At equilibrium, the springs are released, mass M is oscillating under the influence of three springs, k_(1),k_(2)andk_(3) as shown. Its frequency (upsilon) of oscillation is (1)/(2pi)sqrt((k_(eq))/(M)), where k_(eq) I such that

At equilibrium, the springs are released, mass M is oscillating under

1.	At equilibrium, the springs are released, mass M is oscillating under the influence of three springs, k_(1),k_(2)andk_(3) as shown. Its frequency (upsilon) of oscillation is (1)/(2pi)sqrt((k_(eq))/(M)), where k_(eq) I such that
Answer» `k_(eq)=k_(1)+k_(2)+k_(3)` `k_(eq)=(k_(1)+k_(2)+k_(3))/(3)` `(1)/(k_(eq))=(1)/(k_(1))+(1)/(k_(2))+(1)/(k_(3))` `k_(eq)=(k_(1)+k_(2)-k_(3))` Solution :When the mass M is DISPLACED SLIGHTLY to the right, the resultant FORCES acting on the mass M is as shown. `F_("net")=-(k_(1)+k_(2)+k_(3)).x,` Thus, `k_(eq)=k_(1)+k_(2)+k_(3)`