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The system shown in the picture consists of a spring of constant $k$, a pulley (disk) of mass $M$ and radius $R$ and a block of mass $m$ is let free from rest. There is no slipping between the rope and the pulley, and the rope is massless.

Will the motion of the block be harmonic?, if so calculate the period.

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I call the tension of the rope segment joining the block and the pulley $T_1$ and the tension of the other rope joining the spring and the pulley $T_2$. \begin{align} \sum F_{Block} &= m\cdot a_{CM} =m \cdot\frac{d^2x}{dt^2}=mg-T_1\rightarrow (1) T_1=mg-m\cdot a_{CM} \\ \sum M_{Pulley} &= I\cdot \frac{a_{CM}}{R}=\frac{1}{2}R\cdot M \cdot a_{CM}=R\cdot(T_1-T_2)\rightarrow (2) \frac{1}{2}M \cdot a_{CM}=T_1-T_2 \\ \sum F_{Spring}&=0=T_2-kx\rightarrow(3)T_2=kx \end{align} Taking 1 and 3 in 2 gives:

$$\frac{d^2x}{dt^2}+\frac{k}{\frac{1}{2}M+m}\cdot x=\frac{mg}{\frac{1}{2}M+m}$$

Where have I made a mistake? $m\cdot g$ should have simplified so I missed some force (I suspect acting on the spring) that I'm unable to find out.

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closed as off-topic by Aaron Stevens, Jon Custer, John Rennie, M. Enns, ZeroTheHero Jan 11 at 8:24

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  • $\begingroup$ Why do you think the $mg$ should not be there? I'm not going to check everything you did, but it seems reasonable there would be a constant weight force in the problem. Note however that you must have made some mistake since your dimensions don't match up (maybe the right hand side should have $m/(m+1/2M)$ instead of $m$?) $\endgroup$ – jacob1729 Jan 10 at 14:24
  • $\begingroup$ I forgot to divide in the right side of the equation, the problem here is that the simple harmonic equation I have seen in class equals 0 which is not what I got. $\endgroup$ – user605734 MBS Jan 10 at 14:31
  • $\begingroup$ Initial conditions must be taken into account. The spring is initially elongated mehn. $\endgroup$ – Aditya Garg Jan 10 at 14:31
  • $\begingroup$ So $T_2=kx+mg$ because in equilibrium the spring would do a force equal to the weight? $\endgroup$ – user605734 MBS Jan 10 at 14:37
  • $\begingroup$ A simple method to find a frequency of harmonic oscillations in problems of this kind is to introduce generalized coordinate x and express kinetic energy as $a\dot{x}^2/2$ and potential energy as $bx^2/2$. Then $\omega = \sqrt{b/a}$. $\endgroup$ – Gec Jan 10 at 14:38
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Simple Harmonic Motion results from a quadratic potential energy (and a quadratic kinetic energy). So, defining $x$ to be displacement along the rope, the kinetic energy is:

$$ T(\dot x) = \frac 1 2 m \dot x^2 + \frac 1 2 I \cdot (\frac{\dot x}{R})^2 $$

where $I$ is the moment of inertia of the disk, $M$.

The potential energy is:

$$ U(x) = mgx + \frac 1 2 k x^2$$

which is quadratic, meaning S.H.M.; nevertheless, if the linear term bothers you, we can change coordinates, per @Gec:

$$ q = (x - \frac{mg} k) $$

so that

$$ \frac 1 2 kq^2 = \frac k 2[x^2 + 2mgx/k+(\frac{mg}{k})^2] = \frac k 2x^2 + mgx +\frac{(mg)^2}{2k}$$

hence:

$$ U(q) = \frac 1 2 k q^2 - \frac{(mg)^2}{2k}$$

Meanwhile:

$$ T(\dot q) = \frac 1 2 [(m + \frac I {R^2})] \dot q^2$$

Compare that with the standard SHM, and you should be able to read-off both the frequency and the offset (from $x=0$) of the equilibrium position.

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