AP EAMCET202420 May 2024Morning ShiftPhysicsLaws of MotionActual
A 3 kg block is connected as shown in the figure. Spring constants of two springs K ₁ and K ₂ are 50 Nm ⁻¹ and 150 Nm ⁻¹ respectively. The block is released from rest with the springs unstretched, The acceleration of the block in its lowest position is ( g =10 ~ms ⁻² )
Options
- A10 ~ms ⁻²
- B12 ~ms ⁻²
- C8 ~ms ⁻²
- D8.8 ~ms ⁻²
Correct answer
A. 10 ~ms ⁻²
Step-by-step solution
k ₁=50 Nm ⁻¹, k ₂=150 Nm ⁻¹ W = k _ eq m = k ₁+ k ₂ ~m 50+150 3 = 200 3 rad / s Amplitude, A = mg k _ eq = 3 10 200 = 3 20 ~m Acceleration of the block at lowest position, A_ = ^2 A= 200 3 3 20 =10 ~m / s ^2