Calculus IV - HW 4. Due 7/20

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1 Calculus IV - HW 4 Due 7/ Section.7. (Prolems, from B&D) For each of the following, determine ω, R, and δ to write the given expression in the form u = R cos(ω t δ). (a) u = cos(t) + sin(t) () u = 4 cos(t) sin(t) Solution: (a) We can immediately see that ω =. We showed in class that R = A + B = ( ) + ( ) = + =. Finally, tan(δ) = B A =. Thus δ = π or δ = π. However, we want that cos(δ) = and sin(δ) =. Thus we want δ = π. Putting this all together, we get u = cos(t π ). () Firstly, ω =. Next, R = 4 + = 5. Finally tan δ = =. 4 Thus δ = tan ( ).466 or δ = π. However, we want = sin(δ) < so we choose δ =.466. Putting this all together, we B = R 5 get u = 5 cos(t +.466).. (Prolems 7 from B&D) A mass weighing l stretches a spring in. Suppose the mass is pushed upward, contracting the spring a distance of in., and is then set in motion with a downward velocity of ft/s. Assuming there is no damping, find the position u of the mass at any time t. Determine the frequency, period, amplitude, and phase of the motion.

2 Solution: To find the mass, we note that w = mg so l = m ft implying s m = l s. Since the mass weighing l stretches the spring in., we have that ft F S = l = k l ft implying k =. Since there is no damping or external force, 4 ft we have γ = and F (t) =. Thus the equation we get is If we multiply through y, we get u + u = ; y() =, y () = u + 8u = which has characteristic equation r + 8 = which has roots r = ± 8 = ±i 8. Thus the general solution is u(t) = A sin(8 t) + B cos(8 t). Plugging in our initial conditions we see y() = B =. Taking the derivative, we get y (t) = A8 cos(8 t) B8 sin(8 t). In particular y () = A8 =. Therefore A = 4. Thus the solution is u(t) = 4 sin(8 t) cos(8 t). Here ω = 8 ( rad/s and T = π ω = π 4 s. As usual, R = ) ( 4 + ) = + =. Finally δ = ( tan /) /4 = π arctan( ). since we want sin(δ) > and cos(δ) <.. (Prolems from B&D) A string is stretched cm y a force of N. A mass of kg is hung from the spring and is also attached to a viscous damper that exerts a force of N when the velocity of the mass is 5 m/s. If the mass is pulled down 5 cm elow its equilirium position and given an initial downward velocity of cm/s, determine its position u at any time t. Find the quasi frequency µ and the ratio of µ to the natural frequency of the corresponding undamped motion. Solution: We are given m = kg. Furthermore, we know a force of N stretches the spring cm implying F S = N =.k cm. Thus k = N. Furthermore, m F d = N = γ 5 m. Thus γ = N s and our equation is s 5 m u + 5 u + u = ; u() =.5, u () =. Page

3 Dividing y two we get, u + u + 5u = which has characteristic equation r + r + 5 = which has roots r = 9 4()(5) ± =.5 ± i.878 Thus the general solution is u = Ae.5t sin(.878 t) + Be.5t cos(.878 t). We can plug in our initial conditions and see that u() = B =.5. Taking the derivative, we get so u =.5Ae.5t sin(.878 t) +.878Ae.5t cos(.878 t).5be.5t cos(.878 t).878be.5t cos(.878 t) implying A.778. Thus the solution is u () =.878A.5B =.878A.5(.5) =. u(t) =.778e.5t sin(.878 t) +.5e.5t cos(.878 t). Then we see that the quasi frequency is µ =.878 rad/s. Recall that the natural k frequency is ω = = 5 rad/s. Thus the ratio is µ m ω = (Prolems 7 from B&D) A mass weighing 8 l stretches a spring.5 in. The mass is also attached to a damper with coefficient γ. Determine the value of γ for which the system is critically damped; e sure to give the units for γ. Solution: Recall that a system is critically damped when γ 4mk =. To convert from weight to mass, we note w = mg so m = 8 l s. For k, we see that ft the system reaches equilirium after the spring stretches.5 in. Thus F s +w = so F s = w = 8 l. However, F s = kl where here L =.5 = l ft. Thus k = ft Page

4 Thus we have γ = 4mk = 4 ( 8 l s ft ) ( 64 l ) = 64 l s ft ft = 8 l s ft. 5. Optional (Prolems 9 from B&D) Assume that the system descried y the equation mu +γu +ku = is either critically damped or overdamped. Show that the mass can pass through the equilirium position at most once, regardless of the initial conditions. Hint: Determine all possile values of t for which u =. Solution: The characteristic equation of mu + γu + ku = is mr + γr + k = with roots r = γ m ± γ 4mk m Since the system is critically damped or overdamped, y definition γ 4mk. Let α = γ m and β = γ 4mk m. Thus the equation has solutions of the form u(t) = Ae (α+β)t + Be (α β)t For the mass to pass through the equilirium point at time t means that u(t) =. Thus we set Ae (α+β)t + Be (α β)t = = Ae (α+β)t = Be (α β)t = e (α+β)t e (α β)t = B A = e βt = B A which has exactly one solution if B A > and zero solutions if B A. Thus, the mass can pass through the equilirium point at most once. Section.8 6. (B&D # 5) A mass weighing 4 l stretches a spring.5 in. The mass is displaced in. in the positive direction from its equilirium position and released with no initial velocity. Assuming that there is no damping and that the mass is acted on y an external force of cos(t) l, formulate the initial value prolem descriing the motion of the mass. Page 4

5 Solution: We can immediately assume the following. Some further computation gives w = 4 l L =.5 in. = ft. 8 γ = s l. ft. F (t) = cos(t) l u() = in. = ft. 6 u () = ft. s m = w 4 l = l s g m s 8 m k = mg = 4 l l L = ft. ft. It follows that the initial value prolem modeling the motion of the mass is 8 u + u = cos(t); u() = 6, u () = (B&D # 6) A mass of 5 kg stretches a spring cm, The mass is acted on y an external force of sin(t/) N (newtons) and moves in a medium that imparts a viscous force of N when the speed of the mass is 4 cm/s. If the mass is set in motion from its equilirium position with an initial velocity of cm/s, formulate the initial value prolem descriing the motion of the mass. Solution: We can immediately assume the following. Some further computation gives m = 5 kg L = cm =. m γ = N 4 cm s = N.4 m s F (t) = sin(t/) N u() = m u () = cm s =. m s = 5 kg s k = mg L 5 kg 9.8 m s. m = 49 N m It follows that the initial value prolem modeling the motion of the mass is 5u + 5u + 49u = sin(t/); u() =, u () =.. Page 5

6 8. (B&D # 9) If an undamped spring-mass system with a mass that weighs 6 l and a spring constant l/in is suddenly set in motion at t = y an external force of 4 cos(7t) l, determine the position of the mass at any time and draw a graph of the displacement vs t. Solution: We can immediately assume the following. Some further computation gives w = 6 l k = l in. = l ft. γ = s l. ft. F (t) = 4 cos(7t) l u() = ft. u () = ft. s m = w 6 l = l s g m s 6 m It follows that the initial value prolem modeling the motion of the mass is 6 u + u = 4 cos(7t); u() =, u () =. To solve this we start y finding the general solution to the associated homogeneous equation, 6 u + u =, which has a characteristic equation of 6 r + =. The roots of the characteristic equation are Hence we have a general solution of r = ±8i. u = c sin(8t) + c cos(8t). Using the method of undetermined coefficients to find a particular solution to the nohomogeneous equation we let u = A sin(7t) + B cos(7t). Page 6

7 It follows that u = 49A sin(7t) 49B cos(7t). Plugging into the nonhomogeneous differential equation we have ( 47 ) 6 + (A sin(t) + B cos(t)) = 45 (A sin(7t) + B cos(7t)) = 4 cos(7t). 6 It follows that A = and B = and our particular solution is u = cos(7t). Putting the pieces together, we have a general solution of u = c sin(8t) + c cos(8t) cos(7t). Plugging the initial values in and solving for the constants, we have Here s a graph of the solution. u = 64 (cos(7t) cos(8t)) (B&D # ) A spring is stretched 6 in y a mass that weighs 8 l. The mass is attached to a dashpot mechanism that has a damping constant of.5 l s/ft and is acted on y Page 7

8 an external force of 4 cos(t) l. Solution: We can immediately assume the following. Some further computation gives w = 8 l L = 6 in. = ft. γ =.5 s l. ft. F (t) = 4 cos(t) l m = w 8 l =.5 g m l s s m k = mg = 8 l l L = 6 ft. ft. It follows that the differential equation modeling the motion of the mass is.5u +.5u + 6u = 4 cos(t). (a) Determine the steady state response of the system. Solution: The steady state response of the system is the particular solution to the differential equation. We determine this using the method of undetermined coefficients. Let u = A cos(t) + B sin(t) and notice that then Plugging into the equation we have u = A sin(t) + B cos(t), u = 4A cos(t) + 4B sin(t). A cos(t) B sin(t) A sin(t) + B cos(t) + 6A cos(t) + 6B sin(t) and (5A + ) B cos(t) + ( ) A + 5B sin(t) = 4 cos(t). This leaves us with the following system of linear equations. A + B = 8 A + B = Page 8

9 Adding thirty times the second equation to the first we have 9B = 8 and B = 8. Sutracting thirty times the first equation from the second we have 9A = 4 and A = 4 9. It follows that the steady state response of the system is u = 4 9 cos(t) sin(t). () Optional If the given mass is replaced y a mass m, determine the value of m for which the amplitude of the steady state response is maximum. Solution: We proceed as aove with m=m left undetermined. Then plugging u = A cos(t) + sin(t) into the equation gives us 4mA cos(t) 4mB sin(t) A sin(t)+ B cos(t)+6a cos(t)+6b sin(t) and ((6 4m)A + ) B cos(t) + ( ) A + (6 4m)B sin(t) = 4 cos(t). This leaves us with the following system of linear equations. ( 8m)A + B = 8 A + ( 8m)B = Rememering that R = A + B is what we wish to maximize (well really this is the square of the amplitude R) we multiply the top equation y B and the ottom equation y A and add yielding A + B = 8B. Page 9

10 Notice that this means we just need to maximize B (B will always e nonnegative according to this equation). We can eliminate A y multiplying the ottom equation y ( 8m) and adding it to the top equation yielding This implies that ( + ( 8m) )B = 8. B = 8 + ( 8m) which is maximized when + ( 8m) is minimized (which is always positive). It is not hard to see that the minimum occurs when 8m = and m = 4. Note that the unit here is l s ft.. Section 6.. (Prolems 7- B&D) Using the fact that cosh(t) = et +e t and sinh(t) = et e t find the Laplace transform of the following functions (a, R). Note that parts (c) and (d) can e deduced from (a) and (). (a) f(t) = e at cosh(t) () f(t) = e at sinh(t) (c) f(t) = cosh(t) (d) f(t) = sinh(t) Solution: (a) L { e at cosh(t) } } = L {e at et + e t = L { e (a+)t} + L { e (a )t}. Notice that in this step we require s > a + and s > a. In other words, s a >. We proceed using the fact that (as shown in lecture) L {e at } = s a for s > a. L { e (a+)t} + L { e (a )t} = ( ) s (a + ) + s (a ) = s (a ) + [s (a + )] s as + s as + a + a = s a (s a). Page

11 () L { e at sinh(t) } } = L {e at et e t = L { e (a+)t} L { e (a )t}. Notice that in this step we require s > a + and s > a. In other words, s a >. L { e (a+)t} L { e (a )t} = ( ) s (a + ) s (a ) = s (a ) [s (a + )] s as + s as + a + a = (s a). (c) (d) for s >. for s >. L {cosh()} = L { e t cosh() } = L {sinh()} = L { e t sinh() } = s s s. (Prolems -4 B&D) Find the Laplace transform of the following functions (a, R). Note that parts (c) and (d) can e deduced from (a) and (). (a) f(t) = e at sin(t) () f(t) = e at cos(t) (c) f(t) = sin(t) (d) f(t) = cos(t) Solution: (a) L { e at sin(t) } = e st e at sin(t) dt = We will first compute the indefinite integral. e (a s)t sin(t) dt = e(a s)t sin(t) e (a s)t sin(t) dt. e (a s)t cos(t) dt Page

12 () = e(a s)t sin(t) Then we have + () () and e (a s)t sin(t) dt = [ e(a s)t cos(t) + ] e (a s)t sin(t) dt. e (a s)t sin(t) dt = e(a s)t sin(t) () e(a s)t cos(t) + () e(a s)t sin(t) + () e(a s)t cos(t)+c. For convenience (so the rest of this will fit in the ox) say Then F (t) = + () e(a s)t sin(t) + () e(a s)t cos(t). L { e at sin(t) } = F (t) = lim F (t) A = lim F (A) F (). A A We now make the stipulation that < (that is s > a) so that the integral converges. Then we have L { e at cos(t) } = lim F (A) F () = F () = A e st e at cos(t) dt = We will first compute the indefinite integral. e (a s)t cos(t) dt = e(a s)t cos(t) + = e(a s)t cos(t)+ Then we have + () () and e (a s)t cos(t) dt = + (). [ e(a s)t sin(t) e (a s)t cos(t) dt. e (a s)t sin(t) dt ] e (a s)t cos(t) dt. e (a s)t cos(t) dt = e(a s)t cos(t)+ () e(a s)t sin(t) + () e(a s)t cos(t)+ + () e(a s)t sin(t)+c. Page

13 For convenience (so the rest of this will fit in the ox) say Then G(t) = + () e(a s)t cos(t) + + () e(a s)t sin(t). L { e at cos(t) } = G(t) = lim G(t) A = lim G(A) G(). A A We now make the stipulation that < (that is s > a) so that the integral converges. Then we have lim A G(A) G() = G() = + () = s a + (s a). (c) (d) for s >. for s >. L {sin()} = L { e t sin() } = L {cos()} = L { e t cos() } = + s s + s. Optional(8 B&D) Assuming n Z, and a R, find the Laplace transform of f(t) = t n e at (it might e useful to try a couple of cases where n is small ) and use your result to determine the Laplace transform of g(t) = t n. Solution: L { t n e at} = = tn e (a s)t e st t n e at dt = n t n e (a s)t dt t n e (a s)t Page

14 = tn e (a s)t = lim A n ( An e (a s)a e st t n e at dt = tn e (a s)t ) n e (a s) n L { t n e at} n L { t n e at}. Here we must make the stipulation that < (s > a) so that the integral converges. Then if we use LHôpital s rule n times we have ( ) lim A An e (a s)a = and are left with Recalling that L { t n e at} = n L { t n e at} = for s > a we have the following. L { e at} = s a n s a L { t n e at}. L {te at } = L {t e at } = L {t e at } =. s a L { e at} = (s a) s a L { te at} = (s a) s a L { t e at} = (s a) 4.. From the aove we can conclude that L { t n e at} = Furthermore, if we let a = we have n! (s a) n+. for s >. L {t n } = n! s n+ Section 6.. (Prolems,,7 from B&D) For each of the following, use the tale on p. 7 of B&D to find the inverse Laplace transform of the given function. Page 4

15 (a) F (s) = 4 (s ) () F (s) = s +s 4 (c) F (s) = s+ s s+ Solution: (a) Line of the the tale on gives that L{t n e at n! } =. Letting a = (s a) n+ and n =, we have L{t e t } =! so L{t e t } = L{t e t } = 4. Thus { } (s ) (s ) L 4 = t e t. (s ) () With prolems of this form, we have two general cases. If the denominator can e factored, we use partial fractions. If not, we complete the square. Here we can factor the denominator to get F (s) = We do partial fractions as follows s + s 4 = (s )(s + 4) A s + B s + 4 = As + 4A + Bs B (s )(s + 4) = (s )(s + 4) Thus 4A B = and As + Bs = s. Thus A = B so we have 4A + A =. Therefore A = 5 meaning B = 5. Thus F (s) = /5 follows that (s ) /5. Line of the tale gives that L (s+4) {eat } =. It s a L {F (s)} = 5 et 5 e 4t. (c) In this prolem, the denominator cannot e factored so instead we complete the square. We get F (s) = s + (s ) + = s (s ) + + (s ) + s = (s ) + + (s ) +. Line 9 of the tale gives that L {e at sin(t)} = that L {e at cos(t)} = s a. (s a) + (s a) + and Line gives Page 5

16 Thus we have that } { } L {F (s)} = L { s + L (s ) + (s ) + { } { } = L s + L (s ) + (s ) + = e t cos(t) + e t sin(t). 4. (Prolems 5,9, from B&D) For each of the following, use the Laplace transform to solve the given initial value prolem. (a) y y + 4y = ; y() =, y () = () y (4) 4y = ; y() =, y () =, y (), y () = (c) y y + y = cos(t) y() =, y () = Solution: (a) Taking the Laplace of oth sides we get Note that Therefore we have L {y } L {y } + 4L {y} = L {} = L {y } = sl {y} y() = sl {y} L {y } = s L {y} sy() y () = s L {y} s (s s + 4)L {y} s + 4 =. Thus L {y} = s 4 s s + 4. The denominator cannot e factored so we complete the square and get L {y} = s 4 (s ) + Again the tale gives us that L {e at sin(t)} = s a. Therefore it is convenient to write (s a) + L {y} = s (s ) + (s ) + and L {e at cos(t)} = (s a) + Page 6

17 s = (s ) + (s ) + ( ) Thus { } y = L s (s ) + { + } L (s ) + ( ) () Taking the Laplace of oth sides gives Note = e t cos( t) + e t sin( t). L { y (4)} 4L {y} = L {} =. L { y (4)} = s 4 L {y} s y() s y () sy () y () Thus the equation ecomes = s 4 L {y} s + s. (s 4 4)L {y} s s = Thus = L {y} = s s s 4 4 = s(s ) (s )(s + ) = s s + { } s y = L = cos( t). s + (c) Taking the Laplace of oth sides gives = L {y } L {y } + L {y} = L {cos(t)} ) ( ) sl {y} y() ( s L {y} sy() y () + L {y} = s s + Plugging in the initial conditions y() = and y () =, this ecomes ( ) s s + L {y} s + = s s + = ( ) s s + L {y} = s s + + s = s s + s s + = L {y} = s s + s (s s + )(s + ) Page 7

18 = = s s + s (s s + )(s + ) (s s + )(s + ) s(s s + ) (s s + )(s + ) (s s + )(s + ) = s s + (s s + )(s + ) From here we need to do partial fractions on (s s+)(s +) As + B s s + + Cs + D s + as follows: = (A + C)s + (B C + D)s + (A + C D)s + (B + D) (s s + )(s + ) so we get the system of equations A + C = B C + D = A + C D = B + D = Thus we have that A = C which plugged into the third equation gives C = D. Plugging this into the second equation gives B = D. Finally putting this into the last equation gives D = 5. Then B = 6 5, C = 4 5, and A = 4 5. Thus Therefore we have = 5 (s s + )(s + ) = 4s s s + + s s + L {y} = s s + 4s s s + s s + 5 = s 5 s + (4s 4) 5 (s ) + 5 (s ) + s s + 5 s s 5 (s ) + 5 (s ) + Therefore taking the inverse Laplace transform we get y = 5 cos(t) 5 sin(t) et cos(t) 5 et sin(t). Page 8