Question

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency off 1 rad/s and an amplitude of 102. The relative change in the angular frequency of the pendulum is best given by:

Answer 1

Answer:

10^-3

Explanation:

Given  

A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency off 1 rad/s and an amplitude of 10^-2. The relative change in the angular frequency of the pendulum is best given by

So angular frequency is given by

ω = √g/l

Δdω/ω = 1/2 Δg/g

Δg is due to oscillation of support

So Δω =1/2 x 2ω^2/100 x 100

  Δω = 2 ω^2 A / 2g

       = 2 x 1 x 10^-2 / 2 x 10

     Δω = 10^- 3

So The relative change in the angular frequency of the pendulum is given by 10^-3 angstrom