k/2
This section explains how to determine the equivalent stiffness when two springs, each possessing a stiffness constant denoted by k, are connected in series.
When springs are arranged in series, they are linked one after another, forming a chain. In this configuration, the total force applied to the spring system is distributed equally across each individual spring. Conversely, the total elongation (or compression) of the system is the sum of the elongations experienced by each spring separately.
The general formula used to calculate the equivalent stiffness ($k_{eq}$) of multiple springs connected in series is based on the reciprocals of their individual stiffness constants ($k_1, k_2, \dots$):
$$ \frac{1}{k_{eq}} = \frac{1}{k_1} + \frac{1}{k_2} + \dots $$
For this problem, we are given two springs, and both have the same stiffness, k. Therefore:
Substitute these values into the series formula:
$$ \frac{1}{k_{eq}} = \frac{1}{k} + \frac{1}{k} $$
Combine the fractions on the right side by adding them:
$$ \frac{1}{k_{eq}} = \frac{1 + 1}{k} $$
$$ \frac{1}{k_{eq}} = \frac{2}{k} $$
To find the equivalent stiffness ($k_{eq}$), we invert both sides of the equation:
$$ k_{eq} = \frac{k}{2} $$
The calculation shows that the equivalent stiffness of two springs, each with stiffness k, connected in series is k/2.
The resistance of a material to elastic deformation is called _______.
An automotive engine having a mass of 135 kg is supported on 4 springs with linear characteristics. Each of the 2 front springs have stiffness of 3 MN/m while the stiffness of each of 2 rear springs is 4.5 MN/m. The engine speed (RPM) at which resonance is likely to occur is