A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. If the support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s with an amplitude of $10^{- 2} m$ . Find the relative change in the angular frequency of the pendulum.

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$10^{- 5} rad / s$

$10^{- 3} rad / s$

$1 rad / s$

$10^{- 1} rad / s$

answer is B.

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Detailed Solution

Angular Frequency $ω = \sqrt{\frac{g}{ℓ}}$

$\frac{Δ ω}{ω} = \frac{1}{2} \frac{Δg}{g}$

$Δg is due to oscillation of support.$

$Δg = 2 ω_{1}^{2} A_{1}$ Where $ω_{1}$ is the angular frequency of oscillation of support A,

$\frac{Δω}{ω} = \frac{1}{2} \frac{2 ω_{1}^{2} A_{1}}{g}$ (A₁ is the amplitude of oscillation.)

$= \frac{1}{2} \times \frac{2 \times 1^{2} \times 10^{- 2}}{10} = 10^{- 3}$

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