When a mass M hangs from a spring of length l, it stretches the spring by a distance x. Now the spring is cut in two parts of length l/3 and 2l/3, and the two springs thus formed are connected to a straight rod of mass M which is horizontal in the configuration shown in figure. Find the stretch in each of the spring.

When a mass M hangs from a spring of length l, it stretches the spring

1.	When a mass M hangs from a spring of length l, it stretches the spring by a distance x. Now the spring is cut in two parts of length l/3 and 2l/3, and the two springs thus formed are connected to a straight rod of mass M which is horizontal in the configuration shown in figure. Find the stretch in each of the spring.
Answer» Solution : As it given that the mass M stretches the original spring by a distance x, we have `kx = Mg` (or) `k=(Mg)/(x)`......... (1) The new force CONSTANTS of the two springs can be given by using equations `k_1 = (k_(eg) L)/(l_1) and k_2=(k_(eq)l)/(l_2) as , k_1 = 3k and k_2=(3k)/(2)` Let we takethe stretch in the two springs be `x_1 and x_2 ` we have for the equilibrium of the rod `k_1 x_1 + k_2 x_2 = Mg` (or) ` 2kx_1 + (3k)/(2) x_2 = Mg` From equation (1) , we have ` x_1 +(x_2)/(2)=(x)/(3)`......... (2) As the rod is horizontal and in static equilibrium , we have net torque ACTING on the rod about any point on it must be zero. Thus we have torquie on it above end A are `k_2x_2L=Mg(L)/(2)` (or) `x_2=(Mg)/(2k_2)=(Mg)/(3k)=(x)/(3)` Using this value of `x_2 ` in equation (2) , we have `x_1=(x)/(6)` This can also be directly obtained by using torque zero about point B on the rod as `k_1x_1L=Mg(L)/(2)` (or) ` x_1= (Mg)/(2k_1)=(Mg)/(6k)=(x)/(6)`