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When a spring mass system is connected vertically with two massless springs in series whose spring constants are $k_1$ and $k_2$ to a block of mass $m$ we know that equal forces act on both the springs. Let that force during oscillations be $F$.

When we calculate effective spring constant $k_s$, why don't we say the net force acting on the system is $2F$?

Finding net force acting on the above system:

When the block is attached,the system attains equilibrium position through displacements $x'_1$ and $x'_2$.

At equilibrium: $2F'=mg$(Where $F'$ is magnitude of spring force initially by each spring)

So, $k_1x'_1+k_2x'_2=mg$ (equation 1)

When the system is pulled down it makes oscillations,now:

Total elongation be $x$

Elongation in spring 1 be $x_1$ and elongation in spring 2 be $x_2$.

Total spring force $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2$

Total forces acting on the system $= -k_1x'_1-k_2x'_2-k_1x_1-k_2x_2+mg = -mg-k_1x_1-k_2x_2+mg$
(from equation 1)

So, total force $= -k_1x_1-k_2x_2 = F_1+F_2=2F$(as we know that both forces are equal)

So net force acting on the system is $2F$

The way I calculated effective spring constant is:

$x=x_1+x_2$

$2F/k_s = F/k_1 + F/k_2$

$2/k _s = 1/k_1 +1/k_2$

But that is not a correct equation. What's wrong in taking net force acting on system as $2F$.

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ok so the first condition for the series spring system is : The spring force in the entire system is the same i.e the tension in the springs is the same

therefore
$k_1x_1=k_2x_2$

$k_e(x_1+x_2)=k_1x_1$
where $k_e$ is the equivalent spring constant

so if the system is vertical let gravitation force, mg, be $F$

hence $x_1=F/k_1$ ,$x_2=F/k_2$ and $x_1+x_2=F/k_e$

$x_1+x_2=F/k_1+F/k_2=F/k_e$

$\Rightarrow 1/k_e=1/k_1+1/k_2$

Now let's deal with your question, So I recommend You look into the free body diagram again. The tension in the entire system is equal to $F$

If u are facing troubles you can also try relating this with electricity resistor circuits. In a series connection, the current is the same through the resistors similarly in a series spring connection the tension is the same throughout. And in the case of a parallel connection the current splits in the branches but with each branch having the same potential difference, similarly, in a parallel spring system the two springs experience different tensions but extend/ contract equally

For further reference: I have some trouble with springs in series

and

https://physics.stackexchange.com/questions/311111/why-springs-in-series-experience-equal-force#:~:text=In%20series%20circuits%3B%20the%20current,force%20must%20be%20the%20

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  • $\begingroup$ Can you attach a free body diagram of the above case representing all the forces acting on the system? $\endgroup$ – Einstein Apr 4 at 7:27
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When the springs (assumed to be massless) are hung upside down, they will have a zero extension. So in fact $$mg = k_1x+k_2x$$ when the mass is added to the system.

You wrote that the resultant force $$F_r = F_1+F_2$$ and concluded that this should be $2F$ as if the force in both springs were equal.

So what you should have done is write the resultant displacement $$x=x_1+x_2$$ so that $$\frac{mg}{k_s}=\frac{mg}{k_1}+\frac{mg}{k_2}$$ where $k_s$ is a effective spring constant. This will then give $$k_s=(\frac{1}{k_1}+\frac{1}{k_2})^{-1}$$

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The springs are massless, so the tension at any point in either spring is the same value, $F$. For the two individual springs, with extensions $x_1$ and $x_2$, we have#

$F = k_1x_1=k_2x_2$

and when considered as a single spring they have an effective spring constant $k$ such that

$F = k(x_1+x_2) \\\displaystyle \Rightarrow F = k\left(\frac F {k_1} + \frac F {k_2} \right) \\\displaystyle \Rightarrow \frac 1 k = \frac 1 {k_1} + \frac 1 {k_2} $

Note that if we join two identical springs in series (so $k_1=k_2$ and $x_1=x_2$) then the spring constant of the joint spring is only half that of its two components. This is because it produces the same force $F$ with twice the extension of the individual springs, since $x=2x_1$.

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