A simple pendulum of length 1 m is oscillating with an angular frequency `10 rad//s`. The suopport of the down with a small angular frequecy of ` 1 rad//s` and an amplitude of `10^(-2)m` . The relative changes in the angular frequency of the pendulum is best given by.

Options

A. `10 ^(-3)`
B. 1
C. `10 ^(-5)`
D. `10 ^(-1)`

Answer

A

Explanation

Since `omega = sqrt((g _(eff))/(l ))`
Now supports also oscillates with acceleration ‘a’.
`implies omega _(max ) = sqrt((g +a)/(l ) ) and omega _(min ) = sqrt((g -a)/(l))`
`impliesDelta omega = sqrt((g)/(l )) [(1+ (a)/(g) )^(1//2) - (1-(a)/(g) )^(1//2) ]= sqrt((g)/(l ))[ cancel1 + (1)/(2) (a)/(g) cancel1 + (1)/(2) (a)/(g) ]," " Delta omega =omega xx a/g`
Now acceleration of support `a = A _(s ) omega _(s) ^(2)`
` a = 10 ^(-2) xx1 ^(2) = 10 ^(-2) implies Delta omega 10 xx ( 10 ^(-2))/(10)= 10 ^(-2)` Rad/sec.