Question

Three identical springs, each with stiffness 1200 N/m are attached in series (that is, end to end) to make a longer spring to hold up a heavy weight. What is the stiffness of the longer spring?

Answer 1

Answer:

400N/m

Explanation:

When n identical springs of stiffness k, are attached in series, the reciprocal of their equivalent stiffness (1 / m) is given by the sum of the reciprocal of their individual stiffnesses. i.e

$\frac{1}{m}$ = ∑ⁿ₁ [$\frac{1}{k_{i}}$]          -----------------------(i)

That is;

$\frac{1}{m}$ = $\frac{1}{k_{1}}$ + $\frac{1}{k_{2}}$ + $\frac{1}{k_{3}}$ + . . . + $\frac{1}{k_{n}}$      -------------------(ii)

If they have the same value of stiffness say s, then equation (ii) becomes;

$\frac{1}{m}$ = n x $\frac{1}{s}$           -----------------(iii)

Where;

n = number of springs

From the question,

There are 3 identical springs, each with stiffness of 1200N/m and they are attached in series. This implies that;

n = 3

s = 1200N/m

Now, to calculate the effective stiffness,m, (i.e the stiffness of a longer spring formed from the series combination of these springs), we substitute these values into equation (iii) above as follows;

$\frac{1}{m}$ = 3 x $\frac{1}{1200}$

$\frac{1}{m}$ = $\frac{3}{1200}$

$\frac{1}{m}$ = $\frac{1}{400}$

Cross multiply;

m = 400N/m  

Therefore, the stiffness of the longer spring is 400N/m