My thoughts led me to change the question's title. Here's what I've tried in solving the problem and led me to ask the question in the title.

We have the following diagram where the beam has no mass and can possibly rotate with respect to the wall.

enter image description here

I'm trying to find the oscillation equation for the mass $m$. I'm a little confused as to what happens if the beam has no mass. The equations show me that the spring on top plays no part in the oscillation. I'm gonna list my thoughts :

1) Since the beam has no mass, all the forces and torques sum up to zero. Given that the beam is pivoted to the wall we know it makes no movement at all. If it did, then it would be constant which is not acceptable if the mass m is supposed to oscillate.

2) For the beam and its torques with respect to the point on the wall:


If we move the mass $m$ by $x$ then the torques give $(k)x(L)=(k/2)x'(2L)=>x=x'$ Which means the springs are extended by the same length any moment. This is strange given that the beam is not moving. I already think I've made some wrong assumptions.

3) Taking the forces acting on the mass $m$ we have: $$ΣF=m\ddot x=>\\-kx-c\dot x=m\ddot x=>\\m\ddot x+c\dot x+kx=0$$ which is the standard equation for a simple harmonic oscillation and the spring constant k2 doesn't appear which is strange again. Could it be that the spring's elongation somehow depends on something else ( the spring k2 or the beam for example ) .

Note: Every problem in this section of the book I'm following ends up in this equation : $m\ddot x+c\dot x+kx=0$ This one will be no different. If we assume that the beam rotates, we'll have to assume that the angle is minimal and do the following approximation: if the beam rotates an angle $φ$ then its center has moved a distance $φL$ - the right end , similarly, moves a distance $φ2L$. In that case we'll see that $x'=2x$ which contradicts the $x=x'$ relationship we found above. That is because previously I assumed that the beam remains horizontal.

All in all, I keep making assumptions. Please, a force analysis on the mass m would help me a lot. It's no homework, I've been on this for too long and I just can't skip it and go to the next exercise.

I'm gonna give it another go assuming that the beam has an angle after we move the mass m. I will upload when I'm done.

Second effort for solution

Let's say the bottom spring is spring 1 and the top spring is spring 2. When we move the mass m by $x$ the rod rotates by an angle $θ$, thus reducing the extension of spring 1 approximately by $θL$ but extending spring 2 by the same amount.

Newton's law for the mass gives: $$ΣF=m\ddot x=>-F_{spring1}-c\dot x=m\ddot x$$

where $F_{spring1}= k(x-θL)$

We can also use Newton's law for the rotation of the rod taking torques with respect to the point of the wall (approximately, for small angle θ) : $$ΣT=0=>F_{spring1} L =F_{spring2} 2 L=>k(x-θL)L=\frac{k}{2}(θ2L) 2 L\\=>x-θL=2θL=>x=3θL$$ Now we can replace $θ$ in the mass's equation and get : $$m\ddot x+c\dot x+k(x-x/3)=0=>m\ddot x+c\dot x+\frac{2k}{3}x=0$$ The equivalent spring and mass system has a spring constant of 2k/3. Is this suspicious?

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Aug 28 '18 at 23:17
  • $\begingroup$ If you use \Rightarrow in the MathJax, you'll get $\Rightarrow$ which looks way nicer than the => you've used. $\endgroup$ – Kyle Kanos Aug 29 '18 at 10:04
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    $\begingroup$ I'm not sure why this question was closed. It does appear to ask about a specific concept, namely a conceptual difficulty about how "massless rods" should be treated in physics problems. The lengthy attempts to work through the problem all stem from that, and show significant effort to work through the problem. $\endgroup$ – Michael Seifert Sep 17 '18 at 21:44
  • $\begingroup$ @MichaelSeifert I agree that there's a conceptual question in this. The problem I have is the final question "Is this suspicious?" That's a check-my-work question. If there's a way to remove that bit, it would be a good question. I'll vote to reopen anyways; hopefully someone'll be able to think of how to remove that bit without destroying the meaning of the question. $\endgroup$ – user191954 Sep 22 '18 at 10:10
  • $\begingroup$ @Chair From what I can recall when I was writing the question, at first I couldn’t get my solution to the end. I was constantly trying to take it a bit further from things I’ve learnt either from the answers or by searching online . At some point I did get to the solution and I wondered if it is suspicious that the equivalent spring constant is less than 1k. Since we have two springs with spring constant k I wouldn’t expect such a result . More like something between 1k and 2k. For no particular reason , just intuitively as I don’t remember much on this stuff. $\endgroup$ – John Katsantas Sep 22 '18 at 10:19

Since the beam has no mass, all the forces and torques sum up to zero. Given that the beam is pivoted to the wall we know it makes no movement at all. If it did, then it would be constant which is not acceptable if the mass m is supposed to oscillate.

This is your problem here. A "massless" object has a negligible net force or torque on it, but that net force/torque is "negligible" in the same sense that the mass is negligible. And a small quantity divided by a small quantity is not necessarily a small quantity, so even if $F$ and $m$ are negligible, $a = F/m$ is not necessarily negligible.

As a simpler example, consider two boxes of masses $M_1$ and $M_2$ connected by an inextensible rope of mass $m$. We apply a force $F$ to $M_1$ in the $x$-direction. What is the acceleration of the system? It's not too hard to work out the equations of motion for each of the three masses in question; they are: \begin{align} M_1:& & F - T_1 &= M_1 a \\ M_2:& & T_2 &= M_2 a \\ m:& & T_1 - T_2 &= m a, \end{align} where $T_1$ is the force between the rope and $M_1$, while $T_2$ is the force between the rope and $M_2$. Now, if $m$ is negligible, the last equation means that $T_1 \approx T_2$. In other words, the tension is constant throughout the rope. But even though the net force on the rope is zero this does not mean that the acceleration of the rope is zero; it merely means that $m$ is small enough that we can neglect the difference between $T_1$ and $T_2$.

Alternately, we can solve the above equations to yield $$ a = \frac{F}{M_1 + M_2 + m}, $$ and in the limit of $m \ll M_1, M_2$, you get $a = F/(M_1 + M_2)$ — exactly the result you would expect by simply ignoring the mass of the rope and assuming that the rope simply "transmits" the force between the boxes.

Similarly, in your situation, the mass of the rod is "small". If we take the mass of the rod to be negligible, the torque from the upper spring on the rod must be approximately equal to the torque from the lower spring. With this in mind, you can figure out the force exerted by the rod on the lower spring, and (assuming that the lower spring is massless!) figure out the force exerted by the lower spring on the mass $m$. But the two torques are not exactly equal to each other, and that slight difference, while negligible for the purposes of determining the motion of $m$, is still enough to cause the rod to move.

  • $\begingroup$ That was so nice to read after all these hours of thought. Now that this is out of the way, I'd like to ask one more thing about solving the problem. It is clear now that the rod is moving during the oscillation. Is this problem still first order of freedom? Can I find the mass's oscillation without the rod's angle showing up anywhere? Because I don't think there is a relationship between the angle and the mass's position. $\endgroup$ – John Katsantas Aug 28 '18 at 20:34
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    $\begingroup$ @JohnKatsantas: If you treat this system without the mass of the rod being "small", then you have two degrees of freedom and a coupled oscillation problem. The amplitudes of the oscillations in each of the normal modes will depend on the masses of the rod and the block. ... $\endgroup$ – Michael Seifert Aug 28 '18 at 20:46
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    $\begingroup$ My intuition is that in the limit $m_\text{rod} \ll m_\text{block}$, one of the normal modes would reduce to one where the rod & block are moving in sync and the motion would depend only on $m_\text{block}$; this is the mode you're being asked to find in your problem. The other mode would be one where the block remains fixed, while the rod oscillates back & forth at a different frequency. $\endgroup$ – Michael Seifert Aug 28 '18 at 20:46

\begin{align*} &\textbf{Beam Equation}\\ &\Theta\ddot{\varphi}=f_{K1}\,L\cos(\varphi)+f_{K2}\,2\,L\cos{\varphi}\\ &\text{with}\\ &f_{k1}=k\,\left(L\sin(\varphi)-y\right)\\ &f_{k2}=\frac{k}{2}\,\left(2\,L \sin(\varphi)\right)\\ &\text{and}\quad \Theta=0\quad\text{we get}\\ &\varphi=\arcsin\left(\frac{1}{3}\frac{y}{L}\right)&(1)\\\\ &\textbf{Mass Equation}\\ &m\,\ddot{y}=-k\,\left(L\sin(\varphi)-y\right)-c\,\dot{y}\\ &\text{with equation (1)}\\ &m\,\ddot{y}=-\frac{2}{3}\,k\,y-c\dot{y} \end{align*}


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