At what time will its displacement be zero? At what time will it be located at half of its amplitude? f = 0.25 s, T = 1/f = 4.0 s

One complete vibration takes 4 s, Therefore the mass reaches zero in one-fourth of that time, or x= t = 4s/4 = 1.00 s. A = A cos(2 ft ); 2 Now we find the time to reach x = A/2: cos(2 ft) = 0.5; t= (2 ft ) = cos 1 (0.5) = 1.047 rad

2ft = 1.047 rad;

1.047 rad ; 2 (0.25 Hz)

t = 0.677 s

Note that the time to reach A/2 is not equal to one-half the time to reach x = 0. That is because the restoring force is not constant. It reaches x = 0 in 1 s, but it covers half that distance in a time of 0.667 s. 14-7. When a mass of 200-g is hung from a spring, the spring is displaced downward a distance of 1.5 cm. What is the spring constant k? [ F = mg; x = 1.5 cm = 0.015 m ] F (0.200 kg)(9.8 m/s 2 ) ; k= = x 0.015 m k = 131 N/m

14-8. An additional mass of 400 kg is added to the initial 200-g mass in Problem 14-7. What will be the increase in downward displacement? ( F is due only to the added mass.)

188

Chapter 14. Simple Harmonic Motion F ; x F (0.400 kg)(9.8 m/s 2 ) ; = k 131 N/m

Physics, 6th Edition

k=

x =

x = 2.99 cm

*14-9. A mass at the end of a spring vibrates up and down with a frequency of 0.600 Hz and an amplitude of 5 cm. What is its displacement 2.56 s after it reaches a maximum? x = A cos(2ft) = (5 cm) cos [2(0.6 Hz)(2.56 s)]; x = -4.87 cm

*14.10. An object vibrates with an amplitude of 6 cm and a frequency of 0.490 Hz. Starting from maximum displacement in the positive direction, when will be the first time that its displacement is 2 cm? x = A cos(2ft); cos(2 ft ) = x 2 cm = = 0.333 ; A 6 cm 1.23 rad ; 2 (0.490 Hz) (2ft) = cos-1(0.333);

2ft = 1.23 rad;

t=

t = 0.400 s

Velocity in Simple Harmonic Motion14-11. A body vibrates with a frequency of 1.4 Hz and an amplitude of 4 cm. What is the maximum velocity? What is its position when the velocity is zero? v = -2fA sin(2ft); vmax occurs when sin = 1, vmax = 2fA

vmax = 2(1.4 Hz)(4 cm); vmax = 35.2 cm/s When v = 0, x = A or x = 4.00 cm 14-12. An object oscillates at a frequency of 5 Hz and an amplitude of 6 cm. What is the maximum velocity? vmax = -2fA = -2(1.4 Hz)(4 cm); vmax = 35.2 cm/s

189

Chapter 14. Simple Harmonic Motion

Physics, 6th Edition

14-13. A smooth block on a frictionless surface is attached to a spring, pulled to the right a distance of 4 cm and then released. Three seconds later it returns to the point of release. What is the frequency and what is the maximum speed? [ T = 3.00 s ] f = 1/T = 0.333 Hz; vmax = 2fA = 2(0.333 Hz)(4 cm); vmax = 8.38 cm/s 14-14. In Problem 14-13, what are the position and velocity 2.55 s after release? x = A cos(2ft) = (4 cm)cos[2(0.333 Hz)(2.55 s)]; x = 2.35 cm v = -2fA sin(2ft) = -2(0.333 Hz)(4 cm)sin [2(0.333 Hz)(2.55 s); v = 6.78 cm/s The location is 2.35 cm to the right (+) of center,

and the body is moving to the right (+) at 6.78 cm/s. *14-15. Show that the velocity of n object in SHM can be written as a function of its amplitude and displacement: (Look at the reference circle: = 2ft) sin = y A y v = 2 fA sin = 2 fA ; A v = 2fy A2 x 2 A x y

But, y = A2 x 2 from Pythagorass theorem, Thus: v = 2f

NOTE: This expression can also be derived from conservation of energy principles. The equation derived in this example is so important for many applications, that the author strongly suggests it be assigned to every student. 14-16. Use the relation derived in Problem 14-15 to verify the answers obtained for position and velocity in Problem 14-14.

190

Chapter 14. Simple Harmonic Motion v = 2 f

Physics, 6th Edition

A2 x 2 = 2 (0.333 Hz) (4 cm) 2 (2.35 cm) 2 ; v = 6.78 cm/s

14-17. A mass vibrating at a frequency of 0.5 Hz has a velocity of 5 cm/s as it passes the center of oscillation. What are the amplitude and the period of vibration? vmax = -2fA = 5 cm/s A = T= vmax 5 cm/s = ; 2 f 2 (0.5 Hz) T = 2.00 s A = 1.59 cm

1 1 ; f 0.5 Hz

*14-18. A body vibrates with a frequency of 8 Hz and an amplitude of 5 cm. At what time after it is released from x = +5 cm will its velocity be equal to +2.00 m/s? v = -2fA sin(2ft); sin(2 ft ) = v 2 m/s = = 0.796 ; 2 fA -2 (8 Hz)(0.05 m) t= 0.921 ; 2 (8 Hz) t = -18.3 ms

(2ft) = sin-1(-0.796); 2ft = -0.920

NOTE: The negative sign for time, means that the velocity was +2 m/s, 18.3 ms BEFORE its displacement was +5 cm. There will be two times within the first period of the vibration, that the velocity will be + 5 m/s. One is 18.3 ms before reaching the end of the first period (t = -18.3 ms), the second is 18.3 s after reaching half of the period. Thus, to find the first time that the velocity was +2 m/s after release from +5 cm, we need to ADD 18.3 ms to one-half of the period T. T= 1 1 = = 0.125 s , f 8 Hz t= T 125 ms + 18.3 ms = + 18.3 ms 2 2 t = 80.8 ms

t = 62.5 ms + 18.3 ms;

191

Chapter 14. Simple Harmonic Motion

Physics, 6th Edition

Acceleration in Simple Harmonic Motion14-19. A 400-g mass is attached to a spring causing it to stretch a vertical distance of 2 cm. The mass is then pulled downward a distance of 4 cm and released to vibrate with SHM as shown in Fig. 14-10. What is the spring constant? What is the magnitude and direction of the acceleration when the mass is located 2 cm below its equilibrium position? k= (0.400 kg)(9.8 m/s 2 ) ; 0.02 m k = 196 N/m; Consider down as +.

a=

k (196 N/m)(0.02 m) x= m 0.400 kg

a = -9.8 m/s2, directed upward

14-20. What is the maximum acceleration for the system described in Problem 14-19 and what is its acceleration when it is located 3 cm above its equilibrium position? (down + ) amax = k (196 N/m)(0.04 m) A= m 0.400 kg a = -19.6 m/s2, directed upward

a=

k (196 N/m)(-0.03 m) x= m 0.400 kg

a = 14.7 m/s2, directed downward

14-21. A body makes one complete oscillation in 0.5 s. What is its acceleration when it is displaced a distance of x = +2 cm from its equilibrium position? (T = 0.5 s) f = 1 1 = = 2.00 Hz ; a = -42f2x = -42(2 Hz)2(0.02 m); a = -3.16 m/s2 T 0.5 s

14-22. Find the maximum velocity and the maximum acceleration of an object moving in SHM of amplitude 16 cm and frequency 2 Hz.

192

Chapter 14. Simple Harmonic Motion vmax = 2 fA = 2 (2 Hz)(0.16 m); amax = -42f2A = -42(2 Hz)2(0.16 m) vmax = 2.01 m/s amax = 25.3 m/s2

Physics, 6th Edition

*14-23. A object vibrating with a period of 2 seconds is displaced a distance of x = +6 cm and released. What are the velocity and acceleration 3.20 s after release? [f = 1/T= 0.5 Hz ] v = 2 fA sin(2 ft ) = 2 (0.5 Hz)(0.06 m) sin[(2 (0.5 Hz)(3.2 s)]; v = +0.111 m/s; v = +11.1 cm/s, in + direction

a = -42f2A cos(2ft) = -42(0.5 Hz)2(0.06 m) cos [2(0.5 Hz)(3.2 s)] a = +0.479 m/s2, in positive direction 14-24. A body vibrates with SHM of period 1.5 s and amplitude 6 in. What are its maximum velocity and acceleration? [ f = 1/T = (1/1.5 s) = 0.667 Hz ; A = 6 in. = 0.5 ft ] vmax = 2 fA = 2 (0.667 Hz)(0.5 ft); amax = -42f2A = -42(0.667 Hz)2(0.5 ft) vmax = 2.09 ft/s amax = 8.77 ft/s2

*14-25. For the body in Problem 14-24, what are its velocity and acceleration after a time of 7 s? v = 2 fA sin(2 ft ) = 2 (0.667 Hz)(0.5 ft) sin[(2 (0.667 Hz)(7 s)]; v = +1.81 ft/s; v = +1.81 ft/s, in + direction

a = -42f2A cos(2ft) = -42(0.667 Hz)2(0.5 ft) cos [2(0.5 Hz)(7 s)] a = +4.39 ft/s2, in positive direction

Period and Frequency*14-26. The prong of a tuning fork vibrates with a frequency of 330 Hz and an amplitude of 2 mm. What is the velocity when the displacement is 1.5 mm? (See Prob. 14-15) v = 2 f A2 x 2 = 2 (330 Hz) (2.0 mm) 2 (1.5 mm) 2 193

Chapter 14. Simple Harmonic Motion v = 2743 mm/s; v = 2.74 m/s

Physics, 6th Edition

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