Question

A spring mass system consists of a spring with spring constant k and an attached block of mass m is submerged in a liquid that produces a damping force Fᵣ. m=1Kg Fᵣ =11 times the instantaneous velocity of the center of mass of the block. k=18 N/m If the mass is initially released from rest 1 meter below the equilibrium position.

a. Give a second degree equation that describe the motion of the center of mass of the attached block

Answer 1

The motion of the center of mass of a damped spring-mass system can be characterized by a second-degree differential equation, which for the given parameters, is represented as 'm d²x/dt² + γ dx/dt + kx = 0' with γ being the damping coefficient.

Step-by-step explanation:

Equation of Motion for a Damped Harmonic Oscillator

The motion of the center of mass of the attached block in a damped spring-mass system can be described by a second-degree differential equation. Since the mass m is 1 kg, the damping force Fₖ is 11 times the instantaneous velocity, and the spring constant k is 18 N/m, we can write the equation of motion as:

m d²x/dt² + γ dx/dt + kx = 0

Where x is the displacement from the equilibrium position, t is the time, γ is the damping coefficient which is 11 in this case. Given that the mass is released from rest 1 meter below the equilibrium position, the initial conditions for the system are: x(0) = -1 meter and v(0) = dx/dt(0) = 0 m/s. The negative sign indicates that the mass is below the equilibrium position.