Example: A (1.84 N) body is suspended by a spring which is stretched (15.3 cm) when it is loaded. If the body is drawn down (10 cm) from the position of time (1) if: equilibrium; find the position of the spring as a function of 1. e-1.5 2. c= 3.75 3. c=3 Solution: M y"+c y'+k y = 0 W 1.84 9.8 M 0.188 Ag 1.84 - 15.3 cm 5. 0.153 10 cm 1) for e-1.5 L84 0.188y"-1.5y'+12 y-0 y"+8y'+64 y = 0 m +8m + 64 = 0 – 87 v64 – 4× 64 Dashpot 44/31 o Under Damping 2 v)"(Acos4/3 t+ Bsin 4/3t) Initial Conamo at 1=0 y'=o y=0.lm y(0) = A+0=0.1 = A=0.1 y0)--45 Asin 4/31+4/5 Beos45 ] [tcos4/3+ Bsin4/5-4e") y(0) = 45 B-4A= 4/3 B -4x0.1=0 = B-
Example: A (1.84 N) body is suspended by a spring which is stretched (15.3 cm) when it is loaded. If the body is drawn down (10 cm) from the position of time (1) if: equilibrium; find the position of the spring as a function of 1. e-1.5 2. c= 3.75 3. c=3 Solution: M y"+c y'+k y = 0 W 1.84 9.8 M 0.188 Ag 1.84 - 15.3 cm 5. 0.153 10 cm 1) for e-1.5 L84 0.188y"-1.5y'+12 y-0 y"+8y'+64 y = 0 m +8m + 64 = 0 – 87 v64 – 4× 64 Dashpot 44/31 o Under Damping 2 v)"(Acos4/3 t+ Bsin 4/3t) Initial Conamo at 1=0 y'=o y=0.lm y(0) = A+0=0.1 = A=0.1 y0)--45 Asin 4/31+4/5 Beos45 ] [tcos4/3+ Bsin4/5-4e") y(0) = 45 B-4A= 4/3 B -4x0.1=0 = B-
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
How solve this step?!
![(15.3
em) when it is loaded. If the body is drawn down (10 cm) from the position of
time (t) if:
Example: A (1.84 N) body is suspended by a spring which is stretched
equilibrium; find the position of the spring as a function of
1. c 1.5
2. c= 3.75
3. c= 3
Solution:
My"tc y'+k y = 0
1.84
=0.188 kg
9.8
1.84
=12
0.153
k =
= 15.3 cm
= 10 cm
1) for e 1.5
184
0.188 y"+1.5 y'+12 y=0
y"+8 y'+64 y = 0
m' +8m+64 0
Dashpot
-87 V64 - 4x 64
4743i e Under Damping
m =
2
v)=e"(Acos4,/3 t+ Bsin 4,/3 t)
Initial Condiuons
at 1=0 = y'=0 =y=0,1m
y(0) = A+0=0.1 = A=0.1
y) =-45 Asin 43 +4/3 Bcos4 31-[Acos4,3+Bsin 4/3](4e)
y'(0) = 43 B-44=4/3 B-4x0.1=0 = B=
103](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe865e4d1-8009-48f3-8b5e-f72c4b053baa%2F43df54ad-62f8-45e8-b3d2-96d785f415ab%2Ftiz0hsk_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(15.3
em) when it is loaded. If the body is drawn down (10 cm) from the position of
time (t) if:
Example: A (1.84 N) body is suspended by a spring which is stretched
equilibrium; find the position of the spring as a function of
1. c 1.5
2. c= 3.75
3. c= 3
Solution:
My"tc y'+k y = 0
1.84
=0.188 kg
9.8
1.84
=12
0.153
k =
= 15.3 cm
= 10 cm
1) for e 1.5
184
0.188 y"+1.5 y'+12 y=0
y"+8 y'+64 y = 0
m' +8m+64 0
Dashpot
-87 V64 - 4x 64
4743i e Under Damping
m =
2
v)=e"(Acos4,/3 t+ Bsin 4,/3 t)
Initial Condiuons
at 1=0 = y'=0 =y=0,1m
y(0) = A+0=0.1 = A=0.1
y) =-45 Asin 43 +4/3 Bcos4 31-[Acos4,3+Bsin 4/3](4e)
y'(0) = 43 B-44=4/3 B-4x0.1=0 = B=
103
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