A mass weighing $16$ pounds stretches a spring $\frac{8}{3}$ feet

PROBLEM STATEMENT:

A mass weighing $$16$$ pounds stretches a spring $$\frac{8}{3}$$ feet. The mass if initially released from rest from a point $$4$$ feet below the equilibrium position and the subsequent motion takes place in a medium that offers a damping force is $$\frac{1}{2}$$ times $$x^{'}$$ where $$x=x(t)$$. The mass is driven by external force given by $$f(t)=20cos(3t)$$. Convert mass = $$\frac{weight}{32}$$, Hook's constant Weight = k (length of stretch). Also the position below equilibrium is positive, and the downward motion is positive. Find the equation of motion.

My Working

Since the position below equilibrium is positive and the downward motion is positive we have:

$$x(0)=4$$ $$x'(0)=0$$

Now I have figured out the above initial conditions but can't seem to form the resulting differential equation of this problem. Once I do it, I can solve for the $$x(t).$$ Can anyone help in it. I will really appreciate that.

• Hint: This is a damped and driven harmonic oscillator. The solution to such is $m\ddot{x}+A\dot{x}+kx=C$, where A is the damping force, k is the spring constant, and C is the driving force Commented Jul 20, 2021 at 9:09
• I have slight problem. According to you $A$ is damping force then from the statement of the problem we see that $A=2\dot{x}$, now When we put the value of $A$ the result is the non-linear DE. Isn't that? Commented Jul 20, 2021 at 9:38
• The unit of $x'$ would be something like feet/fortnight, which is certainly not a unit of force?! And $f(t)$ carries no unit at all -- I don't know how to interpret this problem statement Commented Jul 20, 2021 at 9:43
• How do you approach $A$ - the damping force? Commented Jul 20, 2021 at 9:44
• @Ameer786: In my setup, A=2 (the $\dot{x}$ is implied) Commented Jul 20, 2021 at 11:01

After the $$k$$ determination from static considerations, we have
$$m \ddot x = -\mu \dot x - k x + f(t),\ \ \ x(0)=4,\ \dot x(0) = 0$$
here $$\mu = \frac 12$$