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The amplitude and frequency of the steady-state solution for the system. The amplitude is 0.149 _ and frequency is 1 π _ . Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Calculation: Consider the general form m d 2 y d t 2 + b d y d t + k y = F ( t ) . The corresponding homogeneous equation is, m d 2 y d t 2 + b d y d t + k y = 0 . Given that, 8kg is attached to a spring hanging from the ceiling and comes to rest at the equilibrium. That is, m = 8 . The external force is F ( t ) = 2 sin ( 2 t + π 4 ) N , the spring constant is 40 N/m and the damping constant is 3 N-sec/m . It is known that, sin ( x + y ) = sin x cos y + cos x sin y . Thus, F ( t ) = 2 sin ( 2 t + π 4 ) becomes, F ( t ) = 2 [ sin 2 t cos π 4 + cos 2 t sin π 4 ] = 2 [ sin 2 t ( 1 2 ) + cos 2 t ( 1 2 ) ] = 2 sin 2 t + 2 cos 2 t Substitute m = 8 , k = 40 and b = 3 in m d 2 y d t 2 + b d y d t + k y = 0 to obtain as, 8 y ″ + 3 y ′ + 40 y = 0 . The associated auxiliary equation will be, 8 r 2 + 3 r + 40 = 0 . Factorize 8 r 2 + 3 r + 40 = 0 as, r = − 3 ± 3 2 − 4 ( 8 × 40 ) 2 ( 8 ) = − 3 ± − 1271 16 = − 3 ± 35 46 i 16 ( ∵ − 1 = i ) = − 3 16 ± 35 16 46 That is, the roots are r 1 = − 3 16 + 35 16 46 and r 2 = − 3 16 − 35 16 46 . Here, α = − 3 16 and β = 35 16 46 . The RHS of the differential equation m d 2 y d t 2 + b d y d t + k y = F ( t ) is F ( t ) = 2 sin 2 t + 2 cos 2 t . Comparing 2 sin 2 t + 2 cos 2 t with the form P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , it is observed that, m = 0 , n = 0 , α = 0 and β = 2 . Since k is the larger of m = 0 and n = 0 , k = 0 . Also, α + i β = 0 + 2 i is not a root of the associated auxiliary equation, s = 0 . Thus, the solution form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t for s = 0 , k = 0 , α = 0 and β = 2 will be, y p ( t ) = t 0 ( A 0 ) e 0 cos 2 t + t 0 ( B 0 ) e 0 sin 2 t = A 0 cos 2 t + B 0 sin 2 t Differentiate y p ( t ) = A 0 cos 2 t + B 0 sin 2 t with respect to t as, y ′ p ( t ) = − 2 A 0 sin 2 t + 2 B 0 cos 2 t . Differentiate y ′ p ( t ) = − 2 A 0 sin 2 t + 2 B 0 cos 2 t with respect to t as, y ″ p ( t ) = − 4 A 0 cos 2 t − 4 B 0 sin 2 t . Thus, 8 y ″ + 3 y ′ + 40 y for y p will be, 8 y ″ p + 3 y ′ p + 40 y p = [ 8 ( − 4 A 0 cos 2 t − 4 B 0 sin 2 t ) + 3 ( − 2 A 0 sin 2 t + 2 B 0 cos 2 t ) + 40 ( A 0 cos 2 t + B 0 sin 2 t ) ] = [ − 32 A 0 cos 2 t − 32 B 0 sin 2 t − 6 A 0 sin 2 t + 6 B 0 cos 2 t + 40 A 0 cos 2 t + 40 B 0 sin 2 t ] = ( 40 A 0 − 32 A 0 + 6 B 0 ) cos 2 t + ( 40 B 0 − 32 B 0 − 6 A 0 ) sin 2 t = ( 12 A 0 + 6 B 0 ) cos 2 t + ( 12 B 0 − 6 A 0 ) sin 2 t Setting ( 12 A 0 + 6 B 0 ) cos 2 t + ( 12 B 0 − 6 A 0 ) sin 2 t = 2 sin 2 t + 2 cos 2 t and comparing the coefficients yield 12 A 0 + 6 B 0 = 2 and 12 B 0 − 6 A 0 = 2 . Multiply 12 B 0 − 6 A 0 = 2 by 2 to obtain as, 24 B 0 − 12 A 0 = 2 2 . Add the equations 24 B 0 − 12 A 0 = 2 2 and 12 A 0 + 6 B 0 = 2 . 24 B 0 − 12 A 0 = 2 2 + 6 B 0 + 12 A 0 = 2 _ 30 B 0 = 3 2 Simplify 30 B 0 = 3 2 as, B 0 = 3 2 30 = 2 10 Substitute B 0 = 2 10 in 12 A 0 + 6 B 0 = 2 , 12 A 0 + 6 ( 2 10 ) = 2 12 A 0 = 2 − 3 2 5 12 A 0 = 2 2 5 A 0 = 2 2 5 × 12 A 0 = 2 30 That is, A 0 = 2 30 and B 0 = 2 10 . Thus, the solution y p ( t ) = A 0 cos 2 t + B 0 sin 2 t becomes, y p ( t ) = 2 30 cos 2 t + 2 10 sin 2 t . Note that, for the general solution y ( t ) = C 1 cos ω t + C 2 sin ω t , the amplitude is A = C 1 2 + C 2 2 , angular frequency ω = k m and frequency is ω 2 π . Here, C 1 = 2 30 , C 2 = 2 10 and ω = 2 . Thus, the amplitude A = C 1 2 + C 2 2 becomes, A = ( 2 30 ) 2 + ( 2 10 ) 2 = 2 + 18 30 2 = 20 900 = 0.0222 ≈ 0.149 The frequency ω 2 π becomes, Freq = 2 2 π = 1 π Therefore, the amplitude is 0.149 _ and frequency is 1 π _ .

The amplitude and frequency of the steady-state solution for the system. The amplitude is 0.149 _ and frequency is 1 π _ . Procedure used: Method of undetermined coefficients: A particular solution to the differential equation a y ″ + b y ′ + c y = P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , β ≠ 0 , where P m ( t ) and Q n ( t ) is a polynomial of degree m and n respectively, is of the form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t , where k is the larger of m and n : (i) s = 0 , if α + i β is not a root of the associated auxiliary equation; (ii) s = 1 , if α + i β is a root of the associated auxiliary equation. Calculation: Consider the general form m d 2 y d t 2 + b d y d t + k y = F ( t ) . The corresponding homogeneous equation is, m d 2 y d t 2 + b d y d t + k y = 0 . Given that, 8kg is attached to a spring hanging from the ceiling and comes to rest at the equilibrium. That is, m = 8 . The external force is F ( t ) = 2 sin ( 2 t + π 4 ) N , the spring constant is 40 N/m and the damping constant is 3 N-sec/m . It is known that, sin ( x + y ) = sin x cos y + cos x sin y . Thus, F ( t ) = 2 sin ( 2 t + π 4 ) becomes, F ( t ) = 2 [ sin 2 t cos π 4 + cos 2 t sin π 4 ] = 2 [ sin 2 t ( 1 2 ) + cos 2 t ( 1 2 ) ] = 2 sin 2 t + 2 cos 2 t Substitute m = 8 , k = 40 and b = 3 in m d 2 y d t 2 + b d y d t + k y = 0 to obtain as, 8 y ″ + 3 y ′ + 40 y = 0 . The associated auxiliary equation will be, 8 r 2 + 3 r + 40 = 0 . Factorize 8 r 2 + 3 r + 40 = 0 as, r = − 3 ± 3 2 − 4 ( 8 × 40 ) 2 ( 8 ) = − 3 ± − 1271 16 = − 3 ± 35 46 i 16 ( ∵ − 1 = i ) = − 3 16 ± 35 16 46 That is, the roots are r 1 = − 3 16 + 35 16 46 and r 2 = − 3 16 − 35 16 46 . Here, α = − 3 16 and β = 35 16 46 . The RHS of the differential equation m d 2 y d t 2 + b d y d t + k y = F ( t ) is F ( t ) = 2 sin 2 t + 2 cos 2 t . Comparing 2 sin 2 t + 2 cos 2 t with the form P m ( t ) e α t cos β t + Q n ( t ) e α t sin β t , it is observed that, m = 0 , n = 0 , α = 0 and β = 2 . Since k is the larger of m = 0 and n = 0 , k = 0 . Also, α + i β = 0 + 2 i is not a root of the associated auxiliary equation, s = 0 . Thus, the solution form y p ( t ) = t s ( A k t k + ⋯ + A 1 t + A 0 ) e α t cos β t + t s ( B k t k + ⋯ + B 1 t + B 0 ) e α t sin β t for s = 0 , k = 0 , α = 0 and β = 2 will be, y p ( t ) = t 0 ( A 0 ) e 0 cos 2 t + t 0 ( B 0 ) e 0 sin 2 t = A 0 cos 2 t + B 0 sin 2 t Differentiate y p ( t ) = A 0 cos 2 t + B 0 sin 2 t with respect to t as, y ′ p ( t ) = − 2 A 0 sin 2 t + 2 B 0 cos 2 t . Differentiate y ′ p ( t ) = − 2 A 0 sin 2 t + 2 B 0 cos 2 t with respect to t as, y ″ p ( t ) = − 4 A 0 cos 2 t − 4 B 0 sin 2 t . Thus, 8 y ″ + 3 y ′ + 40 y for y p will be, 8 y ″ p + 3 y ′ p + 40 y p = [ 8 ( − 4 A 0 cos 2 t − 4 B 0 sin 2 t ) + 3 ( − 2 A 0 sin 2 t + 2 B 0 cos 2 t ) + 40 ( A 0 cos 2 t + B 0 sin 2 t ) ] = [ − 32 A 0 cos 2 t − 32 B 0 sin 2 t − 6 A 0 sin 2 t + 6 B 0 cos 2 t + 40 A 0 cos 2 t + 40 B 0 sin 2 t ] = ( 40 A 0 − 32 A 0 + 6 B 0 ) cos 2 t + ( 40 B 0 − 32 B 0 − 6 A 0 ) sin 2 t = ( 12 A 0 + 6 B 0 ) cos 2 t + ( 12 B 0 − 6 A 0 ) sin 2 t Setting ( 12 A 0 + 6 B 0 ) cos 2 t + ( 12 B 0 − 6 A 0 ) sin 2 t = 2 sin 2 t + 2 cos 2 t and comparing the coefficients yield 12 A 0 + 6 B 0 = 2 and 12 B 0 − 6 A 0 = 2 . Multiply 12 B 0 − 6 A 0 = 2 by 2 to obtain as, 24 B 0 − 12 A 0 = 2 2 . Add the equations 24 B 0 − 12 A 0 = 2 2 and 12 A 0 + 6 B 0 = 2 . 24 B 0 − 12 A 0 = 2 2 + 6 B 0 + 12 A 0 = 2 _ 30 B 0 = 3 2 Simplify 30 B 0 = 3 2 as, B 0 = 3 2 30 = 2 10 Substitute B 0 = 2 10 in 12 A 0 + 6 B 0 = 2 , 12 A 0 + 6 ( 2 10 ) = 2 12 A 0 = 2 − 3 2 5 12 A 0 = 2 2 5 A 0 = 2 2 5 × 12 A 0 = 2 30 That is, A 0 = 2 30 and B 0 = 2 10 . Thus, the solution y p ( t ) = A 0 cos 2 t + B 0 sin 2 t becomes, y p ( t ) = 2 30 cos 2 t + 2 10 sin 2 t . Note that, for the general solution y ( t ) = C 1 cos ω t + C 2 sin ω t , the amplitude is A = C 1 2 + C 2 2 , angular frequency ω = k m and frequency is ω 2 π . Here, C 1 = 2 30 , C 2 = 2 10 and ω = 2 . Thus, the amplitude A = C 1 2 + C 2 2 becomes, A = ( 2 30 ) 2 + ( 2 10 ) 2 = 2 + 18 30 2 = 20 900 = 0.0222 ≈ 0.149 The frequency ω 2 π becomes, Freq = 2 2 π = 1 π Therefore, the amplitude is 0.149 _ and frequency is 1 π _ .

Solution Summary: The author explains the amplitude and frequency of the steady-state solution for the system.

Fundamentals of Differential Equations (9th Edition)

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

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Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 3E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - Prob. 5E Ch. 4.2 - Prob. 6E Ch. 4.2 - Prob. 7E Ch. 4.2 - Prob. 8E Ch. 4.2 - Prob. 9E Ch. 4.2 - Prob. 10E Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 112, find a general solution to the...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 15E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 17E Ch. 4.2 - Prob. 18E Ch. 4.2 - Prob. 19E Ch. 4.2 - In Problems 1320, solve the given initial value...Ch. 4.2 - Prob. 21E Ch. 4.2 - Prob. 22E Ch. 4.2 - Prob. 23E Ch. 4.2 - Prob. 24E Ch. 4.2 - Prob. 25E Ch. 4.2 - Prob. 26E Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - In Problems 2732, use Definition 1 to determine...Ch. 4.2 - Prob. 31E Ch. 4.2 - Prob. 32E Ch. 4.2 - Prob. 33E Ch. 4.2 - Prob. 34E Ch. 4.2 - Prob. 35E Ch. 4.2 - Using the definition in Problem 35, prove that if...Ch. 4.2 - Prob. 37E Ch. 4.2 - Prob. 38E Ch. 4.2 - Prob. 39E Ch. 4.2 - Prob. 40E Ch. 4.2 - In Problems 3741, find three linearly independent...Ch. 4.2 - Prob. 42E Ch. 4.2 - Prob. 43E Ch. 4.2 - Solve the initial value problem: y 2y y + 2y =...Ch. 4.2 - Prob. 45E Ch. 4.2 - Prob. 46E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - Prob. 5E Ch. 4.3 - Prob. 6E Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 18, the auxiliary equation for the...Ch. 4.3 - In Problems 920, find a general solution. 9. y 8y...Ch. 4.3 - In Problems 920, find a general solution. 10. y +...Ch. 4.3 - Prob. 11E Ch. 4.3 - Prob. 12E Ch. 4.3 - In Problems 920, find a general solution. 13. y ...Ch. 4.3 - Prob. 14E Ch. 4.3 - Prob. 15E Ch. 4.3 - Prob. 16E Ch. 4.3 - In Problems 920, find a general solution. 17. y y...Ch. 4.3 - In Problems 920, find a general solution. 18. 2y +...Ch. 4.3 - Prob. 19E Ch. 4.3 - In Problems 920, find a general solution. 20. y y...Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 22E Ch. 4.3 - Prob. 23E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 25E Ch. 4.3 - In Problems 2127, solve the given initial value...Ch. 4.3 - Prob. 27E Ch. 4.3 - Prob. 28E Ch. 4.3 - Prob. 29E Ch. 4.3 - Prob. 30E Ch. 4.3 - Prob. 31E Ch. 4.3 - Prob. 32E Ch. 4.3 - Prob. 33E Ch. 4.3 - Prob. 34E Ch. 4.3 - Swinging Door. The motion of a swinging door with...Ch. 4.3 - Prob. 36E Ch. 4.3 - Prob. 37E Ch. 4.3 - Prove the sum of angles formula for the sine...Ch. 4.4 - In Problems 18, decide whether or not the method...Ch. 4.4 - Prob. 2E Ch. 4.4 - Prob. 3E Ch. 4.4 - Prob. 4E Ch. 4.4 - Prob. 5E Ch. 4.4 - Prob. 6E Ch. 4.4 - Prob. 7E Ch. 4.4 - Prob. 8E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 16E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 18E Ch. 4.4 - Prob. 19E Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - In Problems 926, find a particular solution to the...Ch. 4.4 - Prob. 23E Ch. 4.4 - Prob. 24E Ch. 4.4 - Prob. 25E Ch. 4.4 - Prob. 26E Ch. 4.4 - Prob. 27E Ch. 4.4 - Prob. 28E Ch. 4.4 - Prob. 29E Ch. 4.4 - Prob. 30E Ch. 4.4 - Prob. 31E Ch. 4.4 - Prob. 32E Ch. 4.4 - Prob. 33E Ch. 4.4 - Prob. 34E Ch. 4.4 - Prob. 35E Ch. 4.4 - In Problems 3336, use the method of undetermined...Ch. 4.5 - Given that y1(t) = cos t is a solution to y y + y...Ch. 4.5 - Prob. 2E Ch. 4.5 - Prob. 3E Ch. 4.5 - Prob. 4E Ch. 4.5 - Prob. 5E Ch. 4.5 - Prob. 6E Ch. 4.5 - Prob. 7E Ch. 4.5 - In Problems 38, a nonhomogeneous equation and a...Ch. 4.5 - Prob. 9E Ch. 4.5 - Prob. 10E Ch. 4.5 - Prob. 11E Ch. 4.5 - Prob. 12E Ch. 4.5 - Prob. 13E Ch. 4.5 - In Problems 916 decide whether the method of...Ch. 4.5 - Prob. 15E Ch. 4.5 - Prob. 16E Ch. 4.5 - Prob. 17E Ch. 4.5 - Prob. 18E Ch. 4.5 - Prob. 19E Ch. 4.5 - In Problems 1722, find a general solution to the...Ch. 4.5 - Prob. 21E Ch. 4.5 - Prob. 22E Ch. 4.5 - Prob. 23E Ch. 4.5 - Prob. 24E Ch. 4.5 - Prob. 25E Ch. 4.5 - Prob. 26E Ch. 4.5 - Prob. 27E Ch. 4.5 - Prob. 28E Ch. 4.5 - Prob. 29E Ch. 4.5 - Prob. 30E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 33E Ch. 4.5 - In Problems 3136, determine the form of a...Ch. 4.5 - Prob. 35E Ch. 4.5 - Prob. 36E Ch. 4.5 - Prob. 37E Ch. 4.5 - Prob. 38E Ch. 4.5 - In Problems 3740, find a particular solution to...Ch. 4.5 - Prob. 40E Ch. 4.5 - Prob. 41E Ch. 4.5 - Prob. 42E Ch. 4.5 - Prob. 43E Ch. 4.5 - Prob. 44E Ch. 4.5 - Prob. 46E Ch. 4.5 - Prob. 47E Ch. 4.5 - Prob. 48E Ch. 4.6 - Prob. 1E Ch. 4.6 - In Exercises 12, rectangles have been drawn to...Ch. 4.6 - Prob. 3E Ch. 4.6 - Prob. 4E Ch. 4.6 - Prob. 5E Ch. 4.6 - Prob. 6E Ch. 4.6 - Prob. 7E Ch. 4.6 - Prob. 8E Ch. 4.6 - Prob. 9E Ch. 4.6 - Prob. 10E Ch. 4.6 - Prob. 11E Ch. 4.6 - Prob. 12E Ch. 4.6 - Prob. 13E Ch. 4.6 - Prob. 14E Ch. 4.6 - Prob. 15E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - Prob. 17E Ch. 4.6 - In Exercises 1318, use a calculator or a computer...Ch. 4.6 - For f(x) = 3x + 2, evaluate the Riemann sums: (a)...Ch. 4.6 - For g(x) = 4x 1, evaluate the Riemann sums: (a)...Ch. 4.6 - Prob. 21E Ch. 4.6 - Use Table 5.9 to evaluate the Riemann sums: Table...Ch. 4.6 - The graph of f(t) is in Figure 5.40. Which of the...Ch. 4.6 - Prob. 24E Ch. 4.6 - Prob. 25E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 3E Ch. 4.7 - In Problems 1 through 4, use Theorem 5 to discuss...Ch. 4.7 - Prob. 5E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 7E Ch. 4.7 - In Problems 5 through 8, determine whether Theorem...Ch. 4.7 - Prob. 9E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 11E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - Prob. 13E Ch. 4.7 - In Problems 9 through 14, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - In Problems 15 through 18, find a general solution...Ch. 4.7 - Prob. 17E Ch. 4.7 - Prob. 18E Ch. 4.7 - Prob. 19E Ch. 4.7 - Prob. 20E Ch. 4.7 - Prob. 21E Ch. 4.7 - Prob. 22E Ch. 4.7 - Prob. 23E Ch. 4.7 - Solve the following CauchyEuler equations by using...Ch. 4.7 - Prob. 25E Ch. 4.7 - Prob. 26E Ch. 4.7 - Prob. 27E Ch. 4.7 - Prob. 28E Ch. 4.7 - Prove that if y1 and y2 are linearly independent...Ch. 4.7 - Prob. 30E Ch. 4.7 - Prob. 31E Ch. 4.7 - Prob. 32E Ch. 4.7 - Prob. 33E Ch. 4.7 - Prob. 34E Ch. 4.7 - Prob. 35E Ch. 4.7 - Prob. 36E Ch. 4.7 - Prob. 37E Ch. 4.7 - Prob. 38E Ch. 4.7 - Prob. 39E Ch. 4.7 - Prob. 40E Ch. 4.7 - Prob. 41E Ch. 4.7 - In Problems 41 through 44, a differential equation...Ch. 4.7 - Prob. 43E Ch. 4.7 - Prob. 44E Ch. 4.7 - Prob. 45E Ch. 4.7 - Prob. 46E Ch. 4.7 - Prob. 47E Ch. 4.7 - Prob. 48E Ch. 4.7 - Prob. 49E Ch. 4.7 - Prob. 50E Ch. 4.7 - Prob. 51E Ch. 4.7 - Prob. 52E Ch. 4.8 - Prob. 1E Ch. 4.8 - Prob. 2E Ch. 4.8 - Try to predict the qualitative features of the...Ch. 4.8 - Prob. 4E Ch. 4.8 - Prob. 5E Ch. 4.8 - Prob. 6E Ch. 4.8 - Prob. 7E Ch. 4.8 - Prob. 8E Ch. 4.8 - Prob. 9E Ch. 4.8 - Prob. 10E Ch. 4.8 - Prob. 11E Ch. 4.8 - Use reduction of order to derive the solution...Ch. 4.8 - Prob. 13E Ch. 4.8 - Prob. 14E Ch. 4.8 - Prob. 15E Ch. 4.8 - Prob. 16E Ch. 4.8 - Prob. 17E Ch. 4.9 - Prob. 1E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 3E Ch. 4.9 - Prob. 4E Ch. 4.9 - Prob. 5E Ch. 4.9 - Prob. 6E Ch. 4.9 - Prob. 7E Ch. 4.9 - Prob. 8E Ch. 4.9 - All problems refer to the massspring configuration...Ch. 4.9 - Prob. 10E Ch. 4.9 - Prob. 11E Ch. 4.9 - Prob. 12E Ch. 4.9 - Prob. 13E Ch. 4.9 - Prob. 14E Ch. 4.9 - Prob. 15E Ch. 4.9 - Prob. 16E Ch. 4.9 - Prob. 17E Ch. 4.9 - Prob. 18E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 2E Ch. 4.10 - Prob. 3E Ch. 4.10 - Prob. 4E Ch. 4.10 - Prob. 5E Ch. 4.10 - Prob. 6E Ch. 4.10 - Prob. 7E Ch. 4.10 - Prob. 8E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 10E Ch. 4.10 - Prob. 11E Ch. 4.10 - Prob. 12E Ch. 4.10 - In the following problems, take g = 32 ft/sec2 for...Ch. 4.10 - Prob. 14E Ch. 4.10 - Prob. 15E Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 2RP Ch. 4 - Prob. 3RP Ch. 4 - Prob. 4RP Ch. 4 - Prob. 5RP Ch. 4 - Prob. 6RP Ch. 4 - Prob. 7RP Ch. 4 - Prob. 8RP Ch. 4 - Prob. 9RP Ch. 4 - Prob. 10RP Ch. 4 - Prob. 11RP Ch. 4 - Prob. 12RP Ch. 4 - Prob. 13RP Ch. 4 - Prob. 14RP Ch. 4 - Prob. 15RP Ch. 4 - Prob. 16RP Ch. 4 - Prob. 17RP Ch. 4 - In Problems 128, find a general solution to the...Ch. 4 - Prob. 19RP Ch. 4 - Prob. 20RP Ch. 4 - Prob. 21RP Ch. 4 - Prob. 22RP Ch. 4 - Prob. 23RP Ch. 4 - Prob. 24RP Ch. 4 - Prob. 25RP Ch. 4 - Prob. 26RP Ch. 4 - Prob. 27RP Ch. 4 - Prob. 28RP Ch. 4 - Prob. 29RP Ch. 4 - Prob. 30RP Ch. 4 - Prob. 31RP Ch. 4 - Prob. 32RP Ch. 4 - Prob. 33RP Ch. 4 - Prob. 34RP Ch. 4 - Prob. 35RP Ch. 4 - Prob. 36RP Ch. 4 - Prob. 37RP Ch. 4 - Prob. 38RP Ch. 4 - A 32-lb weight is attached to a vertical spring,...Ch. 4 - Prob. 1TWE Ch. 4 - Prob. 2TWE Ch. 4 - Prob. 3TWE Ch. 4 - For students with a background in linear algebra:...

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Simple Harmonic Motion - An introduction : ExamSolutions Maths Revision; Author: ExamSolutions;https://www.youtube.com/watch?v=tH2vldyP5OE;License: Standard YouTube License, CC-BY