A body executes S.H.M under influence of a force with a time period of 1.0s. It has a time period of 1.5s under theaction of a force of different magntitude. What will be the period of oscillation of the body when these two force are impressed simultaneously in the same direction upon the same body?

A body executes S.H.M under influence of a force with a time period of

1.	A body executes S.H.M under influence of a force with a time period of 1.0s. It has a time period of 1.5s under theaction of a force of different magntitude. What will be the period of oscillation of the body when these two force are impressed simultaneously in the same direction upon the same body?
Answer» Solution :We know that `Tprop (1)/(sqrta)` where T - period of oscillation and a = acceleration ALSO `F=ma THEREFORE a prop " FORCE"` `therefore prop (1)/(sqrtF),""T_(1)prop (1)/(F_(1))"………….(1)"` `T_(2)prop (1)/(sqrt(F_(2)))"............(2)"` i.e. `(T_(1))/(T_(2))=sqrt((F_(2))/(F_(1)))"i.e. "((1)/(1.5))^(2)=(F_(2))/(F_(1))` `"or"(F_(1))/(F_(2))=2.55""F_(1)=2.25F_(2)` When the combined forces act on the body then, `T prop (1)/(sqrt(F_(1)+F_(2)))".........(4)"` `(4)+(1)" gives "(T)/(T_(1))=sqrt((F_(1))/(F_(1)+F_(2)))` i.e. `(T)/(T_(1))=sqrt((F_(1))/(F_(1)(1+(F_(2))/(F_(1)))))=sqrt((1)/(1+((F_(2))/(F_(1)))))=sqrt((1)/(1+((1)/(2.25))))=sqrt((1)/(1.444))` `therefore""T=0.8322T_(1)` `"i.e."T=0.8322xx1s=0.8322s.`