Time period of a simple pendulum of length L (small oscillations):
AT = 2π√(L/g)
BT = 2π√(g/L)
CT = √(L/g)
DT = 2π/L
Answer & Solution
Correct answer: A. T = 2π√(L/g)
T = 2π√(L/g) for small oscillations. Independent of mass. Doubling length increases T by factor √2.
Related questions
A particle of mass 0.2 kg performs S.H.M. of amplitude 0.05 m with period π s. Its maximumIn damped harmonic motion, the amplitude of oscillationTwo springs of force constants k₁ and k₂ are connected in series and stretched by an exterA particle in S.H.M. of period 2 s starts from rest at the extreme position. Its displacemThe kinetic energy of a particle of mass m in linear S.H.M. of amplitude A at displacementA simple pendulum of length 1 m on the earth's surface (g = π² m/s²) has periodA spring-mass system of mass m and spring constant k oscillates with periodA particle executes S.H.M. with amplitude A and angular frequency ω. Its maximum accelerat