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

400N/m

Step-by-step 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

`(1)/(m)` = ∑ⁿ₁ [
`(1)/(k_(i))`] -----------------------(i)

That is;

`(1)/(m)` =
`(1)/(k_(1))` +
`(1)/(k_(2))` +
`(1)/(k_(3))` + . . . +
`(1)/(k_(n))` -------------------(ii)

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

`(1)/(m)` = n x
`(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;

`(1)/(m)` = 3 x
`(1)/(1200)`

`(1)/(m)` =
`(3)/(1200)`

`(1)/(m)` =
`(1)/(400)`

Cross multiply;

m = 400N/m

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