Question

What is the frequency of a second pendulum in an elevator moving up with an acceleration of g/2 ?

Answer 1

Given :
For seconds pendulum =Frequency=ν=1/2 s⁻¹

When elevator is moving with acceleration a upwards , then total effective acceleration due gravity is
g1=g+a
=g+g/2
=3g/2

v=1/2π √ [g/l]

ν²α g

v1²/v²=g1/g
=3g/2/g=3/2

v1/v=√3/2
=1.225

v1=1.225xv
=1.225x1/2
=0.612s⁻¹

∴The frequency of a second pendulum in an elevator moving up with an acceleration of g/2 is 0.612s⁻¹

Answer 2

t =2π√l/g t'= 2π√l/3g/2
2 =2π√l/g. t' = 2π√2l/3g
√l/g =1/π t' = 2π√2/3 *1/π
t' = 2√2/3seconds
f =π2nt
= π2*2*√2/3
4π*√2/3