Question

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude a and time period T. The speed of the pendulum at x = a/2 will be ___________

(a) (πa√3)/T

(b) (πa√3)/2T

(c) πa/T

(d) (3π^2 a)/T

Answer 1

The correct option is (a) (πa√3)/T

The best I can explain: Speed,

v=ω√(a^2-x^2), x=a/2

v=ω√(a^2-a^2/4)=2π/T×(√3 a)/2=(πa√3)/T.