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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Oscillations

The equivalent spring constant for the system is

$(a)\;K_{1}+K_{2}\qquad(b)\;\large\frac{K_{1}K_{2}}{K_{1}+K_{2}}\qquad(c)\;K_{1}K_{2}\qquad(d)\;\large\frac{K_{2}}{K_{1}}$

1 Answer

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
$F=K_{1}x_{1}=K_{2}x_{2}$
$x_{1}=\large\frac{F}{K_{1}} \qquad \; x_{2}=\large\frac{F}{K_{2}}$
$x_{1}+x_{2}=F\;(\large\frac{1}{K_{1}}+\large\frac{1}{K_{2}})$
let $\;x_{1}+x_{2}\;$ be x
$x=F\;(\large\frac{K_{1}+K_{2}}{K_{1}K_{2}})$
$F=x\;(\large\frac{K_{1}K_{2}}{K_{1}+K_{2}})$
Therefore , $\;\large\frac{K_{1}K_{2}}{K_{1}+K_{2}}\;$ is the equivalent spring constant .
answered Mar 11, 2014 by yamini.v
 

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