JEE Main202125 Feb 2021Evening ShiftPhysicsOscillationsActual
The point A moves with a uniform speed along the circumference of a circle of radius 0 . 36 m and covers 30 ° in 0 . 1 s . The perpendicular projection P from A on the diameter M N represents the simple harmonic motion of P . The restoration force per unit mass when P touches M will be :
Options
- A0 . 49   N
- B9 . 87   N
- C50   N
- D100   N
Correct answer
B. 9 . 87   N
Step-by-step solution
The point A covers 30 ° in 0 . 1   s . Then, we have π 6   rad ⟶ 0 . 1   s . Or 1   rad ⟶ 0 . 1 π 6   s Or 2 π   in →   0 . 1 × 6 π × 2 π Thus, time taken is T = 1 . 2   sec . We know that ω = 2 π T Then, ω = 2 π 1 . 2 Restoration force ( F ) = mω 2 A . Then Restoration force per unit mass F m = ω 2   A F m = 2 π 1 . 2 2 × 0 . 36 ≅ 9 . 87   N