Question

A particle is executing SHM between points -X_m and X_m , as shown in figure-I. The velocity V(t) of the particle is partially graphed and shown in figure-II. Two points A and B corresponding to time t₁ and time t₂ respectively are marked on the V(t) curve.

(A) At time t₁ , it is going towards X_m . 

(B) At time t₁ , its speed is decreasing. 

(C) At time t₂ , its position lies in between –X_m and O. 

(D) The phase difference Δϕ between points A and B must be expressed as 90° < Δϕ < 180°.

Answer 1

Correct option: (B, C)

Explanation:

At time t₁ , velocity of the particle is negative i.e. going towards –X_m . From the graph, at time t₁ , its speed is decreasing. Therefore particle lies in between –X_m and 0.

At time t₂ , velocity is positive and its magnitude is less than maximum i.e. it has yet not crossed O. It lies in between –X_m and 0.

Phase of particle at time t₁ is (180 + θ₁ ). Phase of particle at time t₂ is (270 + θ₂ )

Phase difference is 90 +(θ₂ - θ₁)

θ₂ - θ₁ can be negative making Δϕ < 90° but can not be more than 90°.