A  body  executes S.H.M. under the action of a force F1 with a time period 7/6  seconds.  If the force is changed to F2 it executes S.H.M. with time period 7/8  seconds. If both the forces F1 and F2 act simultaneously in the same direction on  the body, then its time period in seconds is

For a body executing simple harmonic motion, the force can be written as F=mω2y  For the first force F1=m2πT12y  and for the second force  F2=m2πT22y. On combining the two forces  F1+F2=Fnet , where Fnet=m2πTnet2y Substituting this in the equation F1+F2=Fnet, we getm2πTnet2y=m2πT12y+m2πT22y From the question,  T1=76 s and  T2=78 s1Tnet2=672+8721Tnet2=10049Tnet2=49100 Therefore,  Tnet=710s