Deck 5: Modeling With Higher-Order Differential Equations

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سؤال
The eigenvalue problem y+λy=0,y(0)=0,y(π/2)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi / 2 ) = 0 has the solution

A) y=sin(nx),λ=4n2,n=1,2,3,y = \sin ( n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nx),λ=4n2,n=1,2,3,y = \cos ( n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nx),λ=2n,n=1,2,3,y = \sin ( n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
D) y=cos(nx),λ=2n,n=1,2,3,y = \cos ( n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
E) none of the above
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سؤال
A rocket is launched vertically upward with a speed v0v _ { 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)y ( t ) be the distance from the center of the earth at time, tt . the correct differential equation for the position of the rocket is

A) mdydt=v0k/y2m \frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }
B) md2ydt2=k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
C) dydt=v0k/y2\frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }
D) d2ydt2=k/y2\frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
E) none of the above
سؤال
A beam of length LL is simply supported at the left end embedded at right end. The weight density is constant, ω(x)=ω0\omega ( x ) = \omega _ { 0 } . Let y(x)y ( x ) represent the deflection at point xx . The correct form of the boundary value problem for this beam is

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 } ( 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 ( L ) = 0 , y ^ { \prime } ( 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 } ( L ) = 0
D) 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 ) = 0
E) none of the above
سؤال
The eigenvalue problem y+λy=0,y(0)=0,y(1)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 has the solution

A) y=sin(nπx),λ=(nπ)2,n=1,2,3,y = \sin ( n \pi x ) , \lambda = ( n \pi ) ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nπx),λ=(nπ)2,n=1,2,3,y = \cos ( n \pi x ) , \lambda = ( n \pi ) ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nπx),λ=nπ,n=1,2,3,y = \sin ( n \pi x ) , \lambda = n \pi , n = 1,2,3 , \ldots
D) y=cos(nπx),λ=nπ,n=1,2,3,y = \cos ( n \pi x ) , \lambda = n \pi , n = 1,2,3 , \ldots
E) none of the above
سؤال
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. The value of the Hooke's Law spring constant, kk , is

A) 1/4 meter-Newton
B) 4 meter-Newtons
C) 1/4 Newton per meter
D) 16 Newtons per meter
E) none of the above
سؤال
In the previous problem, the solution for the velocity, vv , is

A) v=v0k/y2v = v _ { 0 } - k / y ^ { 2 }
B) v=v0+k/y2v = v _ { 0 } + k / y ^ { 2 }
C) v=c+2k/(my)v = c + 2 k / ( m y )
D) v=c+2k/(my)v = \sqrt { c + 2 k / ( m y ) }
E) none of the above
سؤال
In the previous two problems, how long does it take for the chain to fall completely to the ground?

A) t=(L/g)2t = ( L / g ) ^ { 2 }
B) t=g/Lt = g / L
C) t=L/gt = L / g
D) t=L/gt = \sqrt { L / g }
E) t=g/Lt = \sqrt { g / L }
سؤال
If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is

A) x=0.02e2t+0.04te2tx = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }
B) x=2e2t+4te2tx = 2 e ^ { - 2 t } + 4 t e ^ { - 2 t }
C) x=0.02e2t0.04te2tx = 0.02 e ^ { 2 t } - 0.04 t e ^ { 2 t }
D) x=e2tsintx = e ^ { - 2 t } \sin t
E) x=0.02e2tcostx = 0.02 e ^ { - 2 t } \cos t
سؤال
In the previous problem, the function xx can be written as

A) x=5sin(t+ϕ), where tanϕ=5/12x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
B) x=12sin(t+ϕ), where tanϕ=5/12x = 12 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
C) x=12sin(tϕ), where tanϕ=5/12x = 12 \sin ( t - \phi ) , \text { where } \tan \phi = 5 / 12
D) x=13sin(t+ϕ), where tanϕ=5/12x = 13 \sin ( t + \phi ) , \text { where } \tan \phi = 5 / 12
E) x=13sin(t+ϕ), where tanϕ=5/12x = 13 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
سؤال
The initial value problem (Lx)d2xdt2(dxdt)2=Lg,x(0)=0,x(0)=0( L - x ) \frac { d ^ { 2 } x } { d t ^ { 2 } } - \left( \frac { d x } { d t } \right) ^ { 2 } = L g , x ( 0 ) = 0 , x ^ { \prime } ( 0 ) = 0 is a model of a chain of length LL falling to the ground, where x(t)x ( t ) represents the length of chain on the ground at time tt . The solution for v=dxdtv = \frac { d x } { d t } in terms of xx is

A) v=Lg(L2(Lx)2)v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) }
B) v=(Lg(L2+(Lx)2))/(Lx)v = \left( \operatorname { Lg } \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) \right) / ( L - x )
C) v=(Lg(L2(Lx)2))/(Lx)v = \left( \operatorname { Lg } \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) \right) / ( L - x )
D) v=Lg(L2+(Lx)2)/(Lx)v = \sqrt { L g \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) } / ( L - x )
E) v=Lg(L2(Lx)2)/(Lx)v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) } / ( L - x )
سؤال
In the previous problem, if the mass is set in motion, the natural frequency, ω\omega , is

A) 2sec2 \mathrm { sec }
B) 2sec12 \sec ^ { - 1 }
C) 4sec4 \mathrm { sec }
D) 4sec14 \sec ^ { - 1 }
E) 16sec116 \mathrm { sec } ^ { - 1 }
سؤال
The differential equation d2xdt2+xe0.01x=0\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)=0x ( 0 ) = 0 , x(0)=1x ^ { \prime } ( 0 ) = 1 . The solution of the linearized system is

A) x=(et+et)/2x = \left( e ^ { t } + e ^ { - t } \right) / 2
B) x=(etet)/2x = \left( e ^ { t } - e ^ { - t } \right) / 2
C) x=costx = \cos t
D) x=sintx = \sin t
E) x=costsintx = \cos t - \sin t
سؤال
In the previous problem, the solution for xx as a function of tt is

A) x=LL2Lgtx = L - \sqrt { L ^ { 2 } - L g t }
B) x=LL2+Lgtx = L - \sqrt { L ^ { 2 } + L g t }
C) x=LL2Lgt2x = L - \sqrt { L ^ { 2 } - L g t ^ { 2 } }
D) x=LL2+Lgt2x = L - \sqrt { L ^ { 2 } + L g t ^ { 2 } }
E) x=LLgt2L2x = L - \sqrt { L g t ^ { 2 } - L ^ { 2 } }
سؤال
The solution of the boundary value problem in the previous problem is

A) y=ω0/EI(L3x/48Lx3/16+x4/24)y = \omega _ { 0 } / E I \left( L ^ { 3 } x / 48 - L x ^ { 3 } / 16 + x ^ { 4 } / 24 \right)
B) y=ω0/EI(x2/2Lx)y = \omega _ { 0 } / E I \left( x ^ { 2 } / 2 - L x \right)
C) y=ω0EI(L3x/48Lx3/16+x4/24)y = \omega _ { 0 } E I \left( L ^ { 3 } x / 48 - L x ^ { 3 } / 16 + x ^ { 4 } / 24 \right)
D) y=ω0EI(x2/2Lx)y = \omega _ { 0 } E I \left( x ^ { 2 } / 2 - L x \right)
E) none of the above
سؤال
A pendulum of length 16 feet hangs from the ceiling. Let g=32g = 32 represent the gravitational acceleration. The correct linearized differential equation for the angle, θ\theta , that the swinging pendulum makes with the vertical is

A) dθdt+2θ=0\frac { d \theta } { d t } + 2 \theta = 0
B) d2θdt2+2θ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + 2 \theta = 0
C) dθdt+θ/2=0\frac { d \theta } { d t } + \theta / 2 = 0
D) d2θdt2+θ/2=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \theta / 2 = 0
E) d2θdt22θ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } - 2 \theta = 0
سؤال
The boundary value problem rd2udr2+2dudr=0,u(a)=u0,u(b)=u1r \frac { d ^ { 2 } u } { d r ^ { 2 } } + 2 \frac { d u } { d r } = 0 , u ( a ) = u _ { 0 } , u ( b ) = u _ { 1 } is a model for the temperature distribution between two concentric spheres of radii aa and bb , with a<ba < b .The solution of this problem is

A) u=c2+c1/r, where c1=ab(u1u0)/(ba) and c2=(u1bu0a)/(ba)u = c _ { 2 } + c _ { 1 } / r \text {, where } c _ { 1 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a ) \text { and } c _ { 2 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a )
B) u=c2+c1/r, where c1=(u1bu0a)/(ba) and c2=ab(u1u0)/(ba)u = c _ { 2 } + c _ { 1 } / r \text {, where } c _ { 1 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a )
C) u=c2c1/r, where c1=ab(u1u0)/(ba) and c2=ab(u1bu0a)/(ba)u = c _ { 2 } - c _ { 1 } / r , \text { where } c _ { 1 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a )
D) u=c2c1/r, where c1=(u1bu0a)/(ba) and c2=ab(u1u0)/(ba)u = c _ { 2 } - c _ { 1 } / r \text {, where } c _ { 1 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a )
E) none of the above
سؤال
In the previous two problems, if the mass is set into motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position, x(t)x ( t ) , of the mass at a function of time, tt , is

A) d2xdt2+4dxdt+x/4=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + x / 4 = 0
B) d2xdt2+4dxdt+2x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 2 x = 0
C) d2xdt2+4dxdt+4x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 4 x = 0
D) d2xdt2+4dxdt+8x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 8 x = 0
E) d2xdt2+4dxdt+32x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 32 x = 0
سؤال
The solution of a vibrating spring problem is x=5cost12sintx = 5 \cos t - 12 \sin t . The amplitude is

A) 1 77
B) 7- 7
C) 7
D) 13
E) 60
سؤال
In the previous problem, if y(0)=Ry ( 0 ) = R , the escape velocity is

A) v0=2k/(mR)v _ { 0 } = 2 k / ( m R )
B) v0=2R/(mk)v _ { 0 } = 2 R / ( m k )
C) v0=2m/(kR)v _ { 0 } = 2 m / ( k R )
D) v0=2m/(kR)v _ { 0 } = \sqrt { 2 m / ( k R ) }
E) none of the above
سؤال
The solution of the differential equation of the previous problem is

A) θ=ce2t\theta = c e ^ { 2 t }
B) θ=ce2t\theta = c e ^ { - 2 t }
C) θ=cet/2\theta = c e ^ { - t / 2 }
D) θ=c1cos(2t)+c2sin(2t)\theta = c _ { 1 } \cos ( 2 t ) + c _ { 2 } \sin ( 2 t )
E) θ=c1cos(2t)+c2sin(2t)\theta = c _ { 1 } \cos ( \sqrt { 2 } t ) + c _ { 2 } \sin ( \sqrt { 2 } t )
سؤال
The boundary value problem Td2ydx2+ρω2=0,y(0)=0,y(L)=0T \frac { d ^ { 2 } y } { d x ^ { 2 } } + \rho \omega ^ { 2 } = 0 , y ( 0 ) = 0 , y ( L ) = 0 is a model of the shape of a rotating string. Suppose TT and ρ\rho are constants. The critical angular rotation speed ω=ωn\omega = \omega _ { n } , for which there exist non-trivial solutions are

A) ωn=(T/ρ)(nπL)2\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { L } \right) ^ { 2 }
B) ωn=T/ρnπL\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { L }
C) ωn=T/ρnπ2L\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { 2 L }
D) ωn=(T/ρ)(nπ2L)2\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { 2 L } \right) ^ { 2 }
E) ωn=ρ/TnπL\omega _ { n } = \sqrt { \rho / T } \frac { n \pi } { L }
سؤال
The moment of inertia of a cross section of a beam is II , and the Young's modulus is EE . Its flexural rigidity is

A) E/IE / I
B) EIE I
C) I/EI / E
D) EIkE I _ { k } , where kk is the curvature
E) k/EIk / E I , where kk is the curvature
سؤال
In the previous problem, the solution for the velocity, vv , is

A) v=k/y2v = - k / y ^ { 2 }
B) v=k/y2v = k / y ^ { 2 }
C) v=c+2k/(my)v = \sqrt { c + 2 k / ( m y ) }
D) v=c+2k/(my)v = c + 2 k / ( m y )
E) none of the above
سؤال
The eigenvalue problem y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 has the solution

A) y=sin(nx),λ=n2,n=1,2,3,y = \sin ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nx),λ=n2,n=1,2,3,y = \cos ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nx),λ=n,n=1,2,3,y = \sin ( n x ) , \lambda = n , n = 1,2,3 , \ldots
D) y=cos(nx),λ=n,n=1,2,3,y = \cos ( n x ) , \lambda = n , n = 1,2,3 , \ldots
E) none of the above
سؤال
The solution of the differential equation of the previous problem is

A) θ=cegt/l\theta = c e ^ { - \mathrm { g } t / l }
B) θ=celt/g\theta = c e ^ { - l t / g }
C) θ=cegt/l\theta = c e ^ { g t / l }
D) θ=c1cos(gt/l)+c2sin(gt/l)\theta = c _ { 1 } \cos ( g t / l ) + c _ { 2 } \sin ( g t / l )
E) θ=c1cos(g/lt)+c2sin(g/lt)\theta = c _ { 1 } \cos ( \sqrt { g / l } t ) + c _ { 2 } \sin ( \sqrt { g / l } t )
سؤال
A spring attached to the ceiling is stretched one foot by a four pound weight. The value of the Hooke's Law spring constant, kk , is

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
سؤال
The solution of the problem given in the previous problem is

A) x=1515(1410t/15)2x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) ^ { 2 }
B) x=15/215(1410t/15)/2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) / 2
C) x=15/215(1210t/15)2/2x = 15 / 2 - 15 ( 1 - 2 \sqrt { 10 } t / 15 ) ^ { 2 } / 2
D) x=15/215(1410t/15)2/2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) ^ { 2 } / 2
E) x=1515(1410t/15)x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 )
سؤال
In the previous problem the corresponding non-trivial solutions for yy are

A) y=sin(ωnx)y = \sin \left( \omega _ { n } x \right)
B) y=cos(ωnx)y = \cos \left( \omega _ { n } x \right)
C) y=sin(ωn2x)y = \sin \left( \omega _ { n } ^ { 2 } x \right)
D) y=cos(ωn2x)y = \cos \left( \omega _ { n } ^ { 2 } x \right)
E) none of the above
سؤال
A pendulum of length ll hangs from the ceiling. Let gg represent the gravitational acceleration. The correct linearized differential equation for the angle, θ\theta , that the swinging pendulum makes with the vertical is

A) dθdt+gθ/l=0\frac { d \theta } { d t } + g \theta / l = 0
B) d2θdt2+gθ/l=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + g \theta / l = 0
C) dθdt+lθ/g=0\frac { d \theta } { d t } + l \theta / g = 0
D) d2θdt2+lθ/g=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + l \theta / g = 0
E) d2θdt2gθ/l=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } - g \theta / l = 0
سؤال
A beam of length LL is simply supported at one end and free at the other end. The weight density is constant, ω(x)=ω0\omega ( x ) = \omega _ { 0 } . Let y(x)y ( x ) represent the deflection at point xx . The correct form of the boundary value problem for this beam is

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 ) = 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 ) = 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 ) = 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 ) = 0
E) none of the above
سؤال
The differential equation d2xdt2+xcosx=0\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)=0.1x ( 0 ) = 0.1 , x(0)=0.1x ^ { \prime } ( 0 ) = - 0.1 . The solution of the linearized system is

A) x=0.1costx = 0.1 \cos t
B) x=0.1sintx = 0.1 \sin t
C) x=0.1cost0.1sintx = 0.1 \cos t - 0.1 \sin t
D) x=0.1etx = 0.1 e ^ { - t }
E) x=0.1etx = 0.1 e ^ { t }
سؤال
In the previous problem, if the mass is set in motion, the natural frequency, ω\omega ,is

A) 42sec4 \sqrt { 2 } \mathrm { sec }
B) 42sec14 \sqrt { 2 } \sec ^ { - 1 }
C) 32sec32 \mathrm { sec }
D) 32sec132 \mathrm { sec } ^ { - 1 }
E) sec1\sec ^ { - 1 }
سؤال
The solution of a vibrating spring problem is x=4cost3sintx = 4 \cos t - 3 \sin t . The amplitude is

A) 77
B) 11
C) 2525
D) 55
E) 1- 1
سؤال
The eigenvalue problem y+λy=0,y(0)=0,y(π/2)=0y ^ { \prime \prime } + \lambda y = 0 , y ^ { \prime } ( 0 ) = 0 , y ^ { \prime } ( \pi / 2 ) = 0 has the solution

A) y=sin(2nx),λ=4n2,n=1,2,3,y = \sin ( 2 n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(2nx),λ=4n2,n=1,2,3,y = \cos ( 2 n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(2nx),λ=2n,n=1,2,3,y = \sin ( 2 n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
D) y=cos(2nx),λ=2n,n=1,2,3,y = \cos ( 2 n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
E) none of the above
سؤال
In the previous problem, the function xx can be written as

A) x=3sin(t+ϕ), where tanϕ=4/3x = 3 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
B) x=4sin(t+ϕ), where tanϕ=4/3x = 4 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
C) x=5sin(t+ϕ), where tanϕ=4/3x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
D) x=5sin(t+ϕ), where tanϕ=3/4x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 3 / 4
E) x=4sin(t+ϕ), where tanϕ=3/4x = 4 \sin ( t + \phi ) , \text { where } \tan \phi = - 3 / 4
سؤال
In the previous two problems, the correct differential equation for the position, x(t)x ( t ) , of the mass at a function of time, tt ,is

A) d2xdt2+x/4=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + x / 4 = 0
B) d2xdt2+2x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0
C) d2xdt2+4x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 x = 0
D) d2xdt2+8x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 8 x = 0
E) d2xdt2+32x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 32 x = 0
سؤال
A rocket with mass mm is launched vertically upward from the surface of the earth with a velocity v0v _ { 0 } . Let y(t)y ( t ) be the distance of the rocket from the center of the earth at time tt . Assuming 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, the correct differential equation for the position of the rocket is

A) dydt=k/y2\frac { d y } { d t } = - k / y ^ { 2 }
B) d2ydt2=k/y2\frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
C) mdydt=k/y2m \frac { d y } { d t } = - k / y ^ { 2 }
D) md2ydt2=k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
E) none of the above
سؤال
In the previous problem, if y(0)=Ry ( 0 ) = R , what is the escape velocity?

A) v0=2k/(mR)v _ { 0 } = \sqrt { 2 k / ( m R ) }
B) v0=2k/(mR)v _ { 0 } = 2 k / ( m R )
C) v0=2m/(kR)v _ { 0 } = 2 m / ( k R )
D) v0=2m/(kR)v _ { 0 } = \sqrt { 2 m / ( k R ) }
E) none of the above
سؤال
If the mass in the previous problem is pulled down two feet and released, the solution for the position is

A) x=2cos(42t)+2sin(42t)x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )
B) x=2sin(42t)x = 2 \sin ( 4 \sqrt { 2 } t )
C) x=2cos(42t)x = 2 \cos ( 4 \sqrt { 2 } t )
D) x=2sin(4t)x = 2 \sin ( 4 t )
E) x=2cos(4t)x = 2 \cos ( 4 t )
سؤال
A 10 foot chain of weight density 2 pounds per foot is coiled on the ground. One end is pulled upward by a force of 10 pounds. The correct differential equation for the height, x(t)x ( t ) , of the end of the chain above the ground at time tt is

A) xdxdtd2xdt2+dx2dt+32x=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
B) dxdtd2xdt2+dx2dt+32x=160\frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
C) xd2xdt2+dx2dt+32x=160x \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
D) xdxdtd2xdt2+dxdt+32x=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x } { d t } + 32 x = 160
E) xdxdtd2xdt2+dx2dt=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } = 160
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Deck 5: Modeling With Higher-Order Differential Equations
1
The eigenvalue problem y+λy=0,y(0)=0,y(π/2)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi / 2 ) = 0 has the solution

A) y=sin(nx),λ=4n2,n=1,2,3,y = \sin ( n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nx),λ=4n2,n=1,2,3,y = \cos ( n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nx),λ=2n,n=1,2,3,y = \sin ( n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
D) y=cos(nx),λ=2n,n=1,2,3,y = \cos ( n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
E) none of the above
none of the above
2
A rocket is launched vertically upward with a speed v0v _ { 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)y ( t ) be the distance from the center of the earth at time, tt . the correct differential equation for the position of the rocket is

A) mdydt=v0k/y2m \frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }
B) md2ydt2=k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
C) dydt=v0k/y2\frac { d y } { d t } = v _ { 0 } - k / y ^ { 2 }
D) d2ydt2=k/y2\frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
E) none of the above
md2ydt2=k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
3
A beam of length LL is simply supported at the left end embedded at right end. The weight density is constant, ω(x)=ω0\omega ( x ) = \omega _ { 0 } . Let y(x)y ( x ) represent the deflection at point xx . The correct form of the boundary value problem for this beam is

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 } ( 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 ( L ) = 0 , y ^ { \prime } ( 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 } ( L ) = 0
D) 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 ) = 0
E) none of the above
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 ( L ) = 0 , y ^ { \prime } ( L ) = 0
4
The eigenvalue problem y+λy=0,y(0)=0,y(1)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( 1 ) = 0 has the solution

A) y=sin(nπx),λ=(nπ)2,n=1,2,3,y = \sin ( n \pi x ) , \lambda = ( n \pi ) ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nπx),λ=(nπ)2,n=1,2,3,y = \cos ( n \pi x ) , \lambda = ( n \pi ) ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nπx),λ=nπ,n=1,2,3,y = \sin ( n \pi x ) , \lambda = n \pi , n = 1,2,3 , \ldots
D) y=cos(nπx),λ=nπ,n=1,2,3,y = \cos ( n \pi x ) , \lambda = n \pi , n = 1,2,3 , \ldots
E) none of the above
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5
A spring attached to the ceiling is stretched 2.45 meters by a four kilogram mass. The value of the Hooke's Law spring constant, kk , is

A) 1/4 meter-Newton
B) 4 meter-Newtons
C) 1/4 Newton per meter
D) 16 Newtons per meter
E) none of the above
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6
In the previous problem, the solution for the velocity, vv , is

A) v=v0k/y2v = v _ { 0 } - k / y ^ { 2 }
B) v=v0+k/y2v = v _ { 0 } + k / y ^ { 2 }
C) v=c+2k/(my)v = c + 2 k / ( m y )
D) v=c+2k/(my)v = \sqrt { c + 2 k / ( m y ) }
E) none of the above
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7
In the previous two problems, how long does it take for the chain to fall completely to the ground?

A) t=(L/g)2t = ( L / g ) ^ { 2 }
B) t=g/Lt = g / L
C) t=L/gt = L / g
D) t=L/gt = \sqrt { L / g }
E) t=g/Lt = \sqrt { g / L }
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8
If the mass in the previous problem is pulled down two centimeters and released, the solution for the position is

A) x=0.02e2t+0.04te2tx = 0.02 e ^ { - 2 t } + 0.04 t e ^ { - 2 t }
B) x=2e2t+4te2tx = 2 e ^ { - 2 t } + 4 t e ^ { - 2 t }
C) x=0.02e2t0.04te2tx = 0.02 e ^ { 2 t } - 0.04 t e ^ { 2 t }
D) x=e2tsintx = e ^ { - 2 t } \sin t
E) x=0.02e2tcostx = 0.02 e ^ { - 2 t } \cos t
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9
In the previous problem, the function xx can be written as

A) x=5sin(t+ϕ), where tanϕ=5/12x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
B) x=12sin(t+ϕ), where tanϕ=5/12x = 12 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
C) x=12sin(tϕ), where tanϕ=5/12x = 12 \sin ( t - \phi ) , \text { where } \tan \phi = 5 / 12
D) x=13sin(t+ϕ), where tanϕ=5/12x = 13 \sin ( t + \phi ) , \text { where } \tan \phi = 5 / 12
E) x=13sin(t+ϕ), where tanϕ=5/12x = 13 \sin ( t + \phi ) , \text { where } \tan \phi = - 5 / 12
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10
The initial value problem (Lx)d2xdt2(dxdt)2=Lg,x(0)=0,x(0)=0( L - x ) \frac { d ^ { 2 } x } { d t ^ { 2 } } - \left( \frac { d x } { d t } \right) ^ { 2 } = L g , x ( 0 ) = 0 , x ^ { \prime } ( 0 ) = 0 is a model of a chain of length LL falling to the ground, where x(t)x ( t ) represents the length of chain on the ground at time tt . The solution for v=dxdtv = \frac { d x } { d t } in terms of xx is

A) v=Lg(L2(Lx)2)v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) }
B) v=(Lg(L2+(Lx)2))/(Lx)v = \left( \operatorname { Lg } \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) \right) / ( L - x )
C) v=(Lg(L2(Lx)2))/(Lx)v = \left( \operatorname { Lg } \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) \right) / ( L - x )
D) v=Lg(L2+(Lx)2)/(Lx)v = \sqrt { L g \left( L ^ { 2 } + ( L - x ) ^ { 2 } \right) } / ( L - x )
E) v=Lg(L2(Lx)2)/(Lx)v = \sqrt { L g \left( L ^ { 2 } - ( L - x ) ^ { 2 } \right) } / ( L - x )
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11
In the previous problem, if the mass is set in motion, the natural frequency, ω\omega , is

A) 2sec2 \mathrm { sec }
B) 2sec12 \sec ^ { - 1 }
C) 4sec4 \mathrm { sec }
D) 4sec14 \sec ^ { - 1 }
E) 16sec116 \mathrm { sec } ^ { - 1 }
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12
The differential equation d2xdt2+xe0.01x=0\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)=0x ( 0 ) = 0 , x(0)=1x ^ { \prime } ( 0 ) = 1 . The solution of the linearized system is

A) x=(et+et)/2x = \left( e ^ { t } + e ^ { - t } \right) / 2
B) x=(etet)/2x = \left( e ^ { t } - e ^ { - t } \right) / 2
C) x=costx = \cos t
D) x=sintx = \sin t
E) x=costsintx = \cos t - \sin t
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13
In the previous problem, the solution for xx as a function of tt is

A) x=LL2Lgtx = L - \sqrt { L ^ { 2 } - L g t }
B) x=LL2+Lgtx = L - \sqrt { L ^ { 2 } + L g t }
C) x=LL2Lgt2x = L - \sqrt { L ^ { 2 } - L g t ^ { 2 } }
D) x=LL2+Lgt2x = L - \sqrt { L ^ { 2 } + L g t ^ { 2 } }
E) x=LLgt2L2x = L - \sqrt { L g t ^ { 2 } - L ^ { 2 } }
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14
The solution of the boundary value problem in the previous problem is

A) y=ω0/EI(L3x/48Lx3/16+x4/24)y = \omega _ { 0 } / E I \left( L ^ { 3 } x / 48 - L x ^ { 3 } / 16 + x ^ { 4 } / 24 \right)
B) y=ω0/EI(x2/2Lx)y = \omega _ { 0 } / E I \left( x ^ { 2 } / 2 - L x \right)
C) y=ω0EI(L3x/48Lx3/16+x4/24)y = \omega _ { 0 } E I \left( L ^ { 3 } x / 48 - L x ^ { 3 } / 16 + x ^ { 4 } / 24 \right)
D) y=ω0EI(x2/2Lx)y = \omega _ { 0 } E I \left( x ^ { 2 } / 2 - L x \right)
E) none of the above
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15
A pendulum of length 16 feet hangs from the ceiling. Let g=32g = 32 represent the gravitational acceleration. The correct linearized differential equation for the angle, θ\theta , that the swinging pendulum makes with the vertical is

A) dθdt+2θ=0\frac { d \theta } { d t } + 2 \theta = 0
B) d2θdt2+2θ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + 2 \theta = 0
C) dθdt+θ/2=0\frac { d \theta } { d t } + \theta / 2 = 0
D) d2θdt2+θ/2=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + \theta / 2 = 0
E) d2θdt22θ=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } - 2 \theta = 0
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16
The boundary value problem rd2udr2+2dudr=0,u(a)=u0,u(b)=u1r \frac { d ^ { 2 } u } { d r ^ { 2 } } + 2 \frac { d u } { d r } = 0 , u ( a ) = u _ { 0 } , u ( b ) = u _ { 1 } is a model for the temperature distribution between two concentric spheres of radii aa and bb , with a<ba < b .The solution of this problem is

A) u=c2+c1/r, where c1=ab(u1u0)/(ba) and c2=(u1bu0a)/(ba)u = c _ { 2 } + c _ { 1 } / r \text {, where } c _ { 1 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a ) \text { and } c _ { 2 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a )
B) u=c2+c1/r, where c1=(u1bu0a)/(ba) and c2=ab(u1u0)/(ba)u = c _ { 2 } + c _ { 1 } / r \text {, where } c _ { 1 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a )
C) u=c2c1/r, where c1=ab(u1u0)/(ba) and c2=ab(u1bu0a)/(ba)u = c _ { 2 } - c _ { 1 } / r , \text { where } c _ { 1 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a )
D) u=c2c1/r, where c1=(u1bu0a)/(ba) and c2=ab(u1u0)/(ba)u = c _ { 2 } - c _ { 1 } / r \text {, where } c _ { 1 } = \left( u _ { 1 } b - u _ { 0 } a \right) / ( b - a ) \text { and } c _ { 2 } = a b \left( u _ { 1 } - u _ { 0 } \right) / ( b - a )
E) none of the above
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17
In the previous two problems, if the mass is set into motion in a medium that imparts a damping force numerically equal to 16 times the velocity, the correct differential equation for the position, x(t)x ( t ) , of the mass at a function of time, tt , is

A) d2xdt2+4dxdt+x/4=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + x / 4 = 0
B) d2xdt2+4dxdt+2x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 2 x = 0
C) d2xdt2+4dxdt+4x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 4 x = 0
D) d2xdt2+4dxdt+8x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 8 x = 0
E) d2xdt2+4dxdt+32x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 \frac { d x } { d t } + 32 x = 0
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18
The solution of a vibrating spring problem is x=5cost12sintx = 5 \cos t - 12 \sin t . The amplitude is

A) 1 77
B) 7- 7
C) 7
D) 13
E) 60
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19
In the previous problem, if y(0)=Ry ( 0 ) = R , the escape velocity is

A) v0=2k/(mR)v _ { 0 } = 2 k / ( m R )
B) v0=2R/(mk)v _ { 0 } = 2 R / ( m k )
C) v0=2m/(kR)v _ { 0 } = 2 m / ( k R )
D) v0=2m/(kR)v _ { 0 } = \sqrt { 2 m / ( k R ) }
E) none of the above
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20
The solution of the differential equation of the previous problem is

A) θ=ce2t\theta = c e ^ { 2 t }
B) θ=ce2t\theta = c e ^ { - 2 t }
C) θ=cet/2\theta = c e ^ { - t / 2 }
D) θ=c1cos(2t)+c2sin(2t)\theta = c _ { 1 } \cos ( 2 t ) + c _ { 2 } \sin ( 2 t )
E) θ=c1cos(2t)+c2sin(2t)\theta = c _ { 1 } \cos ( \sqrt { 2 } t ) + c _ { 2 } \sin ( \sqrt { 2 } t )
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21
The boundary value problem Td2ydx2+ρω2=0,y(0)=0,y(L)=0T \frac { d ^ { 2 } y } { d x ^ { 2 } } + \rho \omega ^ { 2 } = 0 , y ( 0 ) = 0 , y ( L ) = 0 is a model of the shape of a rotating string. Suppose TT and ρ\rho are constants. The critical angular rotation speed ω=ωn\omega = \omega _ { n } , for which there exist non-trivial solutions are

A) ωn=(T/ρ)(nπL)2\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { L } \right) ^ { 2 }
B) ωn=T/ρnπL\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { L }
C) ωn=T/ρnπ2L\omega _ { n } = \sqrt { T / \rho } \frac { n \pi } { 2 L }
D) ωn=(T/ρ)(nπ2L)2\omega _ { n } = ( T / \rho ) \left( \frac { n \pi } { 2 L } \right) ^ { 2 }
E) ωn=ρ/TnπL\omega _ { n } = \sqrt { \rho / T } \frac { n \pi } { L }
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22
The moment of inertia of a cross section of a beam is II , and the Young's modulus is EE . Its flexural rigidity is

A) E/IE / I
B) EIE I
C) I/EI / E
D) EIkE I _ { k } , where kk is the curvature
E) k/EIk / E I , where kk is the curvature
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23
In the previous problem, the solution for the velocity, vv , is

A) v=k/y2v = - k / y ^ { 2 }
B) v=k/y2v = k / y ^ { 2 }
C) v=c+2k/(my)v = \sqrt { c + 2 k / ( m y ) }
D) v=c+2k/(my)v = c + 2 k / ( m y )
E) none of the above
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24
The eigenvalue problem y+λy=0,y(0)=0,y(π)=0y ^ { \prime \prime } + \lambda y = 0 , y ( 0 ) = 0 , y ( \pi ) = 0 has the solution

A) y=sin(nx),λ=n2,n=1,2,3,y = \sin ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(nx),λ=n2,n=1,2,3,y = \cos ( n x ) , \lambda = n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(nx),λ=n,n=1,2,3,y = \sin ( n x ) , \lambda = n , n = 1,2,3 , \ldots
D) y=cos(nx),λ=n,n=1,2,3,y = \cos ( n x ) , \lambda = n , n = 1,2,3 , \ldots
E) none of the above
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25
The solution of the differential equation of the previous problem is

A) θ=cegt/l\theta = c e ^ { - \mathrm { g } t / l }
B) θ=celt/g\theta = c e ^ { - l t / g }
C) θ=cegt/l\theta = c e ^ { g t / l }
D) θ=c1cos(gt/l)+c2sin(gt/l)\theta = c _ { 1 } \cos ( g t / l ) + c _ { 2 } \sin ( g t / l )
E) θ=c1cos(g/lt)+c2sin(g/lt)\theta = c _ { 1 } \cos ( \sqrt { g / l } t ) + c _ { 2 } \sin ( \sqrt { g / l } t )
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26
A spring attached to the ceiling is stretched one foot by a four pound weight. The value of the Hooke's Law spring constant, kk , is

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
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27
The solution of the problem given in the previous problem is

A) x=1515(1410t/15)2x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) ^ { 2 }
B) x=15/215(1410t/15)/2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) / 2
C) x=15/215(1210t/15)2/2x = 15 / 2 - 15 ( 1 - 2 \sqrt { 10 } t / 15 ) ^ { 2 } / 2
D) x=15/215(1410t/15)2/2x = 15 / 2 - 15 ( 1 - 4 \sqrt { 10 } t / 15 ) ^ { 2 } / 2
E) x=1515(1410t/15)x = 15 - 15 ( 1 - 4 \sqrt { 10 } t / 15 )
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28
In the previous problem the corresponding non-trivial solutions for yy are

A) y=sin(ωnx)y = \sin \left( \omega _ { n } x \right)
B) y=cos(ωnx)y = \cos \left( \omega _ { n } x \right)
C) y=sin(ωn2x)y = \sin \left( \omega _ { n } ^ { 2 } x \right)
D) y=cos(ωn2x)y = \cos \left( \omega _ { n } ^ { 2 } x \right)
E) none of the above
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29
A pendulum of length ll hangs from the ceiling. Let gg represent the gravitational acceleration. The correct linearized differential equation for the angle, θ\theta , that the swinging pendulum makes with the vertical is

A) dθdt+gθ/l=0\frac { d \theta } { d t } + g \theta / l = 0
B) d2θdt2+gθ/l=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + g \theta / l = 0
C) dθdt+lθ/g=0\frac { d \theta } { d t } + l \theta / g = 0
D) d2θdt2+lθ/g=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } + l \theta / g = 0
E) d2θdt2gθ/l=0\frac { d ^ { 2 } \theta } { d t ^ { 2 } } - g \theta / l = 0
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30
A beam of length LL is simply supported at one end and free at the other end. The weight density is constant, ω(x)=ω0\omega ( x ) = \omega _ { 0 } . Let y(x)y ( x ) represent the deflection at point xx . The correct form of the boundary value problem for this beam is

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 ) = 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 ) = 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 ) = 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 ) = 0
E) none of the above
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31
The differential equation d2xdt2+xcosx=0\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)=0.1x ( 0 ) = 0.1 , x(0)=0.1x ^ { \prime } ( 0 ) = - 0.1 . The solution of the linearized system is

A) x=0.1costx = 0.1 \cos t
B) x=0.1sintx = 0.1 \sin t
C) x=0.1cost0.1sintx = 0.1 \cos t - 0.1 \sin t
D) x=0.1etx = 0.1 e ^ { - t }
E) x=0.1etx = 0.1 e ^ { t }
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32
In the previous problem, if the mass is set in motion, the natural frequency, ω\omega ,is

A) 42sec4 \sqrt { 2 } \mathrm { sec }
B) 42sec14 \sqrt { 2 } \sec ^ { - 1 }
C) 32sec32 \mathrm { sec }
D) 32sec132 \mathrm { sec } ^ { - 1 }
E) sec1\sec ^ { - 1 }
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33
The solution of a vibrating spring problem is x=4cost3sintx = 4 \cos t - 3 \sin t . The amplitude is

A) 77
B) 11
C) 2525
D) 55
E) 1- 1
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34
The eigenvalue problem y+λy=0,y(0)=0,y(π/2)=0y ^ { \prime \prime } + \lambda y = 0 , y ^ { \prime } ( 0 ) = 0 , y ^ { \prime } ( \pi / 2 ) = 0 has the solution

A) y=sin(2nx),λ=4n2,n=1,2,3,y = \sin ( 2 n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
B) y=cos(2nx),λ=4n2,n=1,2,3,y = \cos ( 2 n x ) , \lambda = 4 n ^ { 2 } , n = 1,2,3 , \ldots
C) y=sin(2nx),λ=2n,n=1,2,3,y = \sin ( 2 n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
D) y=cos(2nx),λ=2n,n=1,2,3,y = \cos ( 2 n x ) , \lambda = 2 n , n = 1,2,3 , \ldots
E) none of the above
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35
In the previous problem, the function xx can be written as

A) x=3sin(t+ϕ), where tanϕ=4/3x = 3 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
B) x=4sin(t+ϕ), where tanϕ=4/3x = 4 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
C) x=5sin(t+ϕ), where tanϕ=4/3x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 4 / 3
D) x=5sin(t+ϕ), where tanϕ=3/4x = 5 \sin ( t + \phi ) , \text { where } \tan \phi = - 3 / 4
E) x=4sin(t+ϕ), where tanϕ=3/4x = 4 \sin ( t + \phi ) , \text { where } \tan \phi = - 3 / 4
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36
In the previous two problems, the correct differential equation for the position, x(t)x ( t ) , of the mass at a function of time, tt ,is

A) d2xdt2+x/4=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + x / 4 = 0
B) d2xdt2+2x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 2 x = 0
C) d2xdt2+4x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 4 x = 0
D) d2xdt2+8x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 8 x = 0
E) d2xdt2+32x=0\frac { d ^ { 2 } x } { d t ^ { 2 } } + 32 x = 0
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37
A rocket with mass mm is launched vertically upward from the surface of the earth with a velocity v0v _ { 0 } . Let y(t)y ( t ) be the distance of the rocket from the center of the earth at time tt . Assuming 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, the correct differential equation for the position of the rocket is

A) dydt=k/y2\frac { d y } { d t } = - k / y ^ { 2 }
B) d2ydt2=k/y2\frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
C) mdydt=k/y2m \frac { d y } { d t } = - k / y ^ { 2 }
D) md2ydt2=k/y2m \frac { d ^ { 2 } y } { d t ^ { 2 } } = - k / y ^ { 2 }
E) none of the above
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38
In the previous problem, if y(0)=Ry ( 0 ) = R , what is the escape velocity?

A) v0=2k/(mR)v _ { 0 } = \sqrt { 2 k / ( m R ) }
B) v0=2k/(mR)v _ { 0 } = 2 k / ( m R )
C) v0=2m/(kR)v _ { 0 } = 2 m / ( k R )
D) v0=2m/(kR)v _ { 0 } = \sqrt { 2 m / ( k R ) }
E) none of the above
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39
If the mass in the previous problem is pulled down two feet and released, the solution for the position is

A) x=2cos(42t)+2sin(42t)x = 2 \cos ( 4 \sqrt { 2 } t ) + 2 \sin ( 4 \sqrt { 2 } t )
B) x=2sin(42t)x = 2 \sin ( 4 \sqrt { 2 } t )
C) x=2cos(42t)x = 2 \cos ( 4 \sqrt { 2 } t )
D) x=2sin(4t)x = 2 \sin ( 4 t )
E) x=2cos(4t)x = 2 \cos ( 4 t )
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40
A 10 foot chain of weight density 2 pounds per foot is coiled on the ground. One end is pulled upward by a force of 10 pounds. The correct differential equation for the height, x(t)x ( t ) , of the end of the chain above the ground at time tt is

A) xdxdtd2xdt2+dx2dt+32x=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
B) dxdtd2xdt2+dx2dt+32x=160\frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
C) xd2xdt2+dx2dt+32x=160x \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } + 32 x = 160
D) xdxdtd2xdt2+dxdt+32x=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x } { d t } + 32 x = 160
E) xdxdtd2xdt2+dx2dt=160x \frac { d x } { d t } \frac { d ^ { 2 } x } { d t ^ { 2 } } + \frac { d x ^ { 2 } } { d t } = 160
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