First slide

Applications of SHM

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 $10^{-2} m$ . The relative change in the angular frequency of the pendulum is best given by(take acceleration due to gravity g=10m/ $s^{2}$ )

Moderate

A

$10^{-3} {rad.s}^{-1}$
B

$10^{-2} {rad.s}^{-1}$
C

$10^{-1} {rad.s}^{-1}$
D

$10^{-4} {rad.s}^{-1}$

Solution

$ω = 10 {rad s}^{-1}, g = 10 {m s}^{-2}$
$g_{\max} = g + a g_{\min} = g - a d i f f e r e n c e o f a b o v e t w o e q u a t i o n s i s Δ g, Δ g = 2 a = (2) a c c e l e r a t i o n, a n d a = ω_{s u p p o r t}^{2} A, h e r e A i s a m p l i t u d e, Δ g = 2 ω_{s}^{2} A - - - (1) ω_{support} = ω_{s} = 1 {rad.s}^{-1} A = 10^{-2} m ω = \sqrt{\frac{g_{e f f}}{l}} d i f f e r e n t i a t e a b o v e e q u a t i o n \Rightarrow \frac{Δ ω}{ω} = \frac{1}{2} \frac{Δ g}{g} Δ ω = \frac{1}{2} \frac{Δ g}{g} \times ω s u b s t i t u t e e q u a t i o n (1), Δ ω = \frac{1}{2} \times \frac{2 ω_{s}^{2} A}{g} \times ω ∴ Δ ω = \frac{1 \times 10^{- 2}}{10} \times 10 = 10^{- 2} {rad s}^{-1} Δ ω c h a n g e i n a n g u l a r f r e q u e n c y o f p e n d u l u m$

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