Question

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 $K _ { 2 }$. 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. $\left[ \frac { K _ { 1 } K _ { 2 } } { m \left( K _ { 1 } + K _ { 2 } \right) } \right] ^ { 1 / 2 }$
• C. $\left[ \frac { K _ { 1 } K _ { 2 } } { \left( K _ { 1 } - K _ { 2 } \right) m } \right] ^ { 1 / 2 }$
• D. $\left[ \frac { K _ { 1 } ^ { 2 } + K _ { 2 } ^ { 2 } } { \left( K _ { 1 } + K _ { 2 } \right) m } \right] ^ { 1 / 2 }$

Answer 1

Answer: (A)

$\left( \frac { K _ { 1 } + K _ { 2 } } { m } \right) ^ { 1 / 2 }$

Answer 2

In this case springs are in parallel, so

$\omega = \sqrt { \frac { k _ { e q } } { m } } = \sqrt { \frac { k _ { 1 } + k _ { 2 } } { m } }$