Question

A spring with a 2-kg mass and a damping constant 5 can be held stretched 1.5 meters beyond its natural length by a force of 7.5 newtons. Suppose the spring is stretched 3 meters beyond its natural length and then released with zero velocity, In the notation of the text, what is the value c² −4mk ?

Answer 1

The value of c² −4mk for the spring mass system with damping, given the damping constant of 5, mass of 2 kg, and calculated spring constant of 5 N/m, is −15.

Step-by-step explanation:

The student's question pertains to the dynamics of a damped harmonic oscillator, which in this case is a mass-spring system with damping. To find the value of c² −4mk, we first recognize that the damping constant is given as 5 (usually denoted by 'c'), the mass (m) is 2 kg, and we need to determine the spring constant (k).

Since a force of 7.5 newtons stretches the spring 1.5 meters, the spring constant (k) can be calculated using Hooke's Law, F = kx, where F is the force and x is the displacement. Therefore,
k = F/x = 7.5 N / 1.5 m = 5 N/m.

The equation c² −4mk then becomes (5²) − (4 × 2 kg × 5 N/m), which calculates to 25 − 40 = −15. So, the value of c² −4mk for the given spring mass system with damping is −15.