A constant force produces maximum velocity v on the block connected to the spring of force constant k as shown in the figure. When the force constant of spring becomes 4k,the maximum velocity of the block is (block is at rest when spring is relaxed):

By work energy theorem;Fx1−12kx12=12mv2     .......(1)and Fx2−12k′x22=12mv′2   .........(2)where x1,x2 are initial and final extensions and v,t' are initialand final velocities.In both cases, force applied is same, and velocity becomes maximum when F=kx. (after which the mass will decelerate)∴ F=kx1=(4k)x2⇒ x2=x14Substituting in (2):Fx14−12(4K)x142=12mv′2⇒ 14Fx1−12kx12=12mv′2     .......(3)Dividing (3)/(1), we get:14=v′2v2⇒v′=v2Hence, (c).