JEE Main20249 Apr 2024Evening ShiftPhysicsOscillationsActual
A particle of mass 0.50 ~kg executes simple harmonic motion under force F=-50 ( Nm ⁻¹ ) x . The time period of oscillation is x 35 ~s . The value of x is _______ (Given = 22 7 )
Correct answer
0
Step-by-step solution
aligned & m =0.5 ~kg & ~F =-50( x ) & ma =(-50 x ) & 0.5 a =-50 x & a =(-100 x ) & W ^2=100 ( w =10) & T = 2 10 = ( 5 )= 22 7 15 = ( 22 35 ) & 35 = 22 35 x =22 aligned