JEE Main 2022 · 29 July · morning · Single correct · +4 / −1
The time period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle, which moves without friction down an inclined plane of inclination α, is given by :
- (a) 2 π √(L /(g cos α))
- (b) 2 π √(L /(g sin α))
- (c) 2 π √(L / g)
- (d) 2 π √(L /(g tan α))
Answer and solution
Answer (a) 2 π √(L /(g cos α))
Solution
Physics sir · 4 steps
Go into the vehicle's frame: the slide down the slope cancels part of gravity, leaving only g cos α.
- Vehicle's acceleration. With no friction, the vehicle slides down the incline with acceleration a = g sin α, directed along the slope.
- Pseudo force. Inside the vehicle, the bob feels its weight mg and a pseudo force mg sin α pointing up the slope. Split mg into mg sin α down the slope and mg cos α perpendicular to it.
- Effective gravity. The two components along the slope cancel. Only mg cos α is left, perpendicular to the incline, so the pendulum hangs perpendicular to the roof with g_(eff) = g cos α.
T = 2π√(L/g_(eff)) = 2π√(L/(g cos α))
- So. The time period is 2π√(L/(g cos α)), which is option (a).
The idea to keep. In an accelerating frame, use g⃗_(eff) = g⃗ - a⃗ in T = 2π√(L/g_(eff)). On a smooth incline, the slope component cancels and g cos α survives.
Answer from the NTA final answer key. Solution by PrepWiser.