Problem 2: A mass of 2 kg stretches a spring by 0.2 m. At time t = 0, the mass is released from rest from a position 0.5 meters below equilibrium. Assume a damping force numerically equal to 8 times the velocity, and assume the mass is driven by a external force of f(t) = 2e-t. a) Find the spring constant. b) Set up an IVP modeling the situation. c) Find the equation of motion for the mass on the spring.
Problem 2: A mass of 2 kg stretches a spring by 0.2 m. At time t = 0, the mass is released from rest from a position 0.5 meters below equilibrium. Assume a damping force numerically equal to 8 times the velocity, and assume the mass is driven by a external force of f(t) = 2e-t. a) Find the spring constant. b) Set up an IVP modeling the situation. c) Find the equation of motion for the mass on the spring.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please help answer question 2. Spring-mass system. Thanks so much.
Transcribed Image Text:Magnitude of acceleration of gravity in the units of ft/sec^2: 32
Magnitude of acceleration of gravity in the units of m/sec^2: 9.8
(Optional) Problem 1: A mass of 1 kg stretches one spring by 0.1 m and another spring by 0.2 m. The
mass is then attached to the two springs in series. At time t= 0, the mass is released from rest from a
position 0.1 meter below equilibrium. Assume no damping forces.
a) Find the effective spring constant.
b) Set up an IVP modeling the situation.
c) Find the equation of motion for the mass on the spring.
Problem 2: A mass of 2 kg stretches a spring by 0.2 m. At time t = = 0, the mass is released from rest from
a position 0.5 meters below equilibrium. Assume a damping force numerically equal to 8 times the
velocity, and assume the mass is driven by a external force of f(t) = 2e¯t.
a) Find the spring constant.
b) Set up an IVP modeling the situation.
c) Find the equation of motion for the mass on the spring.
Transcribed Image Text:Answer key:
1c) x(t): = -0.1 cos
12
2c) x (t) = -3e-2t cos(3 √5 t) - 135e-2t sin(3 √5 t) + et
47
138√5
46
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