Solutions More Practice with Modeling

Solutions: More Practice with Modeling 1. A mass weighing 4 lb stretches a spring 2 in. Suppose that the mass is displaced an additional 6 in the positive direction and released. The mass is in a medium that exerts a viscous resistance of 6 lb when the mass has a velocity of 3 ft/s. Formulate the IVP that governs the motion of the mass. ( Hint: All units should be consistent. When working with US units, use pounds, feet and seconds.) SOLUTION: Remember to change units to pounds, feet and second. The first piece of information is given to find the spring constant. Use the relationship: mg − kL = 0, and the fact that pounds is already mg to get: 1 = 0 ⇒ k = 24 6 Next, we need the damping constant, γ. We know that damping is proportional to the velocity, and the value of γ is non-negative, so we take the signs to be the appropriate values: Fdamping = −γy 0 (t) ⇒ 6 = γ3 ⇒ γ = 2 4−k For the mass, remember that “pounds” is mg, and g = 32 (I would give you this value on an exam/quiz). For the mass, ⇒ mg = 4 m= 4 4 1 = = g 32 8 To finish, we have an initial position of 1/2 feet and zero initial velocity: 1 00 y + 2y 0 + 24y = 0 8 y(0) = 1 2 y 0 (0) = 0 2. Suppose we consider a mass-spring system with no damping (the damping constant is then 0), so that the differential equation expressing the motion of the mass can be modeled as mu00 + ku = 0 Find value(s) of β so that A cos(βt) and B sin(βt) are each solutions to the homogeneous equation (for arbitrary values of A, B). SOLUTION: We subsitute u = A cos(βt) + B sin(βt) into the differential equation: m(−β 2 A cos(βt) − β 2 B sin(βt) + k(A cos(βt) + B sin(βt)) = 0 We need the coefficient in front on cos(βt) to be zero: −mβ 2 A + kA = 0 and the coefficient in front of sin(βt) to be zero: −mβ 2 B + kB = 0 Take either equation to see that β = q k m 1 (we’ll only take the positive root). 3. A mass weighing 2 lb stretches a spring 6 in. If the mass is pulled down an additional 3 in and then released, and if there is no damping, determine the IVP that governs the motion of the mass. (You might re-read the hint for (1)) SOLUTION (remember units): Given the first line, we can find the value of the spring constant. 1 mg − kL = 0 ⇒ 2 − k = 0 ⇒ k = 4 2 For the mass, 2 1 2 = mg = 2 ⇒ m = = g 32 16 Now we have the IVP: 1 00 y + 4y = 0 16 1 y(0) = , 4 y 0 (0) = 0 4. A mass of 20 grams stretches a spring 5 cm. Suppose tha the mass is attached to a viscous damper with a constant damping constant of 400 dyn-s/cm (note: a dyne is a unit of force using centimeters-grams- seconds for units). If the mass is pulled down an additional 2 cm then released, find the IVP that governs the motion of the mass. SOLUTION: The additional note tells us we can stick with the given units, but beware of g, which is normally 9.8 m/s2 , now it would be 980 cm/s2 . Given that, mg − kL = 0 ⇒ (20)(980) − k(5) = 0 ⇒ k = 3920 The damping constant is given as γ = 400, so now we put the equation together: 20y 00 + 400y 0 + 3920y = 0 y(0) = 2, 2 y 0 (0) = 0 5. For exercises 1, 3 and 4 above, convert the mass-spring second order DE into a system of first order equations, then analyze the solutions using the HPGSystemSolver software (in DETools). SOLUTION: When converting to a linear system Y 0 = AY , where A is: " (1) 0 1 −192 −16 # " (3) 0 1 −64 0 # " (4) 0 1 −196 −20 # 6. A mass weighing 8 lbs stretches a spring 32 in. The mass is attached to a damper with coefficient γ. Using the ansatz to get the characteristic equation for the second order linear, homogeneous, DE, find the value of γ at which we get one single real solution (to the characteristic equation). SOLUTION: As usual, the first line is used for both the spring constant k and the value of m: 1 mg − kL = 0 ⇒ 8 − k = 0 ⇒ k = 64 8 We also have: 1 8 = 8 = mg ⇒ m = 32 4 The DE is now: 1 00 y + γy 0 + 64y = 0 ⇒ y 00 + 4γy 0 + 256 = 0 4 The ansatz is y = est . Substituting this into the DE, we get s2 + 4γs + 256 = 0 We can use the quadratic formula to solve for s, but note that we want only ONE real solution- That happens when the discriminant is 0: b2 − 4ac = 0 ⇒ 16γ 2 − 4(256) = 0 so that γ = 8 (only the positive one). 3 ⇒ γ 2 = 64