Oscillations numerical: springs arranged in series

Question

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?

Answer 1

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

Answer 2

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

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