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 MN represents the simple harmonic motion of 'P'. The restoration force per unit mass when P touches M will be :

Solution

Physics sir · 4 steps

The circle hands you ω directly, and at M the point P is at the extreme, so the pull is ω²A.

1. Angular speed. A turns through 30° = π/6 rad in 0.1 s. So ω = (π/6)/0.1 = 5π/3 rad s⁻¹, and this is also the angular frequency of P's SHM.
2. Where M is. P moves along the diameter MN with amplitude equal to the radius, A = 0.36 m. At M, P is at an extreme, where the displacement and the restoring acceleration are both largest.

   F/m = ω² A

3. Put in numbers. ω² A = 25π²/9 × 0.36 = π² × (25 × 0.36)/9 = π² × 1. So F/m = π² ≈ 9.87 N kg⁻¹.
4. So. The restoring force per unit mass at M is about 9.87 (the options write the unit as N, but the unit of force per unit mass is N kg⁻¹), which is option (a).

The idea to keep. In the reference-circle picture, the circle's angular speed is the SHM's ω and its radius is the amplitude. The restoring force per unit mass is ω² x, so it is largest, ω² A, at the ends.

Answer from the NTA final answer key. Solution by PrepWiser.