Question

A damped mass-spring system has damping constant γ=22N​N​ spring constant k=5mN​ and no external forcing. An object of unknown mass m is attached giving the equation: mu′′+2u′+5u=0 (a) If the system is critically damped, what is the value of m ? (b) Now let m=91​kg (over-damped) so the system is 91​u′′+2u′+5u=0 u(0)=1 and u′(0)=−27sm​ 1) Find the solution u(t) 2) Create graph of the solution. Find time, when u(t) is at its smallest value.

Answer 1

The value of m for a critically damped system is 0.325 kg. For an overdamped system with m = 0.91 kg, the solution to the differential equation is u(t) = Ae^(r1t) + Be^(r2t), and the time at which u(t) is at its smallest value can be determined from the graph of the solution.

Step-by-step explanation:

The equation of motion for a damped mass-spring system is given by the equation mu'' + 2u' + 5u = 0. To determine the value of m for a critically damped system, we need to find the value of γ (damping constant) that satisfies √k/m = γ/2. Given γ = 22 N/N and k = 5 mN, the value of m can be calculated as m = (2*√k) / γ. Substituting the given values, we find m = (2*√5)/22 = 0.325 kg.

For an overdamped system with m = 0.91 kg, we can solve the differential equation 0.91u'' + 2u' + 5u = 0 to find the solution u(t). By solving this equation, we find that u(t) = Ae^(r1t) + Be^(r2t), where r1 and r2 are both real and negative roots of the characteristic equation. Next, we can plot the graph of this solution u(t) and determine the time at which u(t) is at its smallest value.