0
$\begingroup$

My question is:

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

enter image description here

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?

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

This is the diagram,

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

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

$\endgroup$
  • $\begingroup$ What does x¨2 and x¨1 stand for?and how can Feq be equal to F (individual force given by a spring)? $\endgroup$ – Physkiz Dec 31 '16 at 3:21
  • $\begingroup$ @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. $\endgroup$ – AHB Dec 31 '16 at 5:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.