Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Oscillations
0 votes

The equivalent spring constant for the system is


Can you answer this question?

1 Answer

0 votes
Answer : (b) $\;\large\frac{K_{1}K_{2}}{K_{1}+K_{2}}$
Explanation :
Let the spring $\;K_{1} \;$ move by $\;x_{1}\;$ and $\;K_{2}\;$ move by $\;x_{2}\;$ . Therefore , The system finally moves $\;x_{1}+x_{2}$
As the spring remains in equilibrium
$x_{1}=\large\frac{F}{K_{1}} \qquad \; x_{2}=\large\frac{F}{K_{2}}$
let $\;x_{1}+x_{2}\;$ be x
Therefore , $\;\large\frac{K_{1}K_{2}}{K_{1}+K_{2}}\;$ is the equivalent spring constant .
answered Mar 11, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App