Question

A spring with a 10-kg mass and a damping constant 15 can be held stretched 1 meters beyond its natural length by a force of 2 newtons. Suppose the spring is stretched 2 meters beyond its natural length and then released with zero velocity. In the notation of the text, what is the value c^2 - 4mk? _________ m^2 kg^2/sec^2.

Find the position of the mass, in meters, after t seconds. Your answer should be a function of the variable t of the form c_1 e^alpha t + c_2 e^beta t where alpha = (the larger of the two) beta = (the smaller of the two) c_1 = _______c_2 =______.

Answer 1

Answer:7.5=−k(2.5)−7.52.5=kma=−7.5x2.5my′′+9y′+−3y=0,y(0)=2.5,y(5)=0−9±81+4(m)(3)−−−−−−−−−−−√2m−92m±81+12m−−−−−−−−√2my=Ae−(9/2)xcos(81+12m−−−−−−−−√2mx)+Be−(9/2)xsin(81+12m−−−−−−−−√2mx)2.5=A+B⋅00=(2.5)e−45/2cos(81+12m−−−−−−−−√52m)+Be−45/2sin(81+12m−−−−−−−−√52m)

Explanation: