Damped Vibrations

Consider a simple spring-mass-damper system as shown in the figure. One end of the spring and damper is fixed to a wall while the other is fixed to a body of mass M M . In this question, the effect of gravity is ignored.

M = 20 k g M = 20 kg

k = 800 N / m k = 800 N/m

c = 40 N s / m c = 40 Ns/m

The mass is given a massive displacement of 10 m 10 m downward (from its equilibrium position) and released from rest at time t = 0 t = 0 . The goal of the question is to compute the rate at which the system loses energy. Enter your answer as the magnitude of the rate of total energy loss of the system in J / s J/s at time t = 2 t = 2 seconds.

Bonus 1: Is the system over, under or critically damped? Why?

Bonus 2: If the value of the damping coefficient c c happened to be negative, how would one interpret the situation physically? Is such a situation possible in reality?

Note: The damping force is the product of the damping coefficient c c and the instantaneous velocity of the mass. Its direction is opposite to that of the velocity of the mass. The spring is linear and obeys Hooke's law.


The answer is 17.4981.

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1 solution

Karan Chatrath
Apr 13, 2019

I will briefly lay out the steps of the problem.

  • Solve the following differential equation for x x as a function of time: M x ¨ + c x ˙ + k x = 0 M\ddot{x} + c\dot{x} + kx = 0 , where, x ( 0 ) = 10 x(0) = 10 and x ˙ ( 0 ) = 0 \dot{x}(0) = 0 .
  • Having obtained the particular solution x ( t ) x(t) , differentiate with respect to time to obtain x ˙ ( t ) \dot{x}(t) .
  • The total energy of the system is the sum of the kinetic energy of the mass and potential energy stored in the spring: E = 1 2 k x 2 + 1 2 M x ˙ 2 E = \frac{1}{2}kx^2 + \frac{1}{2}M\dot{x}^2

Differentiating the energy with respect to time: d E d t = x ˙ ( k x + M x ¨ ) \frac{dE}{dt} = \dot{x}(kx + M\ddot{x}) .

We know that: M x ¨ + c x ˙ + k x = 0 M\ddot{x} + c\dot{x} + kx = 0 . This implies: M x ¨ + k x = c x ˙ M\ddot{x} + kx = -c\dot{x}

Replacing the above result in the 'rate of energy change' equation gives us: d E d t = c x ˙ 2 \frac{dE}{dt} = -c\dot{x}^2

  • Replacing x ˙ \dot{x} with its particular solution, gives us rate of energy change as a function of time. The negative sign indicates that energy is being dissipated from the system. Substitute t = 2 t = 2 and obtain the answer.

I am, for now, refraining to answer the bonus questions posed. These are conceptual questions worth thinking about.

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