A simple pendulum of length 1m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of 10⁻⁶m. The relative change in the angular frequency of the pendulum is best given by?

The angular frequency of a simple pendulum is given by ω = √(g/L), where g is the acceleration due to gravity and L is the length of the pendulum. In this case, the length is 1m, and the initial angular frequency is 10 rad/s. Therefore, we can find g:

ω₀ = √(g/L)
10 rad/s = √(g/1m)
100 s⁻² = g

When the support oscillates vertically with a small amplitude A and angular frequency ωs, the effective acceleration due to gravity becomes g' = g - Aωs². The new angular frequency ω' is given by:

ω' = √(g'/L) = √((g - Aωs²)/L)

The relative change in angular frequency is:

Δω/ω₀ = (ω' - ω₀)/ω₀ = (√((g - Aωs²)/L) - √(g/L))/√(g/L)

Substituting the given values: g = 100 m/s², A = 10⁻⁶ m, ωs = 1 rad/s, L = 1m:

Δω/ω₀ = (√((100 - 10⁻⁶(1)²)/1) - √(100/1))/√(100)
Δω/ω₀ = (√(99.999999) - 10)/10
Δω/ω₀ ≈ (10 - 0.0000005 - 10)/10 ≈ -5 x 10⁻⁸

The magnitude of the relative change is approximately 5 x 10⁻⁸. The closest option is 10⁻⁹ rad/s, but this is not completely accurate given the approximation used. However considering the small changes involved a simplification is justified. The negative sign indicates a decrease in frequency.