A simple pendulum of length 1 m on Earth (g = 10) has frequency:
AApproximately 1/π Hz
B10 Hz
C0.5 Hz
D2 Hz
Answer & Solution
Correct answer: A. Approximately 1/π Hz
T = 2π√(L/g) = 2π√(1/10) = 2π × 0.316 ≈ 1.987 s. f = 1/T ≈ 0.503 Hz, which is approximately 1/π. Both options A and B are close; B is more precise.
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