3.7 Spring Systems 253

Transcription

1 3.7 Spring Systems 253 The resulting amplification of vibration eventually becomes large enough to destroy the mechanical system. This is a manifestation of resonance discussed further in Section??. Exercises Assume forces are in pounds or Newtons and lengths are in feet or meters Assume Hooke s Law. 1. A body weighing 16 lbs stretches a spring 6 inches. Find the spring constant. Solution: The force is 16 lbs. A length of 6 inches is 1/2 ft. The spring constant is k = 16/(1/2) = 32 lbs/ft. Answer: k = 32 lbs/ft 2. The spring constant of a certain spring isk = 20 lbs/ft. If a body stretches the spring 9 inches how much does it weigh? Solution: The equation for the spring constant gives 20 = x 9/12. So x = 240/9 = 80/3 lbs. Answer: 80/3 lbs. 3. A mass of 40 kilograms stretches a spring 80 centimeters. Find the spring constant. Solution: The force exerted by the mass is = 392 N. Thus k = 392/.8 = 490 N/m. Answer: k = 490 N/m 4. The spring constant of a certain spring is k = 784 N/m. How far will a mass of 20 kilograms stretch the spring?

2 254 3 Second Order Constant Coefficient Linear Differential Equations Solution: The force exerted by the mass is = 196 N. Let x be the distance stretched. Thus 784 = 196 x. Solving for x gives x = 196/784 = 1/4 m or 250 centimeters. Answer: 250 centimeters 5 8. Assume the damping force of a dashpot is proportional to the velocity of the body. 5. A dashpot exerts a damping force of 4 lbs when the velocity of the mass is 6 inches per second. Find the damping constant. Solution: The force is 4 lbs and the velocity is 1/2 ft per second. So μ = Force/velocity= 4 1/2 = 8 lbs s/ft. Answer: μ = 8 lbs s/ft 6. A dashpot exerts a damping force of 40 Newtons when the velocity is 30 centimeters per second. Find the damping constant. Solution: The force is 40 Newtons and the velocity is 30 centimeters per second which is.3 m per second. So μ = 40/.3 = N s/m. Answer: μ = N s/m 7. A dashpot has a damping constant μ = 100 lbs s/ft and decelerates a body by 4 ft per second. What was the force exerted by the body? Solution: Let x be the force. Then 100 = x/4 so x = 400 lbs. Answer: 400 lbs 8. A force of 40 N is applied on a body connected to a dashpot having a damping constant μ = 200 N s/m. By how much will the dashpot decelerates the body? Solution: Let v be the velocity of deceleration. Then 200 = 40/v and so v = 1/5 m/s or 20 centimeters per second. Answer: 20 centimeters per second Find a mathematical model that represents the motion of the body and determine the resulting motion. 9. A spring is stretched 10 centimeters by a force of 2 Newtons. A body of mass 6 kilogram is attached to such a spring with an accompanying dashpot. At t = 0 the mass is pulled down from its equilibrium position a distance of 10 centimeters and released. Suppose there is no damping.

3 3.7 Spring Systems 255 Solution: The mass is m = 6. The spring constant is given by k = 2/.1 = 20. The damping constant is μ = 0. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) =.1 m and y (0) = 0. The following equation 6y (t)+20y = 0, y(0) =.1, y (0) = 0 represents the model for the motion of the body. The motion is undamped free or simple harmonic motion. Answer: 6y (t)+20y = 0, simple harmonic motion y(0) =.1, y (0) = 0, undamped free or 10. A body of mass4kg will stretch a spring 80 centimeters. This same body damping force is 98 N when the velocity of the body is 2 m/s. At t = 0 the mass is pulled down from its equilibrium position a distance of 50 centimeters and released with an upward velocity of 10 centimeters per second. Solution: The mass is m = 4. The spring constant k is given by k = 4 9.8/.8 = 49. The damping constant is given by μ = 98/2 = 49. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) =.5 and y (0) =.1. The equation 4y (t)+49y (t)+ 49y = 0, y(0) =.5, y (0) = 11 models the motion of the body. The discriminant of the characteristic equation isd = = 1617 > 0 so the motion is overdamped Answer: 4y (t)+49y (t)+49y = 0, y(0) =.5, y (0) = 11, overdamped 11. A body weighing16 pounds will stretch a spring6inches. This same body damping force is 4 pounds when the velocity of the body is 2 feet per second. At t = 0 the mass is pulled down from its equilibrium position a distance of 1 foot and released with a downward velocity of 1 foot per second. Solution: The mass is m = 16/32 = 1/2 slugs. The spring constant k is given by k = 16/(6/12) = 32. The damping constant is given by μ = 4/2 = 2. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = 1 and y (0) = 1. The following equation 1 2 y (t) + 2y (t) + 32y = 0, y(0) = 1, y (0) = 1 models the motion of the body. The discriminant of the characteristic equation is D = (1/2) 32 = 60 < 0 so the motion is underdamped free motion.

4 256 3 Second Order Constant Coefficient Linear Differential Equations Answer: 1 2 y (t) +2y (t) +32y = 0, y(0) = 1, y (0) = 1, underdamped 12. A body weighing 32 pounds will stretch a spring 2 foot. This same body damping constant is 8 lbs s/ft. At t = 0 the mass is pulled down from its equilibrium position a distance of 1 foot and released. Solution: The mass is m = 32/32 = 1 slug. The spring constant k is given by k = 32/2 = 16. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) = 1 and y (0) = 0. The equation y (t) + 8y (t) + 16y = 0, y(0) = 1, y (0) = 0 models the motion of the body. The discriminant of the characteristic equation is D = = 0 so the motion is critically damped Answer: y (t)+8y (t)+16y = 0, y(0) = 1, y (0) = 0, critically damped 13. A body weighing 2 pounds will stretch a spring 4 inches. This same body is attached to such a spring with no accompanying dashpot. At t = 0 the body is pushed downward from equilibrium with a velocity of 8 inches per second. Solution: The mass is m = 2/32 = 1/16 slug. The spring constant k is given by k = 2/(4/12) = 6 and the damping constant is μ = 0. The initial conditions are y(0) = 0 and y (0) = 8/12 = 2/3. The equation 1 16 y (t)+6y = 0, y(0) = 0, y (0) = 2/3 models the motion of the body. The motion is undamped free or simple harmonic motion. Answer: y (t) + 96y = 0, y(0) = 0, y (0) = 2/3, undamped free or simple harmonic motion 14. A spring is stretched 1 m by a force of 5 N. A body of mass 2 kg is attached to the spring with accompanying dashpot. Suppose the damping force of the dashpot is 6 N when the velocity of the body is 1 m/s. At t = 0 the mass is pulled down from its equilibrium position a distance of 10 centimeters and given an initial downward velocity of 10 centimeters per second. Solution: The mass is m = 2. The spring constant k is given by k = 5/1 = 5. The damping constant is given byμ = 6/1 = 6. Since no external force is mentioned we may assume it is zero. The initial conditions are y(0) =.1 and y (0) =.1. The equation 2y (t)+6y (t)+5y = 0, y(0) =.1, y (0) =.1 models the motion of the body. The discriminant of the characteristic equation is D = = 4 < 0 so the motion is underdamped

5 3.7 Spring Systems 257 Answer: 2y (t) +6y (t) +5y = 0, y(0) =.1, y (0) =.1, underdamped