Most Important Selected Qs for JEE AdvancedPhysicsOscillations
A small block of mass m=0.5 ~kg is attached to two springs each of force constant k =10 ~N / m as shown in figure. The block is executive SHM with amplitude A = 1 8 ~m . When the block is at equilibrium position one of the spring breaks without changing momentum of block. New amplitude of oscillation is e 16 ~m . Find the value of e ?
Correct answer
8
Step-by-step solution
= k m = 40 when spring breaks new = 20 Equilibrium position of original system (2k) x₀=m g or x₀= 1 4 ~m New equilibrium is at kx = mg ; x = 1 2 ~m thus v _ = A =( 40 ) ( 1 4 ) 10 2 = 20 [A^2- ( 1 4 )^2 ]^ 1 / 2 ; 10 4 =20 [A^2- 1 16 ] ; 1 8 +A^2=A^ 2 ; A^ = 1 2 n=50 ~cm