JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Elasticity & SHM

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 of

1 rad / s

and an amplitude of

1 0^{- 2} m

. The relative change in the angular frequency of the pendulum is best given by:

Answer: (A)

Solution

For a simple pendulum

ω = g / L

, so

ω^{2} = g / L

. The vertical oscillation of the support adds a maximum acceleration

a = ω_{s}^{2} A = (1)^{2} (1 0^{- 2}) = 1 0^{- 2} m/ s^{2}

, i.e.

Δ g = a

. Differentiating

ω^{2} = g / L

2 ω Δ ω = Δ g / L

, so

Δ ω = \frac{a}{2 ω L} = \frac{1 0 ^{- 2}}{2 ( 10 ) ( 1 )} = 5 \times 1 0^{- 4} rad/s

. To the nearest listed order of magnitude this is

1 0^{- 3} rad/s

JEE Main · 2019 · 11 Jan 2019 · Shift 2 · Single correct option