Question

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 10⁻² m. The relative change in the angular frequency of the pendulum is best given by:
(A) 10⁻³
rad/s (B) 1 rad/s
(C) 10⁻¹
rad/s (D) 10⁻⁵
rad/s

Answer 1

Explanation:

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 10⁻² m. The relative change in the angular frequency of the pendulum is best given by:

(A) 10⁻³

rad/s (B) 1 rad/s

(C) 10⁻¹ rad/s ✓

(D) 10⁻⁵

rad/s

Answer 2

Answer:

10^-3

Explanation:

we know

T = 2× 3.14 × root(l/ g) [ w= 2 × 3.14 /T]

w = root (l/g)

♢w /w = 1/2 ♢ g/g

♢ g= 2 w'^2 A

[ ♢g= ( g+ w'^2 A)-(g-w'^2A) ]

{w' = angular frequency of support , A = amplitude of oscillation of support}

♢ w/w = 1/2 × 2 w'^2 A / g (w' = 1 rad /s) (A=10^-2)

= 10^-2 / 10

=10^-3 . (being a relative term it should not have unit I guss).