# Oscillations numerical: springs arranged in series

My question is:

Springs $k_1,k_2,k_3$ are arranged in series with a mass $m$ as given in the image:

We have to find $k_{\text{eq}}$ if the time period is given as $T= 2\pi\sqrt{\dfrac{m}{k_{\text{eq}}}}$.

I saw the solution and there the book has written that $X=x_1+ x_2 + x_3$. Can someone explain this?

Also force given by each spring is the same ($=F$) and $F$ is the net force experienced by the mass. Shouldn't the forces add?

Since the springs are connected in series the total displace of the mass would just be the sum of displacements of the individual springs.

This is the diagram,

For the points $A$, and $B$; I apply newton's second law:

$$k_2(\Delta x)_2-k_1(\Delta x)_1=m_A \frac{d^2x_1}{dt^2}=0 \implies k_2(\Delta x)_2=k_1(\Delta x)_1$$

$$k_3(\Delta x)_3-k_2(\Delta x)_2=m_B \frac{d^2x_2}{dt^2}=0 \implies k_3(\Delta x)_3=k_2(\Delta x)_2$$

So,

$$k_3(\Delta x)_3=k_2(\Delta x)_2=k_1(\Delta x)_1$$

The equivalent spring force must be of the form,

$$F_{eq}=k_{eq}(\Delta X)$$

We know that $\Delta X=(\Delta x)_1+(\Delta x)_2+(\Delta x)_3$,

$$F_{eq}=k_{eq}((\Delta x)_1+(\Delta x)_2+(\Delta x)_3)$$ $$F_{eq}=k_{eq}(\frac{k_1(\Delta x)_1}{k_1}+\frac{k_2(\Delta x)_2}{k_2}+\frac{k_3(\Delta x)_3}{k_3})$$

According to $k_3(\Delta x)_3=k_2(\Delta x)_2=k_1(\Delta x)_1$,

$$k_{eq}=\frac1{\frac1{k_1}+\frac1{k_2}+\frac1{k_3}}$$

• What does x¨2 and x¨1 stand for?and how can Feq be equal to F (individual force given by a spring)? – Physkiz Dec 31 '16 at 3:21
• @Physkiz see the edit please. I didn't use any force this time. It was nonsense. That double dot means second derivative with respect to time. – AHB Dec 31 '16 at 5:10