When two springs (each having stiffness constant K) are connected in series, the equivalent stiffness will be

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UKPSC JE Mechanical 2013 Official Paper I
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  1. K
  2. 2K
  3. \(\frac{K}{2}\)
  4. \(\frac{1}{K}\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{K}{2}\)
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Detailed Solution

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Explanation:

In mechanics, two or more springs are said to be in series when they are connected end-to-end, and in parallel when they are connected side-by-side. 

SSCJE ME SOM 57

Equivalent spring constant

In Series connection:

\(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\)

\({k_{eq}} = \frac{{{k_1} \times {k_2}}}{{{k_1} + {k_2}}}\)

Here, k1 = k2 = k

\(\therefore {k_{eq}} = \frac{{{k_1} \times {k_2}}}{{{k_1} + {k_2}}}=\frac{{{k} \times {k}}}{{{k} + {k}}}=\frac{k^2}{2k}=\frac k2\)

Additional InformationIn Parallel connection:

\({k_{eq}} = {k_1} + {k_2}\)

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