0
$\begingroup$

A mass $M$ is suspended using two springs having spring constant $k_{1}$ & $k_{2}$ with distance from mass as $a$ & $b$ respectively. Find equivalent spring constant of system.

So I first found out the equivalent extension at the point where mass is suspended using similar triangles. Let $x_{1}$ be the extension in spring having spring constant $k_{1}$ and $x_{2}$ be the extension in other spring. $$ x_{eq}=\frac{x_{1}b+x_{2}a}{a+b} $$

Taking moment about point of suspension $$ k_{1}x_{1}a=k_{2}x_{2}b $$ Now $$ k_{eq}x_{eq}= k_{1}x_{1}+k_{2}x_{2} $$ Putting the value of $x_{eq}$ in above equation and replacing $x_{1}$ from moments equation we get $$ k_{eq}=\frac{k_{1}k_{2}(a+b)^2}{k_{2}b^2+k_{1}a^2} $$

But now if we put $a=b$ in the above equation we get $$ k_{eq}= \frac{4k_{1}k_{2}}{k_{2}+k_{2}} $$ which should have been $$ k_{eq}=k_{1}+k_{2} $$

$\endgroup$
0

2 Answers 2

2
$\begingroup$

Your system has 2 degrees of freedom, but using $x_1$ and $x_2$ will not be helpful in determining the effective spring rate. To get the spring rate you need the extension $x$ of the connection point with mass $M$ and the tilt angle $\theta$.

Do the substitution:

$$ \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} $$

The tension in each spring is

$$ \begin{align} F_1 & = k_1 x_1 = k_1 ( x - a \theta) \\ F_2 & = k_2 x_2 = k_2 ( x + b \theta) \end{align} $$

The total tension is

$$ F = F_1 + F_2 = (k_1 + k_2) x + (b k_2 - a k_1 ) \theta $$

So the effective spring rate is $k_{eff} = \frac{\partial F}{\partial x} = k_1 + k_2$, but only when $\theta = 0$. If $\theta$ is free and the torque at the attachment point is zero then

$$ \tau = a F_1 - b F_2 = (a k_1 - b k_2) x - (a^2 k_1 + b^2 k_2) \theta = 0$$ which is used to find the angle $\theta = \frac{a k_1 - b k_2}{a^2 k_1 + b^2 k_2} x$ for each extension $x$.

This yields the effective spring rate as

$$ F = k_{eff} x = \left( \frac{k_1 k_2 (a+b)^2}{a^2 k_1 + b^2 k_2} \right) x $$

Summary:

When the springs are in parallel with equal extensions the effective spring rate is the sum of the two springs. When the lever is free to rotate then the effective spring rate is less.

$\endgroup$
0
$\begingroup$

The "should have been proof" relies in the extensions of the springs being the same. If you put a=b that condition is not satisfied. Find a spot where the two extensions are the same and you will get "should have been proof" answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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