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

Question

Toppr Admin · Accepted Answer

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-

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

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 $$\omega =\sqrt{\dfrac{k_{eq}}{m}}=\sqrt{\dfrac{k_1+k_2}{m}}$$

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

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 $$\omega =\sqrt{\dfrac{k_{eq}}{m}}=\sqrt{\dfrac{k_1+k_2}{m}}$$

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 $$\omega =\sqrt{\dfrac{k_{eq}}{m}}=\sqrt{\dfrac{k_1+k_2}{m}}$$