Chapter 22 physics solutions
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Publicado: 1 de febrero de 2011
91962_13_s22_p0988-1046
6/8/09
5:39 PM
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© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•22–1. A spring is stretched 175 mm by an 8-kg block. Ifthe block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s. + T ©Fy = may ; k Am $ mg - k(y + yst) = my k $ y + y = 0 m Where k = 8(9.81) = 448.46 N>m 0.175 where kyst = mgHence p =
=
A
448.46 = 7.487 8 ‹ $ y + (7.487)2y = 0 $ y + 56.1y = 0 Ans.
The solution of the above differential equation is of the form: y = A sin pt + B cos pt # y = y = Ap cos pt - Bp sin pt At t = 0, y = 0.1 m and y = y0 = 1.50 m>s From Eq. (1) 0.1 = A sin 0 + B cos 0 From Eq. (2) y0 = Ap cos 0 - 0 Hence At t = 0.22 s , B = 0.1 m A y0 1.50 = = 0.2003 m p 7.487 (1) (2)
y = 0.2003sin 7.487t + 0.1 cos 7.487t y = 0.2003 sin C 7.487(0.22) D + 0.1 cos C 7.487(0.22) D = 0.192 m Ans.
22–2. When a 2-kg block is suspended from a spring, the spring is stretched a distance of 40 mm. Determine the frequency and the period of vibration for a 0.5-kg block attached to the same spring. 2(9.81) F = = 490.5 N>m y 0.040 490.5 k = = 31.321 A 0.5 Am p 31.321 = = 4.985 Hz 2p 2p 1 1 = =0.201 s f 4.985
k =
p =
f =
Ans.
t =
Ans.
988
91962_13_s22_p0988-1046
6/8/09
5:40 PM
Page 989
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
22–3. A block having a weight of 8 lb is suspended from a spring having a stiffness k = 40 lb>ft. If the block is pushed y = 0.2 ft upward from its equilibrium position and then released from rest, determine the equation which describes the motion. What are the amplitude and the natural frequency of the vibration? Assume that positive displacement is downward. + T ©Fy = may ; $ mg - k(y+ yst) = my k $ y + y = 0 m Hence p = where kyst = mg
40 k = = 12.689 A 8>32.2 Am p 12.689 = = 2.02 Hz 2p 2p Ans.
f =
The solution of the above differential equation is of the form: v = A sin pt + B cos pt # y = y = Ap cos pt - Bp sin pt At t = 0, y = -0.2 ft and y = y0 = 0 From Eq. (1) -0.2 = A sin 0° + B cos 0° From Eq. (2) Hence Amplitude y0 = Ap cos 0° - 0 y = -0.2 cos 12.7t C = 0.2ft A = B = -0.2 ft y0 0 = = 0 p 12.689 Ans. Ans. (1) (2)
*22–4. A spring has a stiffness of 800 N>m. If a 2-kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward. k 800 = = 20 Am A 2
p =
y = A sin pt + B cos pt y = -0.05 m when t = 0,-0.05 = 0 + B; B = -0.05
v = Ap cos pt - Bp sin pt v = 0 when t = 0, 0 = A(20) - 0; Thus, y = -0.05 cos (20t) Ans. A = 0
989
91962_13_s22_p0988-1046
6/8/09
5:40 PM
Page 990
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in anyform or by any means, without permission in writing from the publisher.
•22–5. A 2-kg block is suspended from a spring having a stiffness of 800 N>m. If the block is given an upward velocity of 2 m>s when it is displaced downward a distance of 150 mm from its equilibrium position, determine the equation which describes the motion. What is the amplitude of the motion? Assume that positive...
6/8/09
5:39 PM
Page 988
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
•22–1. A spring is stretched 175 mm by an 8-kg block. Ifthe block is displaced 100 mm downward from its equilibrium position and given a downward velocity of 1.50 m>s, determine the differential equation which describes the motion. Assume that positive displacement is downward. Also, determine the position of the block when t = 0.22 s. + T ©Fy = may ; k Am $ mg - k(y + yst) = my k $ y + y = 0 m Where k = 8(9.81) = 448.46 N>m 0.175 where kyst = mgHence p =
=
A
448.46 = 7.487 8 ‹ $ y + (7.487)2y = 0 $ y + 56.1y = 0 Ans.
The solution of the above differential equation is of the form: y = A sin pt + B cos pt # y = y = Ap cos pt - Bp sin pt At t = 0, y = 0.1 m and y = y0 = 1.50 m>s From Eq. (1) 0.1 = A sin 0 + B cos 0 From Eq. (2) y0 = Ap cos 0 - 0 Hence At t = 0.22 s , B = 0.1 m A y0 1.50 = = 0.2003 m p 7.487 (1) (2)
y = 0.2003sin 7.487t + 0.1 cos 7.487t y = 0.2003 sin C 7.487(0.22) D + 0.1 cos C 7.487(0.22) D = 0.192 m Ans.
22–2. When a 2-kg block is suspended from a spring, the spring is stretched a distance of 40 mm. Determine the frequency and the period of vibration for a 0.5-kg block attached to the same spring. 2(9.81) F = = 490.5 N>m y 0.040 490.5 k = = 31.321 A 0.5 Am p 31.321 = = 4.985 Hz 2p 2p 1 1 = =0.201 s f 4.985
k =
p =
f =
Ans.
t =
Ans.
988
91962_13_s22_p0988-1046
6/8/09
5:40 PM
Page 989
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from thepublisher.
22–3. A block having a weight of 8 lb is suspended from a spring having a stiffness k = 40 lb>ft. If the block is pushed y = 0.2 ft upward from its equilibrium position and then released from rest, determine the equation which describes the motion. What are the amplitude and the natural frequency of the vibration? Assume that positive displacement is downward. + T ©Fy = may ; $ mg - k(y+ yst) = my k $ y + y = 0 m Hence p = where kyst = mg
40 k = = 12.689 A 8>32.2 Am p 12.689 = = 2.02 Hz 2p 2p Ans.
f =
The solution of the above differential equation is of the form: v = A sin pt + B cos pt # y = y = Ap cos pt - Bp sin pt At t = 0, y = -0.2 ft and y = y0 = 0 From Eq. (1) -0.2 = A sin 0° + B cos 0° From Eq. (2) Hence Amplitude y0 = Ap cos 0° - 0 y = -0.2 cos 12.7t C = 0.2ft A = B = -0.2 ft y0 0 = = 0 p 12.689 Ans. Ans. (1) (2)
*22–4. A spring has a stiffness of 800 N>m. If a 2-kg block is attached to the spring, pushed 50 mm above its equilibrium position, and released from rest, determine the equation that describes the block’s motion. Assume that positive displacement is downward. k 800 = = 20 Am A 2
p =
y = A sin pt + B cos pt y = -0.05 m when t = 0,-0.05 = 0 + B; B = -0.05
v = Ap cos pt - Bp sin pt v = 0 when t = 0, 0 = A(20) - 0; Thus, y = -0.05 cos (20t) Ans. A = 0
989
91962_13_s22_p0988-1046
6/8/09
5:40 PM
Page 990
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in anyform or by any means, without permission in writing from the publisher.
•22–5. A 2-kg block is suspended from a spring having a stiffness of 800 N>m. If the block is given an upward velocity of 2 m>s when it is displaced downward a distance of 150 mm from its equilibrium position, determine the equation which describes the motion. What is the amplitude of the motion? Assume that positive...
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