A simple pendulum of length $$1m$$ is oscillating with an angular frequency $$10$$ rad $$s-1$$. The support of the pendulum starts oscillating up and down with a small angular frequency of $$1$$ rad $$s-1$$ and an amplitude of $$10$$ $$-2m$$ . The relative change in the angular frequency of the pendulum is best given $$by(take$$ acceleration due to gravity $$g=10m/s2)$$

Question

A simple pendulum of length $$1m$$ is oscillating with an angular frequency  $$10$$ rad $$s-1$$. The support of the pendulum starts oscillating up and down with a small angular frequency of $$1$$ rad $$s-1$$ and an amplitude of $$10$$ $$-2m$$ . The relative change in the angular frequency of the pendulum is best given $$by(take$$ acceleration due to gravity $$g=10m/s2)$$

Infinity Learn · Accepted Answer

ω$$=10$$ rad $$s-1$$,$$g=10$$ m $$s-2gmax=g$$ $$+$$ a $$gmin=g$$ $$-$$ a difference of above two equations is Δg, Δ$$g=2a=($$ $$2)acceleration$$,and $$a=$$ω$$support2A$$, here A is amplitude, Δ$$g=2$$ω$$s2A---(1)$$ ω$$support=$$ω$$s=1$$ rad.$$s-1$$ $$A=10-2m$$ ω$$=geffl$$ differentiate above equation ⇒Δωω$$=12$$Δgg Δω$$=12$$Δgg×ω substitute equation (1), Δω$$=12$$×$$2$$ω$$s2Ag$$×ω ∴ Δω$$=1$$×$$10$$−$$210$$×$$10=10$$−$$2rad$$ $$s-1$$ Δω change in angular frequency of pendulum

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