Question

A mass of 1.5 kg stretches a spring 0.05 mm. The mass is in a medium that exerts a viscous resistance of 240 NNwhen the mass has a velocity of 6 msms. The viscous resistance is proportional to the speed of the object.

Suppose the object is displaced an additional 0.06 mm and released.

Find an function to express the object's displacement from the spring's natural position, in mm after tt seconds. Let positive displacements indicate a stretched spring, and use 9.8 ms2ms2 as the acceleration due to gravity.

Answer 1

Taking note of the data given by the problem, we have:

DATA

W=1.5kg

X=0.05m

X(0)=0.06m

X'(0)=0

a=9.8m/s^2

Viscous resistance = 240N

Now, with all the data collected of the problem, establish the differential equation for the damped harmonic oscillator, like follows:

Calculate the value of m, B and k like follows:

• Mass:

• Damping coefficient

• Force constant

Now, making the corresponding substitution with the values of m, k and B;

Dividing all the equation by the mass, we have:  

Solve the corresponding differential equation as an homogeneous linear differential equation.

The solutions for this quadratic function are:

So, the solutions are:

Therefore, with the solutions, the function of the object´s displacement is:

Finally, in order to determine the values of the constants C1 and C2, apply the conditions in the beginning like follows:

• First condition: X(0)=0.06m

• Second condition: X'(0)=0

Now, with the values C1 and C2,the final expression for the object's displacement from the spring's natural position is: