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Difficult Exam paper question that I can't seem to grasp. I'm finding it hard to visualise the problem and understand what is actually going on.

A particle of mass $m$ moves in a straight line on a smooth horizontal table, and is connected to two points $A$ and $B$ by light elastic springs of natural lengths $2l_0$ and $3l_0$ respectively, and modulus of elasticity $\lambda$. The points $A$ and $B$ are a distance $6l_0$ apart. Show that the equation of motion is: $$m\ddot{x} =\frac{\lambda}{6l_0}(12l_0 - 5x)$$

(where $x$ is the displacement of the particle from $A$ measured positive towards $B$.

What I know is, the two spring have the same modulus of elasticity λ but different equilibrium length. λ is a property of the material, not the springs and it take in account the relative deformation for each spring. I plugged in the formula I found online for modulus of elasticity and incorporated into the general equation of motion for a spring.

The whole length between A and B is $6l_0$ and the combined length of the springs is $5l_0$.

$$\begin{align} \mathbf{F_n} & = \mathbf{F_A} + \mathbf{F_b} \\ & = k_A \Delta L + k_B \Delta L \\ & = k_A(L-L_0) + k_B(L-L_0) \\ & = k_A(6l_0 - 2x) + k_B(6l_0 - 3x) \\ & = \lambda\Big(\frac{6l_0-2x}{6l_0}\Big) + \lambda\Big(\frac{6l_0-3x}{6l_0}\Big) \\ & = \frac{\lambda}{6l_0}(12l_0-5x) \end{align} $$

Find the equilibrium position and obtain the frequency $\omega$ and period $T$ of oscillations.

The equilibrium position occurs when the net force acting on the mass is $0$. i.e when the acceleration is zero.

$$\mathbf{F_n} = m\ddot{x} = 0$$

Using the definition of equilibrium and setting $x= x_e$ we get

$$ \begin{align} \frac{\lambda}{6l_0}(12l_0 - 5x) & = 0 \\ x_e & = \frac{2}{5}\lambda \end{align} $$

This $x_e$ is the equilibrium position.

Using the formula for the frequency

$$ \begin{align} \omega & = 2\pi f \\ f & = \frac{1}{2\pi}\sqrt{\frac{k}{m}} \\ \end{align} $$

The formula that related the spring constand and the modulus of elasticity is $$k= \frac{\lambda}{L}$$

there is two springs so I really don't know how to take that into consideration. I found the equilibrium position(I think), do I just use that result to find the frequency and period? Because the equilibrium position is related to $\lambda$ can I relate that to $\omega$ and $T$?

The particle, at the equilibrium position, is suddenly given a velocity $U$ towards $B$. Obtain its subsequent displacement $x(t)$. How close to $B$ does it approach in the subsequent motion?

I imagine that this is asking me to solve the differential equation that was given in the question. To find a particular solution we will need two initial condition won't we? The starting position $x_0$ and the starting velocity $\dot{x_0}$

$x_0 = x_e$ The system is in equilibrium before the mass gets pushed.

$\dot{x_0} = U$ The initial velocity is actually given to us in the question.

I think the equation is non-homogeneous. Does the equation need to be changed to solve it? I just started second order differential equations and finding it slightly tricky here.

I appreciate all the feed back thank you.

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Everything is fine until your calculation of $x_e$. The solution to $$ \frac{\lambda}{6l_0}(12l_0-5x_e) = 0 $$ is $x_e = \frac{12}{5}l_0$: it can't depend on $\lambda $ because only the bracket must be zero, the constant outside is irrelevant.


For the frequency, we have to look at the differential equation, which becomes $$ \ddot{x} = \frac{5\lambda}{6ml_0} (x_e-x) $$ if we insert $12l_0 = 5x_e$ in the bracket. The specific value of the equilibrium position itself is not important (we chose our coordinates for $x$ with the origin in a particular place, but we can equally well choose anywhere else as the origin. Hence the way the particle moves does not vary depending on the actual value of $x_e$, so the path is of the form $(x(t)-x_e)+x_e$. We can set $y=x-x_e$, and then the differential equation becomes the standard one for simple harmonic motion, $$ \ddot{y} = -\frac{5\lambda}{6ml_0} y. $$ The general solution to an equation of the form $\ddot{y}=-\omega^2y$ is $$A\cos{\omega t}+B\sin{\omega t},$$ and $\omega$ (normally assumed positive) is therefore called the frequency; the period $T$ is then just $2\pi/\omega$. Therefore in this case, $\omega = \sqrt{5\lambda/(6ml_0)}$.


You have the right initial conditions for the motion; we now have to find the right solution out of the general form I wrote down earlier. Since $x=x_e+y$, and $x(0)=x_e$, $y(0)=0$, and $\dot{x}(0)=\dot{y}(0)=U$. Thus $A=0$, and differentiating, $$ \dot{y} = B\omega \cos{\omega t}, $$ where $\omega$ has the value we found above. Hence $B\omega=U$, so the motion is given by $$ x(t) = x_e + \frac{U}{\omega} \sin{\omega t}. $$


The maximum distance towards $B$ occurs when the sine has its maximum value of $1$; thus the lowest distance to $B$ is $$ 6l_0-x_e-\frac{U}{\omega}, $$ which I leave to you to calculate in terms of the original constants.

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  • $\begingroup$ Thank you very much. I see where I went wrong with the equilibrium position. Did you use a seperate coordinate system when you used $y= x -x_e$, to make it easier to solve the differential equation? $\endgroup$ – Patrick Moloney Feb 22 '17 at 0:46
  • $\begingroup$ Yes: one could equally say that the solution to $ \ddot{x} = \omega^2(x_e-x) $ is $x = x_e + A\cos{\omega t}+B\sin{\omega t}$: the particular integral here is just a constant since constants in $x$ disappear on differentiation (this is actually equivalent to invariance under changing coordinates). $\endgroup$ – Chappers Feb 22 '17 at 0:49
  • $\begingroup$ Thanks a million for the insight. Very well constructed answer. $\endgroup$ – Patrick Moloney Feb 22 '17 at 12:36

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