The velocity of a particle executing SHM varies with displacement ( x ) as 4v^2 = 50 - x^2 . The time period of oscillations is x/7 s. The value of x is (Take = 22/7 )

Step-by-step solution

The given equation of velocity is 4v^2 = 50 - x^2 Rearranging the terms, we get: v^2 = 1 4 (50 - x^2) Comparing this with the standard equation of SHM v^2 = ^2(A^2 - x^2) , we get: ^2 = 1 4 = 1 2 rad/s The time period of oscillation is given by T = 2 T = 2 1/2 = 4 Substituting = 22 7 , we get: T = 4 22 7 = 88 7 s Comparing this with the given time period x 7 s, we get x = 88 . Answer: 88