Question 1

The differential equation $\frac { d ^ { 2 } x } { d t ^ { 2 } } + x \cos x = 0$ is a model for an undamped spring-mass system with a nonlinear forcing function. The initial conditions are $x ( 0 ) = 0.1$ , $x ^ { \prime } ( 0 ) = - 0.1$ . The solution of the linearized system is

Accepted Answer

A)    x=0.1cos⁡tx = 0.1 \cos tx=0.1cost 
B)    x=0.1sin⁡tx = 0.1 \sin tx=0.1sint 
C)    x=0.1cos⁡t−0.1sin⁡tx = 0.1 \cos t - 0.1 \sin tx=0.1cost−0.1sint 
D)    x=0.1e−tx = 0.1 e ^ { - t }x=0.1e−t 
E)    x=0.1etx = 0.1 e ^ { t }x=0.1et C

Question 2

In the previous problem, if $y ( 0 ) = R$ , what is the escape velocity?

Accepted Answer

A)    v0=2k/(mR) v _ { 0 } = \sqrt { 2 k / ( m R )  }v0​=2k/(mR) c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
B)    v0=2k/(mR) v _ { 0 } = 2 k / ( m R ) v0​=2k/(mR)  
C)    v0=2m/(kR) v _ { 0 } = 2 m / ( k R ) v0​=2m/(kR)  
D)    v0=2m/(kR) v _ { 0 } = \sqrt { 2 m / ( k R )  }v0​=2m/(kR) c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
E)   none of the above  A

Question 3

If the mass in the previous problem is pulled down two feet and released, the solution for the position is

Accepted Answer

A)    x=2cos⁡(42t) +2sin⁡(42t) x = 2 \cos ( 4 \sqrt { 2 } t )  + 2 \sin ( 4 \sqrt { 2 } t ) x=2cos(42c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t) +2sin(42c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t)  
B)    x=2sin⁡(42t) x = 2 \sin ( 4 \sqrt { 2 } t ) x=2sin(42c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t)  
C)    x=2cos⁡(42t) x = 2 \cos ( 4 \sqrt { 2 } t ) x=2cos(42c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t)  
D)    x=2sin⁡(4t) x = 2 \sin ( 4 t ) x=2sin(4t)  
E)    x=2cos⁡(4t) x = 2 \cos ( 4 t ) x=2cos(4t)  C

Question 4

A spring attached to the ceiling is stretched one foot by a four pound weight. The value of the Hooke's Law spring constant, $k$ , is

Accepted Answer

A)   4 pounds per foot
B)   1/4 pound per foot
C)   1/4 foot-pound
D)   4 foot-pounds
E)   none of the above

Question 5

The solution of the differential equation of the previous problem is

Accepted Answer

A)    θ=ce2t	heta = c e ^ { 2 t }θ=ce2t 
B)    θ=ce−2t	heta = c e ^ { - 2 t }θ=ce−2t 
C)    θ=ce−t/2	heta = c e ^ { - t / 2 }θ=ce−t/2 
D)    θ=c1cos⁡(2t) +c2sin⁡(2t) 	heta = c _ { 1 } \cos ( 2 t )  + c _ { 2 } \sin ( 2 t ) θ=c1​cos(2t) +c2​sin(2t)  
E)    θ=c1cos⁡(2t) +c2sin⁡(2t) 	heta = c _ { 1 } \cos ( \sqrt { 2 } t )  + c _ { 2 } \sin ( \sqrt { 2 } t ) θ=c1​cos(2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t) +c2​sin(2c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t)

Question 6

In the previous two problems, how long does it take for the chain to fall completely to the ground?

Accepted Answer

A)    t=(L/g) 2t = ( L / g )  ^ { 2 }t=(L/g) 2 
B)    t=g/Lt = g / Lt=g/L 
C)    t=L/gt = L / gt=L/g 
D)    t=L/gt = \sqrt { L / g }t=L/gc34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
E)    t=g/Lt = \sqrt { g / L }t=g/Lc34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​

Question 7

In the previous problem, the function $x$ can be written as

Accepted Answer

A)    x=3sin⁡(t+ϕ) , where tan⁡ϕ=−4/3x = 3 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 4 / 3x=3sin(t+ϕ) , where tanϕ=−4/3 
B)    x=4sin⁡(t+ϕ) , where tan⁡ϕ=−4/3x = 4 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 4 / 3x=4sin(t+ϕ) , where tanϕ=−4/3 
C)    x=5sin⁡(t+ϕ) , where tan⁡ϕ=−4/3x = 5 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 4 / 3x=5sin(t+ϕ) , where tanϕ=−4/3 
D)    x=5sin⁡(t+ϕ) , where tan⁡ϕ=−3/4x = 5 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 3 / 4x=5sin(t+ϕ) , where tanϕ=−3/4 
E)    x=4sin⁡(t+ϕ) , where tan⁡ϕ=−3/4x = 4 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 3 / 4x=4sin(t+ϕ) , where tanϕ=−3/4

Question 8

In the previous problem, the solution for $x$ as a function of $t$ is

Accepted Answer

A)    x=L−L2−Lgtx = L - \sqrt { L ^ { 2 } - L g t }x=L−L2−Lgtc34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
B)    x=L−L2+Lgtx = L - \sqrt { L ^ { 2 } + L g t }x=L−L2+Lgtc34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
C)    x=L−L2−Lgt2x = L - \sqrt { L ^ { 2 } - L g t ^ { 2 } }x=L−L2−Lgt2c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
D)    x=L−L2+Lgt2x = L - \sqrt { L ^ { 2 } + L g t ^ { 2 } }x=L−L2+Lgt2c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​ 
E)    x=L−Lgt2−L2x = L - \sqrt { L g t ^ { 2 } - L ^ { 2 } }x=L−Lgt2−L2c34,79.3,68.167,158.7,102.5,238c34.3,79.3,51.8,119.3,52.5,120c340,-704.7,510.7,-1060.3,512,-1067l0 -0c4.7,-7.3,11,-11,19,-11H40000v40H1012.3s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60zM1001 80h400000v40h-400000z"/>​

Question 9

The solution of the problem given in the previous problem is

Accepted Answer

A)    x=15−15(1−410t/15) 2x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 )  ^ { 2 }x=15−15(1−410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t/15) 2 
B)    x=15/2−15(1−410t/15) /2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 )  / 2x=15/2−15(1−410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t/15) /2 
C)    x=15/2−15(1−210t/15) 2/2x = 15 / 2 - 15 ( 1 - 2 \sqrt { 10 } t / 15 )  ^ { 2 } / 2x=15/2−15(1−210c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t/15) 2/2 
D)    x=15/2−15(1−410t/15) 2/2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 )  ^ { 2 } / 2x=15/2−15(1−410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t/15) 2/2 
E)    x=15−15(1−410t/15) x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) x=15−15(1−410c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5,-221l0 -0c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47zM834 80h400000v40h-400000z"/>​t/15)

Question 10

A beam of length $L$ is simply supported at one end and free at the other end. The weight density is constant, $\omega ( x ) = \omega _ { 0 }$ . Let $y ( x )$ represent the deflection at point $x$ . The correct form of the boundary value problem for this beam is

Accepted Answer

A)    d2ydx2=ω0/EI,y(0) =0,y′′′(L) =0\frac { d ^ { 2 } y } { d x ^ { 2 } } = \omega _ { 0 } / E I , y ( 0 )  = 0 , y ^ { \prime \prime \prime } ( L )  = 0dx2d2y​=ω0​/EI,y(0) =0,y′′′(L) =0 
B)    d4ydx4=ω0/EI,y(0) =0,y′′(0) =0,y′′(L) =0,y′′′(L) =0\frac { d ^ { 4 } y } { d x ^ { 4 } } = \omega _ { 0 } / E I , y ( 0 )  = 0 , y ^ { \prime \prime } ( 0 )  = 0 , y ^ { \prime \prime } ( L )  = 0 , y ^ { \prime \prime \prime } ( L )  = 0dx4d4y​=ω0​/EI,y(0) =0,y′′(0) =0,y′′(L) =0,y′′′(L) =0 
C)    d2ydx2=ω0EI,y(0) =0,y′′′(L) =0\frac { d ^ { 2 } y } { d x ^ { 2 } } = \omega _ { 0 } E I , y ( 0 )  = 0 , y ^ { \prime \prime \prime } ( L )  = 0dx2d2y​=ω0​EI,y(0) =0,y′′′(L) =0 
D)    d4ydx4=ω0EI,y(0) =0,y′′(0) =0,y′′(L) =0,y′′′(L) =0\frac { d ^ { 4 } y } { d x ^ { 4 } } = \omega _ { 0 } E I , y ( 0 )  = 0 , y ^ { \prime \prime } ( 0 )  = 0 , y ^ { \prime \prime } ( L )  = 0 , y ^ { \prime \prime \prime } ( L )  = 0dx4d4y​=ω0​EI,y(0) =0,y′′(0) =0,y′′(L) =0,y′′′(L) =0 
E)   none of the above

Question 11

If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is

Accepted Answer

A)    x=0.02e−2t+0.04te−2tx = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }x=0.02e−2t+0.04te−2t 
B)    x=2e−2t+4te−2tx = 2 e ^ { - 2 t } + 4 t e ^ { - 2 t }x=2e−2t+4te−2t 
C)    x=0.02e2t−0.04te2tx = 0.02 e ^ { 2 t } - 0.04 t e ^ { 2 t }x=0.02e2t−0.04te2t 
D)    x=e−2tsin⁡tx = e ^ { - 2 t } \sin tx=e−2tsint 
E)    x=0.02e−2tcos⁡tx = 0.02 e ^ { - 2 t } \cos tx=0.02e−2tcost

Question 12

The moment of inertia of a cross section of a beam is $I$ , and the Young's modulus is $E$ . Its flexural rigidity is

Accepted Answer

A)    E/IE / IE/I 
B)    EIE IEI 
C)    I/EI / EI/E 
D)    EIkE I _ { k }EIk​  , where  kkk  is the curvature
E)    k/EIk / E Ik/EI  , where  kkk  is the curvature

Question 13

In the previous problem, if the mass is set in motion, the natural frequency, $\omega$ , is

Accepted Answer

A)    2sec2 \mathrm { sec }2sec 
B)    2sec⁡−12 \sec ^ { - 1 }2sec−1 
C)    4sec4 \mathrm { sec }4sec 
D)    4sec⁡−14 \sec ^ { - 1 }4sec−1 
E)    16sec−116 \mathrm { sec } ^ { - 1 }16sec−1

Question 14

A pendulum of length $l$ hangs from the ceiling. Let $g$ represent the gravitational acceleration. The correct linearized differential equation for the angle, $\theta$ , that the swinging pendulum makes with the vertical is

Accepted Answer

A)    dθdt+gθ/l=0\frac { d 	heta } { d t } + g 	heta / l = 0dtdθ​+gθ/l=0 
B)    d2θdt2+gθ/l=0\frac { d ^ { 2 } 	heta } { d t ^ { 2 } } + g 	heta / l = 0dt2d2θ​+gθ/l=0 
C)    dθdt+lθ/g=0\frac { d 	heta } { d t } + l 	heta / g = 0dtdθ​+lθ/g=0 
D)    d2θdt2+lθ/g=0\frac { d ^ { 2 } 	heta } { d t ^ { 2 } } + l 	heta / g = 0dt2d2θ​+lθ/g=0 
E)    d2θdt2−gθ/l=0\frac { d ^ { 2 } 	heta } { d t ^ { 2 } } - g 	heta / l = 0dt2d2θ​−gθ/l=0

Question 15

The solution of a vibrating spring problem is $x = 5 \cos t - 12 \sin t$ . The amplitude is

Accepted Answer

A)   1  777 
B)    −7- 7−7 
C)   7
D)   13
E)   60

Question 16

The differential equation $\frac { d ^ { 2 } x } { d t ^ { 2 } } + x e ^ { 0.01 x } = 0$ is a model for an undamped spring-mass system with a nonlinear restoring force. The initial conditions are $x ( 0 ) = 0$ , $x ^ { \prime } ( 0 ) = 1$ . The solution of the linearized system is

Accepted Answer

A)    x=(et+e−t) /2x = \left( e ^ { t } + e ^ { - t } \right)  / 2x=(et+e−t) /2 
B)    x=(et−e−t) /2x = \left( e ^ { t } - e ^ { - t } \right)  / 2x=(et−e−t) /2 
C)    x=cos⁡tx = \cos tx=cost 
D)    x=sin⁡tx = \sin tx=sint 
E)    x=cos⁡t−sin⁡tx = \cos t - \sin tx=cost−sint

Question 17

In the previous problem, the function $x$ can be written as

Accepted Answer

A)    x=5sin⁡(t+ϕ) , where tan⁡ϕ=−5/12x = 5 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 5 / 12x=5sin(t+ϕ) , where tanϕ=−5/12 
B)    x=12sin⁡(t+ϕ) , where tan⁡ϕ=−5/12x = 12 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 5 / 12x=12sin(t+ϕ) , where tanϕ=−5/12 
C)    x=12sin⁡(t−ϕ) , where tan⁡ϕ=5/12x = 12 \sin ( t - \phi )  , 	ext { where } 	an \phi = 5 / 12x=12sin(t−ϕ) , where tanϕ=5/12 
D)    x=13sin⁡(t+ϕ) , where tan⁡ϕ=5/12x = 13 \sin ( t + \phi )  , 	ext { where } 	an \phi = 5 / 12x=13sin(t+ϕ) , where tanϕ=5/12 
E)    x=13sin⁡(t+ϕ) , where tan⁡ϕ=−5/12x = 13 \sin ( t + \phi )  , 	ext { where } 	an \phi = - 5 / 12x=13sin(t+ϕ) , where tanϕ=−5/12

Question 18

In the previous two problems, the correct differential equation for the position, $x ( t )$ , of the mass at a function of time, $t$ ,is

Accepted Answer

A)    d2xdt2+x/4=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + x / 4 = 0dt2d2x​+x/4=0 
B)    d2xdt2+2x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0dt2d2x​+2x=0 
C)    d2xdt2+4x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 x = 0dt2d2x​+4x=0 
D)    d2xdt2+8x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 8 x = 0dt2d2x​+8x=0 
E)    d2xdt2+32x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 32 x = 0dt2d2x​+32x=0

Question 19

The eigenvalue problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0$ has the solution

Accepted Answer

A)    y=sin⁡(nx) ,λ=n2,n=1,2,3,…y = \sin ( n x )  , \lambda = n ^ { 2 } , n = 1,2,3 , \ldotsy=sin(nx) ,λ=n2,n=1,2,3,… 
B)    y=cos⁡(nx) ,λ=n2,n=1,2,3,…y = \cos ( n x )  , \lambda = n ^ { 2 } , n = 1,2,3 , \ldotsy=cos(nx) ,λ=n2,n=1,2,3,… 
C)    y=sin⁡(nx) ,λ=n,n=1,2,3,…y = \sin ( n x )  , \lambda = n , n = 1,2,3 , \ldotsy=sin(nx) ,λ=n,n=1,2,3,… 
D)    y=cos⁡(nx) ,λ=n,n=1,2,3,…y = \cos ( n x )  , \lambda = n , n = 1,2,3 , \ldotsy=cos(nx) ,λ=n,n=1,2,3,… 
E)   none of the above

Question 20

A rocket is launched vertically upward with a speed $v _ { 0 }$ . Take the upward direction as positive and let the mass be m. Assume that the only force acting on the rocket is gravity, which is inversely proportional to the square of the distance from the center of the earth. Let $y ( t )$ be the distance from the center of the earth at time, $t$ . the correct differential equation for the position of the rocket is

Accepted Answer

A)    mdydt=v0−k/y2m \frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }mdtdy​=v0​−k/y2 
B)    md2ydt2=−k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }mdt2d2y​=−k/y2 
C)    dydt=v0−k/y2\frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }dtdy​=v0​−k/y2 
D)    d2ydt2=−k/y2\frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }dt2d2y​=−k/y2 
E)   none of the above

Exam 5: Modeling With Higher-Order Differential Equations

In the previous problem, if y(0) =Ry ( 0 ) = Ry(0) =R , what is the escape velocity?

If the mass in the previous problem is pulled down two feet and released, the solution for the position is

A spring attached to the ceiling is stretched one foot by a four pound weight. The value of the Hooke's Law spring constant, kkk , is

The solution of the differential equation of the previous problem is

In the previous two problems, how long does it take for the chain to fall completely to the ground?

In the previous problem, the function xxx can be written as

In the previous problem, the solution for xxx as a function of ttt is

The solution of the problem given in the previous problem is

A beam of length LLL is simply supported at one end and free at the other end. The weight density is constant, ω(x) =ω0\omega ( x ) = \omega _ { 0 }ω(x) =ω0​ . Let y(x) y ( x ) y(x) represent the deflection at point xxx . The correct form of the boundary value problem for this beam is

If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is

The moment of inertia of a cross section of a beam is III , and the Young's modulus is EEE . Its flexural rigidity is

In the previous problem, if the mass is set in motion, the natural frequency, ω\omegaω , is

A pendulum of length lll hangs from the ceiling. Let ggg represent the gravitational acceleration. The correct linearized differential equation for the angle, θ\thetaθ , that the swinging pendulum makes with the vertical is

The solution of a vibrating spring problem is x=5cos⁡t−12sin⁡tx = 5 \cos t - 12 \sin tx=5cost−12sint . The amplitude is

In the previous problem, the function xxx can be written as

In the previous two problems, the correct differential equation for the position, x(t) x ( t ) x(t) , of the mass at a function of time, ttt ,is

The eigenvalue problem y′′+λy=0,y(0) =0,y(π) =0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0y′′+λy=0,y(0) =0,y(π) =0 has the solution

In the previous problem, if $y ( 0 ) = R$ , what is the escape velocity?

A spring attached to the ceiling is stretched one foot by a four pound weight. The value of the Hooke's Law spring constant, $k$ , is

In the previous problem, the function $x$ can be written as

In the previous problem, the solution for $x$ as a function of $t$ is

A beam of length $L$ is simply supported at one end and free at the other end. The weight density is constant, $\omega ( x ) = \omega _ { 0 }$ . Let $y ( x )$ represent the deflection at point $x$ . The correct form of the boundary value problem for this beam is

The moment of inertia of a cross section of a beam is $I$ , and the Young's modulus is $E$ . Its flexural rigidity is

In the previous problem, if the mass is set in motion, the natural frequency, $\omega$ , is

A pendulum of length $l$ hangs from the ceiling. Let $g$ represent the gravitational acceleration. The correct linearized differential equation for the angle, $\theta$ , that the swinging pendulum makes with the vertical is

The solution of a vibrating spring problem is $x = 5 \cos t - 12 \sin t$ . The amplitude is

In the previous problem, the function $x$ can be written as

In the previous two problems, the correct differential equation for the position, $x ( t )$ , of the mass at a function of time, $t$ ,is

The eigenvalue problem $y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0$ has the solution