JEE Main202330 Jan 2023Evening ShiftPhysicsOscillationsActual
The velocity of a particle executing SHM varies with displacement ( x ) as 4 v 2 = 50 – x 2 . The time period of oscillations is x 7 s . The value of x is ______. [Take π = 22 7 ]
Correct answer
0
Step-by-step solution
Given: 4 v 2 = 50 - x 2 ⇒ v = 1 2 50 - x 2 Comparing with standard equation of SHM, v = ω A 2 - x 2 , we get ω = 1 2 Now, T = 2 π ω = 4 π = 88 7 Hence, required x = 88 .