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A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 9 times the instantaneous velocity. Determine the initial conditions and equations of motion if the following is true. (a) the mass is initially released from rest from a point 1 meter below the equilibrium position

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User Bratsche
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1 Answer

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Thus the initial conditions and equations of motion are given as;

x(0) = -1 mx'(0) = 0mx'' + 9x' + 14x = -10mx''' + 9x'' + 14x' = 0, which can be written as;m d²x/dt² + 9dx/dt + 14x = -10.

Given: mass of 1kg, spring constant k = 14 N/m, damping force = 9 v, Initial displacement x=1m (below equilibrium)

From the law of conservation of energy, total energy of the system is constant. At the equilibrium point the entire energy is stored in the spring in the form of potential energy. This potential energy is given by U = ½ kx².At the position x the gravitational potential energy of the system is mgx. Therefore, at position x, the total energy of the system is given by;

E = U + K + GPE, where K is the kinetic energy and GPE is gravitational potential energy.

At position x, GPE = 0, K = 0 and U = ½ kx².

So, the total energy of the system is;

E = ½ kx², E = ½ × 14 × 1² = 7 Joule.

Since the system is submerged in a liquid that imparts a damping force numerically equal to 9 times the instantaneous velocity, the damping force is 9v.

By Newton's second law of motion, F = ma, where m is the mass and a is the acceleration of the mass.The acceleration of the mass is given by;

ma = net force = restoring force - damping force - weight

Force acting on the mass is;

F = -kx - bv - mg,

where b is the damping constant, and v is the velocity of the mass.

Therefore, the equation of motion of the mass is given by the following second-order differential equation:

mx'' + bx' + kx = -mgwhere x" and x' are first and second derivatives of x with respect to time respectively.

Substituting the given values of k, b, m and g into the above equation;

1x'' + 9x' + 14x = -10 (note that g = 10 m/s²).

The initial condition of the mass is that the mass is initially released from rest from a point 1 meter below the equilibrium position. Hence, x(0) = -1 m and x'(0) = 0.

Differentiating the above equation w.r.t time we get;

1x''' + 9x'' + 14x' = 0

Thus the initial conditions and equations of motion are given as;

x(0) = -1 mx'(0) = 0mx'' + 9x' + 14x = -10mx''' + 9x'' + 14x' = 0, which can be written as;m d²x/dt² + 9dx/dt + 14x = -10.

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User McNab
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