# 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?

• 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

## 1 Answer

Your reasoning is correct up to the point where you compare the accelerations of the mass and the spring. But the effect on the period is actually inverted.

Intuitively, when the spring acceleration is higher (lower), this actually means that the spring is more (less) deformed and pulls back with a force correspondingly stronger (weaker) compared to case 1, behaving as if it were more (less) rigid. But let's equate the values.

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