Two massless springs of spring constant $K_1$ and $K_2$ are connected one after the other forming a single…
Two massless springs of spring constant $K_1$ and $K_2$ are connected one after the other forming a single chain, suspended vertically and certain mass is attached to the free end. If $e_1$ and $e_2$ are their respective extensions and $f$ is their stretching force, the total extension produced is
$f\left(K_1-K_2\right)$
$f\left( rac{1}{K_1}-\frac{1}{K_2}\right)$
$f\left(K_1+K_2\right)$
$f\left(\frac{1}{K_1}+\frac{1}{K_2}\right)$
Solution
The total restoring force is related to extension as follows:
$f=-K x \quad---(1)$
The extension in the springs are related to the restoring force as follows:
$f=-K_1 e_1---(2)$
and
$f=-K_2 e_2---(3)$
The total extension of the springs is:
$x=e_1+e_2---(4)$
Using equation (1),(2) and (3)
$\begin{aligned} & \Rightarrow\left(-\frac{f}{K}\right)=\left(-\frac{f}{K_1}\right)+\left(-\frac{f}{K_2}\right) \\ & \Rightarrow \frac{1}{K}=\left(\frac{1}{K_1}+\frac{1}{K_2}\right)---(5)\end{aligned}$
The total extension using equation (4) and (5)is given by,
$x=\frac{f}{K}=f\left(\frac{1}{K_1}+\frac{1}{K_2}\right)$