Hey guys I'm doing this exam paper question and I know a quesiton like this will come up. My exam is on Monday I just need someone to check over it and tell me if everything is OK. I'm unable to ask anyone for help. Thanks in advance.

A particle of mass $m$ hangs from an elastic string of natural length $l_0$ and modulus of elasticity $\lambda.$ The mass experiences air resistance forces $2mk\dot{x}$, where $x$ is the position of the mass measured from the point of suspension.

(a) Write down the equation of motion.

$${{F}}_{net}= m\ddot{x} = mg - T-R $$

Taking the downwards direction to be positive.

$$m\ddot{x}= mg-\frac{\lambda}{l_0}(x-l_0)-2mk\dot{x}$$

where $T=\dfrac{\lambda l}{l_0}$ and $l=x-l_0$


This is the equation of motion.

(b) Find the equilibrium position of the particle.

The equilibrium position is defined as the place when $F_{net} = 0 \implies m\ddot{x}=0$.

$$\ddot{x}=0 \,\,\, \text{and} \,\,\,\, \dot{x}=0$$ Letting $x=x_e$ we have

$$\begin{align} \ddot{x}+2k\dot{x}+\dfrac{\lambda x_e}{ml_0} & =g-\dfrac{\lambda}{m} \\ x_e & = \dfrac{mgl_0}{\lambda}+l_0 \end{align}$$

This is the equilibrium position of the particle.

Suppose that $k^2<\dfrac{\lambda}{ml_0}$. Derive the general solution to to the equation of motion. (Hint: Consider a suitable coordinate change.)

To solve the equation we need to introduce a new coordinate system to convert the equation of motion to a fully homogeneous equation. Let

$$y=x-x_e, \,\,\,\,\,\, \dot{y}=\dot{x}, \,\,\,\,\,\, \ddot{y}=\ddot{x}$$

Re-write everything in terms of $y$. I skipped the algebra here, but this is my final result.

$$\ddot{y}+ 2k\dot{y} +\dfrac{\lambda}{ml_0}y=0$$

This is our equation of subsequent motion in terms of y. It's a SHM equation. We can substitute $\omega^2 = \dfrac{\lambda}{ml_0}$ to get

$$\ddot{y}+ 2k\dot{y} +\omega^2y=0$$.

For $k^2<\dfrac{\lambda}{ml_0} \implies k^2 < \omega^2 $ it's just the SHM equation.To solve we use the trial function.

$$y(t) = e^{\lambda t} -x_e \,\,\,\,\,\, \dot{y}= \lambda e^{\lambda t} \,\,\,\,\,\, \ddot{y}=\lambda^2 e^{\lambda t}$$

Inserting these properties into our original equation and eliminating $e^{\lambda t}$ we obtain the characteristic equation

$$\lambda^2 + 2k\lambda +\omega^2 = 0$$.

I don't really understand how I got the roots, I read it from another book. The roots are 2 distinct complex conjugates of each other

$$ \lambda_1 = -k+i\sqrt{k^2-\omega^2} \,\,\, \text{and} \,\,\, \lambda_1 = -k-i\sqrt{k^2-\omega^2}$$

The general solution is

$$x(t) = Ae^{\lambda_1 t}+Be^{\lambda_2 t} +x_e$$

This solution was found using the property of superposition for linearly homogeneous equation. Putting $\omega' = \sqrt{\omega^2 - k^2}$ the general solution can be written as

$$x(t) = Ae^{-kt}\sin\omega' t +x_e$$

How is this the case? I only found this cause I read it from an applied mechanics book. Is there another way to do this part of the problem to make it easier?

Identify the type of motion.

The motion is oscillatory, with reduced frequency of oscillation decreasing with exponential amplitude. Because $k^2 < \omega^2$ the case is weak-damping because the effect of the damping term is not sufficient to prevent oscillation.

Suppose that at time $t=0$ the particle is given a velocity $U$ upwards from its equilibrium position. Assuming that the string never gets slack, find the particle's subsequent position $x(t)$.

For $t=0$ we have

$$x(0) = x_e \,\,\,\, \text{and} \,\,\,\, \dot{x}(0) = U$$

$$x(t) = Ae^{-kt}\sin\omega' t$$

$$ \dot{x}= -kAe^{-kt}\sin\omega' t+ \omega' Ae^{-kt}\cos\omega' t$$

Plugging in the I.C we have


$$U = \omega' A$$ So our particular solution is

$$x(t) = \dfrac{U}{\omega'}e^{-kt}\sin\omega' t + x_e$$.

After creating a new coordinate system and using it to simplyify the Differential Equation, I found it quite difficult coming up with a general solution for the case where $k^2<\omega^2$. I'm starting to think there must be an easier way, or a more correct way of doing it. Thanks for looking at the problem.

  • $\begingroup$ After the sentence "we use the trial function...etc" you are solving for $x$ but this should be $y$ $\endgroup$ – David Quinn May 3 '17 at 20:08
  • $\begingroup$ It is easy to see that your final answer is wrong because the centre of the DHM should be the equilibrium position, not the point of suspension $\endgroup$ – David Quinn May 3 '17 at 20:09
  • $\begingroup$ I changed the general and particular solution, how is it now? $\endgroup$ – Patrick Moloney May 3 '17 at 20:17
  • $\begingroup$ Your use of $x$ and $y$ are still somewhat confused. The final answer seems correct, but it looks like you need to check out the methods for solving second order linear differential equations, particularly regarding what complementary functions result from what auxiliary quadratic equations, and how to deal with the change of variable. $\endgroup$ – David Quinn May 3 '17 at 20:38

Technically SHM is only with no damping. This is damped harmonic motion. There are actually three cases: $k^2<\omega^2$ is underdamped, $k^2=\omega^2$ is critically damped, and $k^2>\omega^2$ is overdamped. Intuitively, underdamped causes it to oscillate while the amplitude decreases, overdamped is "too damped" to carry on past the equilibrium point. Critical damping is a bit weird: it returns to equilibrium just slowly enough that it doesn't overshoot.

The roots of the characteristic equation $\lambda^2+2k\lambda + \omega^2 =0$ are $$ \lambda_{\pm} = -k \pm \sqrt{k^2-\omega^2} = -k \pm i\sqrt{\omega^2-k^2}, $$ using the quadratic formula as usual. In the underdamped case, the general solution can be written in one of two real forms, $$ e^{-kt}(A\cos{\omega' t} + B \sin{\omega't}) $$ or $$ Re^{-kt}\sin{\omega'(t-t_0)}. $$ These are equivalent using the trigonometric addition formulae. These two forms come from the formulae $$ \cos{z} = \frac{e^{iz}+e^{-iz}}{2} \qquad \sin{z} = \frac{e^{iz}-e^{-iz}}{2i} ,$$ which allow us to combine the complex exponentials into real functions provided that certain conditions are satisfied (specifically, that $y$ and $y'$ are real at a specific point).

Now, if $y=0$ when $t=0$, the coefficient of the cosine term is zero. Differentiating, $$\dot{y}(0) = B(ke^{-k0}\sin{\omega'0}-\omega'e^{-k0}\cos{\omega'0}) = B\omega',$$ so $B=U/\omega'$. Now, remember that $x=y+x_{e}$, so the solution is $$ x(t) = x_e + \frac{U}{\omega'} \sin{\omega't}. $$ I think that's everything you asked about.

  • $\begingroup$ Thank you very much again. I forgot to add my $x_e$ there at the end. $\endgroup$ – Patrick Moloney May 3 '17 at 20:14

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