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 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 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 equation `k_(1)=(k_(eq)l)/l_(1) "and" k_(2)=(k_(eq)l)/l_(2)`as,`k_(1)=3k "and" k_(2)=(3k)/2` Let we take the 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)/2x_(2)=Mg` From equation (1), we have `x_(1)+x_(2)/2=x/3`...(2) As the rod is horizontal and in STATIC equlibrium , we have net torque acting on the rod about any point on it must be ZERO . Thus we have torque on it above end A are `k_(2)x_(2)L=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 rod as `k_(1)x_(1)L=Mg L/2` (or) `x_(1)=(Mg)/(2k_(1))=(Mg)/(6k)=x/6`