3
$\begingroup$

Hanging Spring.

A 10 kilogram mass suspended from the end of a vertical spring stretches the spring $\frac{49}{90}$ metres. At time t = 0, the mass is started in motion from the equilibrium position with an initial velocity of $1 $ $m/s$ in the upward direction. At the same time, a constant downward force of $360$ Newtons is applied to the system.

Assume that air resistance is equal to $60 $ times the instantaneous velocity and that the acceleration due to gravity is $g = 9.8 $ m/s^2.

(a) Determine the spring constant.

(b) Show that the equation of motion is

$$ \ddot{x} + 6 \dot{x} + 18x = 36 $$

where $x(t)$ is the displacement of the mass below the equilibrium position at time $t$ . In your answer include a diagram of all forces acting on the mass.

(c) Find the position of the mass at any time. Would you describe the motion as overdamped, underdamped or critically damped?

I'm struggling with part (c) and any help would be appreciated.

Solving for $\lambda$ I know that

$$ X_p =A\cdot t\cdot e^{-3t}\cdot\cos(3t)+B\cdot t\cdot e^{-3t}\cdot\sin(3t) $$

since without $t$, using the equation gives a result of zero,

using

$$ \ddot{x} + 6 \dot{x} + 18x = 36, $$

with $X_p$ I get

$$ 6=e^{-3t} (b \cos(3t)-a \sin(3t))$$

but the answer say

$$ x(t) = −2 e^{−3t} \cos (3t) −\frac73 e^{−3t}\sin (3t) + 2 $$

where does this $+2 $ come from, and how is the equation used without multiplying by t if it equals to zero? Sorry if this question is unclear and if any clarification is needed please ask.

$\endgroup$
  • $\begingroup$ You need to format you're questions using latex here is a good link to get you started. $\endgroup$ – BLAZE Sep 26 '15 at 10:39
  • $\begingroup$ Where is $\lambda$ defined in your question? $\endgroup$ – BLAZE Sep 26 '15 at 10:56
  • $\begingroup$ I haven't defined $\lambda$ in the equation and honestly I might have a lack of understanding about what it is, but from what i've done in the past i've used $ \lambda ^{2}+6\lambda +18=0 $ and solved to get $\lambda$=-3$\frac{+}{-}$$3i$ which I used to find my x(t) $\endgroup$ – Sushifish Sep 26 '15 at 13:06
  • $\begingroup$ Thanks for formatting my question $\endgroup$ – Sushifish Sep 26 '15 at 13:07
  • 1
    $\begingroup$ Remember the method of undetermined coefficients: in this case, you assume that $x=k$ (constant) is a solution of the equation, so $18k=36$. The general solution is $x=2\,+$ the general solution of the associated homogeneous equation that you already found. $\endgroup$ – Tony Piccolo Sep 26 '15 at 13:21
1
$\begingroup$

Your $X_p$ without the extra $t$ is the solution of the homogeneous equation, i.e. when you set the RHS equal to zero.

To get the complete solution, you need to add a particular solution of the non-homogeneous equation.

In $$ \ddot{x} + 6 \dot{x} + 18x = 36, $$ if you try a constant solution $x=C$, the equation simplifies to

$$0+0+18x=36$$ and the constant is $2$.

Hence the general solution

$$e^{-3t}(A\cos 3t+B\sin 3t)+2.$$

Remains to determine the constants using the initial conditions.

$\endgroup$
0
$\begingroup$

Notice, we can find out the solutions as follows

a) let $k$ be the spring constant of the spring then the elongation of the spring when it is acted upon by a force $mg=10\times 9.8=98\ N$ is given as $$=\frac{\text{applied force}}{\text{spring constant }}=\frac{mg}{k}$$ $$ \frac{98}{k}=\frac{49}{90}$$$$\implies \color{red}{k=180\ N/m}$$

b) Now, let $x$ be the downward displacement of the mass at time $t$ when a total $F_t=360\ N$ force is applied in the downward direction at the equilibrium state of mass then let's consider all the forces acting on the mass as follows

The spring force acting in the upward direction$$F_s=kx=180 x$$ Air resistance acting in the upward direction $$R_a=60v=60\frac{dv}{dt}=60\dot x$$ Restoring force acting in the upward direction $F_r=ma=m\frac{d^2x}{dt^2}=10\ddot x$

Now, for the dynamic equilibrium of mass, we have $$\text{net force acting in the upward direction}=\text{net force acting in the downward direction}$$ $$F_s+R_a+F_r=F_t$$ $$180x+60\dot x+10\ddot{x}=360$$ $$\ddot{x}+6\dot x+18x=36$$

c) Now, the damping factor $$=\frac{\text{damping coefficient}}{2\sqrt{km}}$$ $$=\frac{60}{2\sqrt{180\cdot 10}}=0.707<1$$ Hence, the vibrations are under-damping.

For finding out the position of the mass solve the differential equation for $x$ in terms of time $t$.

$\endgroup$
0
$\begingroup$

Since this is a linear autonomous differential equation, you can always add a linear combination of the general solutions. Here the general solution solves the homogeneous differential equation, so with the right-hand-side of the differential equation equal to zero.

I believe you did find the correct general solutions, but you are incorrect by multiplying it by $t$. When you have a nonhomogeneous differential equation, you have to find the particular solution, which solves the nonhomogeneity. If the nonhomogeneity is a polynomial of order $n$, then there will always be a polynomial of also order $n$ which solves the nonhomogeneous differential equation, and thus would be the particular solution.

The generalized solution to your differential equation would thus be equal to the particular solution plus a linear combination of the general solutions.

PS: you only have to multiply by $t$ if the right-hand-side of the nonhomogeneous differential equation is equal to a linear combination of the general solutions, of if you have a repeating eigenvalues.

$\endgroup$
0
$\begingroup$

$$ \ddot{x} + 6 \dot{x} + 18x = f(t) $$ $$ x(t) = x_h(t) + x_p(t) $$ $$ x_h(t) : \ddot{x} + 6 \dot{x} + 18x = 0 \rightarrow x_h(t) = C_1 e^{-\lambda_1 t} + C_2 e^{-\lambda_2 t} $$ $$ x_p(t) : \ddot{x} + 6 \dot{x} + 18x = f(t) = 36 $$ since $f(t)$ is constant you can assume a $x(t) =$ constant $$ x(t) = k \\ 0 + 0 + 18k = f(t) = 36 \rightarrow k = 2 $$ $$ x(t) = x_h(t) + x_p(t) = C_1 e^{-\lambda_1 t} + C_2 e^{-\lambda_2 t} + 2 $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.