A weight W is suspended from a rigid support by a hard spring with stiffness constant K . The spring is enclosed in a- hard plastic sleeve, which prevents horizontal motion, but allows vertical oscillations (see figure). A simple pendulum of length $l$ with a bob of mass m($mg<<W$) is suspended from the weight W and is set oscillating in a horizontal line with a small amplitude. After some time has passed, the weight W is observed to be oscillating up and down with a large amplitude (but not hitting the sleeve). It follows that the stiffness constant K must be? enter image description here

I dont even know where to begin. I know the problem is of harmonic motion and small oscillations. What i dont understand is after the weight $W$ starts bouncing up and down with large amplitudes, is that to be understood as the weight $W$ usurping energy from the oscillation of the mass $m$? If so, then how to find the amplitude?


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  • $\begingroup$ This website has examples of a mass on a spring: physicsclassroom.com/class/waves/Lesson-0/… $\endgroup$ – user108787 Oct 4 '16 at 12:36
  • $\begingroup$ I can solve basic problems on SHM. but this one seems a little strange which is why i posted here $\endgroup$ – Prasad Mani Oct 4 '16 at 12:39
  • $\begingroup$ Try this link instead, you have a 2D SHM problem, farside.ph.utexas.edu/teaching/336k/Newtonhtml/node28.html $\endgroup$ – user108787 Oct 4 '16 at 12:49
  • $\begingroup$ To those voting to close: this is not a particularly basic problem in SHM, which may explain why @PrasadSubramanian didn't "know where to begin". $\endgroup$ – Michael Seifert Oct 4 '16 at 15:45

The problem isn't 100% clear, and a full treatment would probably require the use of coupled oscillation techniques that you may or may not have learned yet. But if this is meant to be solved with "basic" techniques, here's how I would think about it:

  • For a normal pendulum, the tension in the string is largest when the string is passing through the horizontal (since its angular speed is largest there.)

  • Thus, if the pendulum has a frequency $f$, the tension in the string will oscillate with a frequency $2f$.

  • The pendulum string is therefore acting as a driving force with frequency $2f$ on the system consisting of the weight $W$ and the spring $k$. What's more, this driving force, at this frequency, causes the weight-spring system to oscillate with a "large amplitude".

Take it from there.

To sketch out a more formal technique, we can use Lagrangian mechanics. Let $Y$ denote the displacement of the upper mass from its equilibrium position, let $\theta$ denote the angle between the string and the vertical, and let $M = W/g$ denote the mass of the upper block. After some geometry, we can show that the Lagrangian for this system is $$ \mathcal{L} = \frac{1}{2} (m+M) \dot{Y}^2 - m \ell \sin \theta \dot{\theta} \dot{Y} + \frac{m}{2} \ell^2 \dot{\theta}^2 + m g \ell \cos \theta - \frac{1}{2} k Y^2, $$ and taking the associated Euler-Lagrange equations, we conclude that \begin{align*} M \ddot{Y} - m \ell \left(\cos \theta \dot{\theta}^2 + \sin \theta \ddot{\theta} \right) &= - k Y \\ - m \ell \sin \theta \ddot{Y} + m \ell^2 \ddot{\theta} &= - m g \ell \sin \theta. \end{align*}

We now can look for a formal power series solution to these equations: \begin{align} Y(t) &= \epsilon Y^{(1)}(t) + \epsilon^2 Y^{(2)}(t) + \dots \\ \theta(t) &= \epsilon \theta^{(1)}(t) + \epsilon^2 \theta^{(2)}(t) + \dots \end{align} We now want to plug these in to the Euler-Lagrange equations and expand them out order by order in $\epsilon$. At $\mathcal{O}(\epsilon)$, we find that $$ M \ddot{Y}^{(1)} = -k Y^{(1)}, \qquad m \ell^2 \ddot{\theta}^{(1)} = - m g \ell \theta^{(1)}, $$ from which we conclude that for small oscillations, we have simple harmonic motion in both $\theta$ and $Y$. Moreover, these oscillations are uncoupled; at this level of approximation, we would not see the behavior described in the problem.

To see the coupling effects between the two coordinates, we have to expand the Euler-Lagrange equations to $\mathcal{O}(\epsilon^2)$; if we do this, we get (after some algebra) \begin{align} M \ddot{Y}^{(2)} = m \ell \left( \left( \dot{\theta}^{(1)} \right)^2 + \theta^{(1)} \ddot{\theta}^{(1)} \right) -k Y^{(2)} \end{align} along with a similar equation for $\theta^{(2)}$. This latter equation can be rearranged to yield an undamped driven oscillator equation, where the function $\theta^{(1)}$ and its derivatives act as the "driving force" for the second-order perturbations $Y^{(2)}$. The fact that these oscillations become "large" allows us to say something about the values of $k$ and $M$.

  • $\begingroup$ i know what coupled oscillations, normal modes etc. are. But i am not perfect :)! This problem appeared in the exam that has a syllabus of 3 years undergrad physics and 2 years grad physics $\endgroup$ – Prasad Mani Oct 4 '16 at 12:54
  • $\begingroup$ Oh, OK. Since you have more physics background than I had thought, I'll see if I can whip up a solution that actually solves the problem quantitatively. I suspect, though, that it's going to be a heck of a problem. In particular, I think the usual approximation of "small oscillations about equilibrium" results in the spring and pendulum oscillations becoming decoupled, and you have to go to non-linear effects to see the coupling between the two parts of the system. $\endgroup$ – Michael Seifert Oct 4 '16 at 13:00
  • $\begingroup$ If it helps, these are the options. $K$= 4w/$l$ OR 2w/$l$ OR w/$l$ OR w/$2l$ $\endgroup$ – Prasad Mani Oct 4 '16 at 13:07
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    $\begingroup$ @PrasadSubramanian: The logic that I described in my original answer (above the break) leads to the same conclusion as the more formal method I just added (below the break), and doesn't require the solution of any differential equations. Actually solving the equations below the break would be much harder, but the problem doesn't actually ask you for that. $\endgroup$ – Michael Seifert Oct 4 '16 at 16:13
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    $\begingroup$ It should not be necessary to solve any differential eqn. The frequency of the pendulum is $\omega=\sqrt{\frac{g}{l}}$ and that of the spring is $2\omega=\sqrt{\frac{K}{M}}=\sqrt{\frac{gK}{W}}$ so then $2\sqrt{\frac{g}{l}}=\sqrt{\frac{gK}{W}}$ therefore $K=4W/l$. $\endgroup$ – sammy gerbil Oct 7 '16 at 1:41

I agree with Michael Seifert's 'basic' solution.

The context of the question usually indicates how much effort is required. A multiple choice question usually requires only simple reasoning based on insight (such as resonance and 2:1 ratio of frequencies), and should take no more than a couple of minutes to solve.

  • $\begingroup$ Oh, I entirely agree that using Lagrangian mechanics is overkill. I just included it to show that one could use more "cookbook"-style techniques to address the problem instead of the physical reasoning that you (and I) described. And I do think that an answer based on physical reasoning is usually a better one. $\endgroup$ – Michael Seifert Oct 4 '16 at 20:35
  • $\begingroup$ Sorry Michael, I was distracted by your math and didn't look closely at your 'basic' solution, which is the same as mine. $\endgroup$ – sammy gerbil Oct 4 '16 at 21:15

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