Question:
Two massless springs of spring constants \(K_1\) and \(K_2\) are connected one after the other forming a single chain, suspended vertically and a certain mass is attached to the free end. If \(x_1\) and \(x_2\) are their respective extensions and 'F' is their stretching force, the total extension produced is
Two massless springs of spring constants \(K_1\) and \(K_2\) are connected one after the other forming a single chain, suspended vertically and a certain mass is attached to the free end. If \(x_1\) and \(x_2\) are their respective extensions and 'F' is their stretching force, the total extension produced is
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Springs in series carry the same force, and extensions add.
Updated On: Oct 1, 2026
- \(F(K_1+K_2)\)
- \(F(1/K_1-1/K_2)\)
- \(F(K_1-K_2)\)
- \(F(1/K_1+1/K_2)\)
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The Correct Option is D
Solution and Explanation
Step 1: Same force
In a series chain the same force \(F\) acts on each spring.
Step 2: Extensions
\(x_1=\frac{F}{K_1}\) and \(x_2=\frac{F}{K_2}\).
Step 3: Total
\(x=x_1+x_2=F\left(\frac{1}{K_1}+\frac{1}{K_2}\right)\). Option (D).
Final Answer:
Total extension is \(F(1/K_1+1/K_2)\), option (D). \[ \boxed{\text{(D)}} \]
In a series chain the same force \(F\) acts on each spring.
Step 2: Extensions
\(x_1=\frac{F}{K_1}\) and \(x_2=\frac{F}{K_2}\).
Step 3: Total
\(x=x_1+x_2=F\left(\frac{1}{K_1}+\frac{1}{K_2}\right)\). Option (D).
Final Answer:
Total extension is \(F(1/K_1+1/K_2)\), option (D). \[ \boxed{\text{(D)}} \]
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