# The period of a mass-spring system including a pulley [closed]

A spring with spring constant $k$ is attached to a mass $m$ as illustrated in the following four set ups: Calculate the period of the motion for each of them. In cases (1) and (2) the period is of course $T=2\pi \sqrt{m/k}$.

Now let's examine cases 3 and 4:

If I pull the mass in (3) down a distance $Δx$, then the spring is stretched more by amount of $2Δx$. In (4), when pulling $m$ down the same $Δx$ the spring is stretched by $\dfrac{Δx}{2}$.

So that in (3) the relation between the accelerations of the mass and some other point on the spring (assuming it stretches equivalently along it) is $$\vec a_\text{mass}=2\vec a_\text{spring}$$. At (4): $$\vec a_\text{mass}=1/2\vec a_\text{spring}$$.

So that I came into the conclusion that the period differs by a factor +2 in (3) i.e $$T_\text{(3)}=4\pi\sqrt{m/k}$$

and $$T_\text{(4)}=\pi\sqrt{m/k}$$

The answer in the back of the back says: $$T_\text{(3)}=\pi\sqrt{m/k}$$ $$T_\text{(4)}=4\pi\sqrt{m/k}$$ Which is just the opposite of my logic.

NOTE: It is possible that the answer of the book are incorrect.

Is anything wrong with my calculations?

## closed as off-topic by John Rennie, ACuriousMind♦, Brandon Enright, Kyle Kanos, JamalSNov 4 '14 at 20:07

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• Stronger spring = higher frequency and lower period – John Rennie Nov 4 '14 at 16:31
• As stated in the question - same spring is used in all set ups. I don't understand your comment. – E Be Nov 4 '14 at 17:06
• Suppose the force constant of the spring is $k$, i.e. extending the spring by a distance $d$ produces a force $F = kd$. Let $x$ be the distance moved by the mass. In systems (1) and (2) $x = d$ so $F = kx$. In (3) moving the mass a distance $x$ extends the spring by $d = 2x$, so $F = 2kx$ i.e. the spring appears to have a force constant twice as high. In (4) moving the mass a distant $c$ extends the spring by $d = x/2$ so $F = \tfrac{1}{2}kx$ i.e. the spring appears to have a force constant half as high – John Rennie Nov 4 '14 at 17:12
• So that the answer in the book is more logic then mine, because it follows the "rule" you state here. But is it correct with these factors $\pi$ and $4\pi$? – E Be Nov 4 '14 at 17:16
• Your equation is $T=2\pi \sqrt{m/k}$, so for (3) you have $T=2\pi \sqrt{m/2k}$ and for (4) $T=2\pi \sqrt{2m/k}$. – John Rennie Nov 4 '14 at 17:19

Mathematically, the equation for case 1 which would be: $$\ddot{x}+\omega^2x=0$$ where for case 3 would be $$\ddot{x}+2\omega^2x=0$$ where for case 4 would be $$\ddot{x}+\frac{\omega^2}{2}x=0$$ where $\omega=\sqrt{\frac{k}{m}}=\frac{2\pi}{T}$ and this leads to $T_3=\frac{T}{\sqrt{2}}$ and $T_4=\sqrt{2}T$. So the rule would be $$\frac{\omega_a}{\omega_b}=\frac{T_b}{T_a}$$
• I don't understand the "rule" noted at the end of your answer. What are $a$ and $b$? – E Be Nov 4 '14 at 17:32
• Say $\omega_a$ is the new frequency ($\omega_3$) and $\omega_b$ the reference one ($\omega_1$) then the ratio of the periods will be inversely proportional to the ratio of the frequencies. – rmhleo Nov 4 '14 at 17:35