Consider a hypothetical field in which force on any particle at position x varies according to the relation F⃗ = F₀(1 − x/4)î, where F is in newton, x is in metre and F₀ is a positive constant. A particle of mass m is released from rest from x = 12 m at t = 0 in this field. If 2π√(m/F₀) = 6 seconds, then find value of t in second when the particle reaches the origin first time.
Physics · Mixed Chapter
JEE Mains

Question

Consider a hypothetical field in which force on any particle at position x varies according to the relation F⃗ = F₀(1 − x/4)î, where F is in newton, x is in metre and F₀ is a positive constant. A particle of mass m is released from rest from x = 12 m at t = 0 in this field. If 2π√(m/F₀) = 6 seconds, then find value of t in second when the particle reaches the origin first time.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1: The equilibrium point is at x=4 m. Writing y=x−4, the force becomes F=−(F₀/4)y. Hence the motion is SHM with ω=√[F₀/(4m)]. Step 2: Initially x=12, so y=8 m and initial velocity is zero. Thus y(t)=8cosωt. Reaching origin means y=−4, so cosωt=−1/2. Step 3: The first positive phase is ωt=2π/3. From 2π√(m/F₀)=6 s, 1/ω=2√(m/F₀)=6/π seconds. Step 4: t=(2π/3)(6/π)=4 s.