A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to T_(M). If the Young's modulus of the material of the wire is Y then 1/Yis equal to: (g = gravitational acceleration)

A pendulum made of a uniform wire of cross sectional area A has time p

1.	A pendulum made of a uniform wire of cross sectional area A has time period T. When an additional mass M is added to its bob, the time period changes to T_(M). If the Young's modulus of the material of the wire is Y then 1/Yis equal to: (g = gravitational acceleration)
Answer» `[ ((T _(M))/(R))^(2) -1 ] (A)/(Mg )` `[ ((T _(M))/(R))^(2) -1 ] (Mg)/(A )` `[-1 +((T _(M))/(T )) ^(2) ] (A)/(Mg)` `[-1 +((T)/(T _(M))) ^(2) ] (A)/( Mg )` Solution :Time period of simple pendulum, `T =2 pi sqrt ((L )/(g)) ""…(i)` If we increase the MASS of bob its length is increased by `DELTA t.` At that time periodic time is `T_(M)` `T _(M) =2pi sqrt ((l + Deltal )/(g )) ""…(ii)` Now Young.s modulus `Y= (F//A)/(Delta l //l ) = (Mg)/(A) xx (1)/( Delta l //l )` `therefore (Deltal )/(l) = (Mg )/(Ay) ""...(iii)` Divide equation (ii) by equation (i) and do square of it `((T _(M))/( T )) ^(2) = (l + Delta l )/(l)= l + (Delta l )/(l)` `therefore ((T _(M))/(T)) ^(2) -1 = (Delta l )/(l) ""...(iv)` PUTTING value of equation (iii) in equation (iv) `((T_(M))/( T)) ^(2) -1 = (Mg )/(AY) = (A)/( Mg ) [ ((T _(M))/( T )) ^(2) -1] (1)/(Y)`