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
$$\left( \dfrac{K_1+K_2}{m}\right)^{1/2}$$
B
$$\left[ \dfrac{K_1K_2}{m(K_1+K_2)}\right]^{1/2}$$
C
$$\left[ \dfrac{K_1K_2}{(K_1-K_2)m}\right]^{1/2}$$
D
$$\left[ \dfrac{K_1^2+K_2^2}{(K_1+K_2)m}\right]^{1/2}$$
Correct option is A. $$\left( \dfrac{K_1+K_2}{m}\right)^{1/2}$$
In this case springs are parallel, so $$k_{eq}=k_1+k_2$$
and $$\omega =\sqrt{\dfrac{k_{eq}}{m}}=\sqrt{\dfrac{k_1+k_2}{m}}$$