Q. A block is placed on a frictionless horizontal table. The mass of the block is $m$ and springs are attached on either side with force constants $K_{1}$ and $K_{2}.$ If the block is displaced a little and left to oscillate, then the angular frequency of oscillation will be
A) $\left( \frac{K_{1} + K_{2}}{m} \right)^{1/2}$
B) $\left\lbrack \frac{K_{1}K_{2}}{m\left( K_{1} + K_{2} \right)} \right\rbrack^{1/2}$
C) $\left\lbrack \frac{K_{1}K_{2}}{\left( K_{1} - K_{2} \right)m} \right\rbrack^{1/2}$
D) $\left\lbrack \frac{K_{1}^{2} + K_{2}^{2}}{\left( K_{1} + K_{2} \right)m} \right\rbrack^{1/2}$
Explanation
(a) In this case springs are in parallel, so $k_{eq} = k_{1} + k_{2}$
$and\ \omega = \sqrt{\frac{k_{eq}}{m}} = \sqrt{\frac{k_{1} + k_{2}}{m}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.