A block of mass m be placed on a frictionless surface of a table. Spring constants of a spring joint on both side are `k_(1)" and "k_(2)` if block. Leave the block by slight displaced it, then its angular frequency will be……….

Options

A. `((k_(1)+k_(2))/(m))^(1/2)`
B. `((k_(1)k_(2))/(m(k_(1)+k_(2))))^(1/2)`
C. `((k_(1)k_(2))/((k_(1)-k_(2))m))^(1/2)`
D. `((k_(1)^(2)+k_(2)^(2))/((k_(1)+k_(2))m))^(1/2)`

Answer

A

Explanation

Here, spring are parallel hence `k_(eq)= k_(1)+ k_(2)`
and `omega = sqrt((k_(eq))/(m))= ((k_(1)+k_(2))/(m))^(1/2)`.